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Okay, we're going to start off by talking about a bunch of limit laws.

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So today we're talking about, we're starting to figure out limits, not based on

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graphs,

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not based on tables or plugging in values, based on just given the function, the

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equation.

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Okay, let me go over some of these laws.

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You don't need to write much notes, but we'll refer back to these many times.

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Just let me give you a rough idea what it is. So the first one,

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the limit of the sum of two functions,

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we can find each limit separately. Limit of f plus the limit of g. So the limit of a

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sum is the sum of limits.

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The first five, this is only true if the actual limits exist.

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They equal some number.

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Equal some other number.

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If one of the two does not exist, the law does not apply.

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And then the limit of the difference is the difference of limits.

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Okay, any time you have a constant in front of a function, we can bring that outside

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the

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limit.

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So for example, limit as x goes to 5 or something of 7 times the square root of x

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plus 3.

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I'm just making some up.

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If we want to we can just bring this guy outside seven times whatever the limit is

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Number four the limit of a product of two different functions is the product of the

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limits

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For example the limit as x goes to zero

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square root of x times e to the x

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Anytime you have a product that's equal to the limit of the first function times the

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limit whatever the second function is.

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Again that's only true if this equals some number and this equals some number.

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The limit of the division of two functions is the division of the limits.

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This is only true if both of them equal some number and the bottom one does not

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equal zero.

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This one, if we have some function to any power, for example limit as x goes to...

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the fifth root of 4 to the power x squared.

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Well, that's not the best example. Let me go with cosine of x.

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The fifth root is what power? One fifth power.

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So that's the fifth root or one fifth power of the limit.

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Number seven, the limit of a constant is just equal to constant.

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Why is that?

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Alright, we have some constant function, no matter what x is.

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Whatever a is, as you approach a from the left and the right, it's always equal to

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whatever

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that constant is, it is whatever that number is.

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Don't worry about 8 for right now.

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Let's skip the next few.

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Let's turn to the back side.

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It says if f is a polynomial or rational function,

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and a is in the domain.

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All you have to do is directly substitute in that number

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into the function.

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We talked a little bit about this last Wednesday.

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Why can we do that?

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Yes.

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This is continuous everywhere.

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This is continuous at every number in its domain.

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Everything except what makes the denominator zero.

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So in fact, we're gonna use direct substitution,

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not only for polynomials and rational functions,

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any continuous function.

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Which are all the functions you know.

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polynomials, rational functions, exponential functions, logarithmic functions,

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all trigonometric functions. As long as that a value is in the domain, we just

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directly substitute it.

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And then we'll go over these last two some other day.

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Alright.

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Let's pull this tricks...

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for...

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and...

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We're gonna have a big list.

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Okay, so leave a lot of room and then maybe start working on the examples on the

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next

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page.

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Number one trick is direct substitution.

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This is almost always how we start, every limit.

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Ok, so we tried plugging in the value we are approaching.

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What type of function is this?

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Rational.

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We really don't even need to identify as rational for right now but

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we always just try plugging in the number 3

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I'd do it on the side somewhere

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3 squared minus 4 over 3 minus 2

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what is that, 9 minus 4

84
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is 5 over 1 is 5. If you end up with a number that is the limit. Okay done easy

85
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peasy. Okay why does this work? As long as this number is in the domain of a

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continuous function just directly substitute the limit from the left

87
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always leave the limit from right will equal the function value. Okay so that's the

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first trick.

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Next example.

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So this is the same function, but now we're approaching it.

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We're approaching as X goes to two.

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Again, on the side, we always try direct substitution.

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2 squared minus 4 over 2 minus 2.

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And we get 0 divided by 0.

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What is 0 divided by 0?

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That's fine.

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It's not a number.

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OK.

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However, this is very important.

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Any time direct substitution produces 0/0 in the context of a limit,

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this is called an indeterminate form.

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Okay, why is it called an indeterminate form?

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Let me take a guess.

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Because we cannot determine what the value of the limit is.

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The limit might exist, the limit might not exist.

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We cannot determine that yet.

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So any time you get 0/0, it is an indeterminate form.

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And it means we have to do more work.

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Anybody know the trick for this one?

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We try factoring.

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In fact, every time you get 0 over 0, that almost always means x minus this number

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is

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both a factor of the numerator and denominator.

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And once you factor it, we should see this common factor.

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We're going to simplify it.

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Oh, before I do that, let me ask you this.

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If you're just given this function, are we allowed to just simplify this function?

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Are they the same function?

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Is this function the same as this function?

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Definitely not.

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They're way different functions.

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Okay, well not way different.

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This function has a hole in it. This function does not have a hole.

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Okay.

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So if you're just given a rational function, you can't just simplify that changes

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the function. We've got rid of the hole.

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Totally different function.

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However,

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As long as x is not equal to 2, they are exactly equal.

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We agree with that?

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No matter what number you plug in, like we did 3 last time.

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3 minus 2 over 3 plus 2 over 3 minus 2.

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All these happen to be 1, but this number always simplifies to this number.

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As long as x is not 2.

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So these are the exact same functions, however, are everywhere except at the value

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of x equals

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2. So why can we simplify inside limits?

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Okay so we are going to simplify and say that this is the limit as x approaches

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2 of x plus 2. The reason we could do that inside of a limit is because when

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we're approaching 2 we don't care what happens at 2. All we care about is

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what's happening from the left and from the right.

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Since we're approaching 2, we're never equal to 2.

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So these are always exactly the same number.

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And we can simplify it.

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So inside limits, common factors, we can simplify it.

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And then we go back to our previous trick, which is what?

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Direct substitution.

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So we plug it in.

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And we get 2 plus 2.

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And the answer is 4.

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Okay, so trick number two is factor and simplify.

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Okay, and we often do this trick when you end up with zero over zero.

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What's the first step we always do?

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Direct substitution.

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And again, I do it on the side.

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Get 5 over 3 minus 3.

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And again, direct substitution, doesn't matter if we're going from the left to the

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right.

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And we get 5 over 0.

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Is this an indeterminate form?

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This is not an indeterminate form.

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OK?

163
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Any time you have any number other than 0, divided by 0, it's not an indeterminate

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form.

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Not indeterminate.

166
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Okay, whenever you get a number over zero, the answer is either going to be infinity

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or negative infinity.

168
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Okay, here this rational function has a vertical asymptote of three.

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Once we know it's a number over zero, we now just have to determine which one it is

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positive

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or negative infinity.

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And how do we figure that out?

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How do we know if it's going up or down or just plug

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in a number very close to 3, but from the left, which

175
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would be what?

176
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2.9.

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2.9 is fine.

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5 divided by 2.9 minus 3.

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And all we care about is the sign.

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Is it positive or negative?

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5 divided by negative 0.1 is this positive or negative?

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Definitely negative.

183
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So what does that tell us? The answer is negative infinity.

184
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So again, any time you direct substitution, you get a number over zero.

185
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The answer is going to be either positive or negative.

186
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Okay, sort of.

187
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That's when it's going from the left to the right.

188
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If it was this question, the limit as x goes to 3 of 5 over x minus 3.

189
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This is not from the left or from the right, it's from both.

190
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And now we have to check both sides.

191
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If they both go to negative infinity, the answer is going to be negative infinity.

192
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So, here we just have to check what is the limit as x goes to 3 from the right.

193
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Again you substitute and you get 5 over 0.

194
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So we just have to determine whether it is positive or negative infinity.

195
00:16:41,610 --> 00:16:45,070
And we just plug in a number to the right of 0.

196
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5 over 3.1 minus 3.

197
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5 over.1 positive.

198
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This is positive.

199
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This limit is equal to positive infinity.

200
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So what is the limit as x goes to 3?

201
00:17:10,610 --> 00:17:12,430
Does not exist.

202
00:17:12,430 --> 00:17:16,986
From the left it went to negative infinity, from the right it went to positive

203
00:17:16,986 --> 00:17:17,736
infinity,

204
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so this does not exist.

205
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So let's write down this trick.

206
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Not really a trick, but if you get by direct substitution a number divided by zero,

207
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non-zero

208
00:17:48,750 --> 00:18:01,360
number, the limit will either equal positive infinity or negative infinity, or does

209
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not

210
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exist. You gotta check both the left and the right.

211
00:18:11,290 --> 00:18:14,590
Zero over zero is an indeterminate form. Any nonzero number

212
00:18:14,590 --> 00:18:19,630
divided by zero is not indeterminate. The answer is either positive or

213
00:18:19,630 --> 00:18:22,830
negative infinity. Or from the left and the right they're opposite.

214
00:18:23,910 --> 00:18:24,810
Does not exist.

215
00:18:25,420 --> 00:18:27,740
What is the first thing we always try?

216
00:18:29,000 --> 00:18:29,900
Direct substitution.

217
00:18:31,260 --> 00:18:34,400
Square root of 9 minus 3, over 9 minus 9.

218
00:18:36,320 --> 00:18:38,080
Zero over zero.

219
00:18:38,760 --> 00:18:40,420
And what do we call that?

220
00:18:41,500 --> 00:18:42,400
Indeterminate form.

221
00:18:43,700 --> 00:18:44,600
Indeterminate form.

222
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So it means we have to do more work.

223
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Okay.

224
00:18:51,800 --> 00:18:53,260
Anybody know the trick for this one?

225
00:18:55,360 --> 00:18:56,760
Multiply by the conjugate.

226
00:18:57,240 --> 00:18:58,140
Okay. So.

227
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We are going to multiply by the conjugate of either

228
00:19:09,880 --> 00:19:12,500
the numerator or the denominator.

229
00:19:15,000 --> 00:19:15,940
When do we do this trick?

230
00:19:16,180 --> 00:19:18,260
Typically, you've got to have two different terms.

231
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OK.

232
00:19:20,680 --> 00:19:24,120
Both of them have two terms, but especially when

233
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it involves a square root.

234
00:19:26,620 --> 00:19:27,520
OK.

235
00:19:26,920 --> 00:19:32,320
So we're going to multiply the numerator and denominator by the conjugate of one of

236
00:19:32,320 --> 00:19:33,070
them.

237
00:19:33,140 --> 00:19:38,480
In this case, use the conjugate of the numerator: square root of x plus 3.

238
00:19:49,140 --> 00:19:51,940
the limit

239
00:19:51,940 --> 00:19:54,740
as x approaches 9

240
00:19:58,140 --> 00:19:59,040
of (square root of x minus 3)

241
00:20:00,700 --> 00:20:02,100
times (square root of x plus 3)

242
00:20:02,100 --> 00:20:07,220
over (x minus 9)

243
00:20:07,220 --> 00:20:09,260
times (square root of x plus 3).

244
00:20:10,520 --> 00:20:11,420
Okay.

245
00:20:12,260 --> 00:20:15,220
The other thing is, don't distribute the denominator.

246
00:20:15,540 --> 00:20:16,760
Just leave it the way it is.

247
00:20:17,900 --> 00:20:19,440
x minus 9

248
00:20:19,440 --> 00:20:22,060
times square root of x plus 3.

249
00:20:23,020 --> 00:20:25,160
We really only distribute the side that,

250
00:20:25,720 --> 00:20:28,540
or that, uh, the numerator and denominator that

251
00:20:28,540 --> 00:20:30,400
we've multi-

252
00:20:30,400 --> 00:20:32,440
uh, who we chose the

253
00:20:32,440 --> 00:20:33,580
conjugate of.

254
00:20:34,300 --> 00:20:36,340
And, when you multiply conjugates,

255
00:20:36,340 --> 00:20:39,600
the two middle terms always add to 0.

256
00:20:43,280 --> 00:20:47,100
So we get the limit as x approaches 9

257
00:20:47,100 --> 00:20:51,340
of (x minus 9) over (x minus 9)(square root of x plus 3).

258
00:20:57,260 --> 00:20:59,260
and now what?

259
00:20:59,300 --> 00:21:03,780
if you try plugging in 9 again

260
00:21:03,780 --> 00:21:06,120
in direct substitution you still get zero over zero.

261
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However, we go back from our previous trick.

262
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Most of the time you get zero over zero,

263
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we have a common factor.

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In this case, we have a common factor of X minus nine.

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We do simplify.

266
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We try direct substitution now.

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I get one sixth.

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00:21:56,140 --> 00:21:58,180
Okay, so,

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Let's call it trick number four.

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So we multiply the numerator and denominator by the conjugate of either the

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numerator or denominator.

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00:22:48,520 --> 00:22:49,860
When do we do this?

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That is typically when there is a square root and two terms in either the numerator

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00:22:58,127 --> 00:22:59,160
or denominator.

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00:23:05,040 --> 00:23:10,744
If we try direct substitution here, we get (3 plus 0) to the negative 1 power minus

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00:23:10,744 --> 00:23:11,494
3

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00:23:11,080 --> 00:23:12,400
to the negative 1 power, all divided by 0.

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00:23:14,940 --> 00:23:16,580
We get 0 over 0.

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00:23:20,880 --> 00:23:23,000
What are we going to do here?

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00:23:30,400 --> 00:23:32,340
What does this mean to the negative one power?

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00:23:35,100 --> 00:23:36,000
One over.

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00:23:35,640 --> 00:23:40,240
one over so this is

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00:23:46,620 --> 00:23:57,400
okay yes wait no no to the negative one power simply means one divided by

284
00:23:57,400 --> 00:24:07,000
whatever it is just the reciprocal whatever is inside that exponent okay so

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00:24:07,000 --> 00:24:10,740
here we have fractions inside of fractions sometimes we call that complex

286
00:24:10,740 --> 00:24:16,180
fractions not like a complex number just fractions inside fractions so our trick

287
00:24:16,180 --> 00:24:22,080
is just to make one fraction we don't want fractions inside of fractions so we

288
00:24:22,320 --> 00:24:33,324
need a common denominator. Multiply one fraction by 3 plus x over 3 plus x, and the

289
00:24:33,324 --> 00:24:36,560
other by 3 over 3.

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00:24:48,550 --> 00:25:00,550
3 minus (3 plus x), over 3 times (3 plus x), all divided by x.

291
00:25:03,310 --> 00:25:07,310
Now we have one fraction divided by another fraction.

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00:25:08,850 --> 00:25:09,770
Let me put that there.

293
00:25:10,730 --> 00:25:11,630
So we take the numerator,

294
00:25:19,370 --> 00:25:23,590
which, after distributing the negative, becomes 3 minus 3 minus x.

295
00:25:24,510 --> 00:25:26,390
And that 3 minus 3 is 0.

296
00:25:27,610 --> 00:25:33,030
So we get negative x over 3 times (3 plus x).

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00:25:33,810 --> 00:25:36,810
And we multiply by the reciprocal of the denominator.

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00:25:37,330 --> 00:25:38,750
That's what division is.

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00:25:46,550 --> 00:25:48,090
And then what happens?

300
00:25:50,670 --> 00:25:52,250
We have a common factor.

301
00:25:52,250 --> 00:25:55,290
So it simplifies.

302
00:26:11,450 --> 00:26:14,290
Now we directly substitute in.

303
00:26:30,330 --> 00:26:34,730
And when do I stop writing the word limit?

304
00:26:36,350 --> 00:26:39,910
We stop right when we actually find the limit or take the limit.

305
00:26:39,910 --> 00:26:46,390
so I did right here now I'm gonna directly substitute it and I got a

306
00:26:46,390 --> 00:26:52,090
number this one we don't write the word limit anymore as we say we took the

307
00:26:52,090 --> 00:26:57,370
limit the limit as it approaches is this everywhere else we got to be writing the

308
00:26:57,370 --> 00:27:05,577
word limit as X approaches whatever so there it is this is negative one ninth so

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00:27:05,577 --> 00:27:06,327
let's

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00:27:06,150 --> 00:27:09,150
Let's call this...

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00:27:11,150 --> 00:27:12,450
Trick 5.

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00:27:15,830 --> 00:27:16,730
Simplify...

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00:27:19,110 --> 00:27:20,010
Complex...

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00:27:21,470 --> 00:27:22,370
Fractions...

315
00:27:22,870 --> 00:27:25,150
Which are...

316
00:27:25,790 --> 00:27:27,170
Fractions within fractions.