Limits from Tables and Graphs

Outputs approach 7 as x approaches 2

1. $$\lim_{x\to 2^-}f(x)=7$$
2. $$\lim_{x\to 2^+}f(x)=7$$
3. $$\lim_{x\to 2}f(x)=7$$

The conclusion does not depend on whether $f(2)$ equals 7—or even exists.

A removable discontinuity

1. $$f(x)=\frac{(x-2)(x^2+3)}{x-2}$$
2. For $x\ne 2$, the expression simplifies to $x^2+3$.
3. $$\lim_{x\to 2}f(x)=2^2+3=7$$

$f(2)$ is undefined in the original expression, but the limit is still 7. The missing point can be “filled in,” so the discontinuity is removable.

Opposite infinite behavior

1. $$h(x)=\frac{x^2-3}{x-2}$$
2. $$\lim_{x\to 2^-}h(x)=-\infty$$
3. $$\lim_{x\to 2^+}h(x)=+\infty$$
4. $\lim_{x\to 2}h(x)$ does not exist.

The one-sided limits do not agree, so there is no two-sided limit.

Limit and function value can differ

At $x=5$, the graph approaches 2 from both sides, but its filled point is at 1.

1. $$\lim_{x\to 5}g(x)=2$$
2. $$g(5)=1$$
3. Because the limit does not equal the function value, $g$ is discontinuous at 5.

Use continuity when it is known

Polynomials are continuous for every real input, so direct substitution works:

1. $$\lim_{x\to 0}(x^5-7x+4)$$
2. $$=0^5-7(0)+4=4$$

Bounded does not guarantee a limit

$\sin\!\left(\frac{\pi}{x}\right)$ stays between −1 and 1, but it oscillates faster and faster near $x=0$.

Therefore, $\lim_{x\to 0}\sin\!\left(\frac{\pi}{x}\right)$ does not exist.