Evaluating Limits Algebraically

Direct substitution finishes the limit

1. $$\lim_{x\to 3}\frac{x^2-4}{x-2}$$
2. $$=\frac{3^2-4}{3-2}=\frac{5}{1}$$
3. $$=5$$

The rational function is continuous at 3 because its denominator is not zero there.

Factor and simplify

1. $\lim_{x\to 2}\frac{x^2-4}{x-2}$ gives $\frac{0}{0}$.
2. $$x^2-4=(x-2)(x+2)$$
3. For nearby inputs, cancel $x-2$.
4. $$\lim_{x\to 2}(x+2)=4$$

Check each side of a vertical asymptote

1. $$\lim_{x\to 3^-}\frac{5}{x-3}=-\infty$$
2. $$\lim_{x\to 3^+}\frac{5}{x-3}=+\infty$$
3. The two-sided limit does not exist.

The numerator stays positive; the denominator changes sign across 3.

Rationalize a radical expression

1. $\lim_{x\to 9}\frac{\sqrt{x}-3}{x-9}$ gives $\frac{0}{0}$.
2. Multiply by $\frac{\sqrt{x}+3}{\sqrt{x}+3}$.
3. $$(\sqrt{x}-3)(\sqrt{x}+3)=x-9$$
4. $$\lim_{x\to 9}\frac{1}{\sqrt{x}+3}=\frac{1}{6}$$

Combine the inner fractions first

1. $$\lim_{x\to 0}\frac{\frac{1}{3+x}-\frac{1}{3}}{x}$$
2. $\frac{3-(3+x)}{3(3+x)}$ simplifies to $\frac{-x}{3(3+x)}$.
3. Cancel the common factor $x$.
4. $$\lim_{x\to 0}\frac{-1}{3(3+x)}=-\frac{1}{9}$$