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# Direct Substitution

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Substitution to Find Limits
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Finding limits for the vast majority of points for a given function is as simple as substituting the number that  $x$ approaches into the function. Since this turns evaluating limits into an algebra-level substitution, most questions involving limits focus on the cases where substituting does not work. How can you decide if substitution is an appropriate analytical tool for finding a limit?

#### Watch This

http://www.youtube.com/watch?v=VLiMfJHZIpk James Sousa: Determine a Limit Analytically

#### Guidance

Finding a limit analytically means finding the limit using algebraic means. In order to evaluate many limits, you can substitute the value that  $x$ approaches into the function and evaluate the result. This works perfectly when there are no holes or asymptotes at that particular  $x$ value. You can be confident this method works as long as you don’t end up dividing by zero when you substitute.

If the function  $f(x)$ has no holes or asymptote at  $x=a$ then: $\lim_{x \to a}f(x)=f(a)$

Occasionally there will be a hole at $x=a$ . The limit in this case is the height of the function if the hole did not exist. In other words, if the function is a rational expression with factors that can be canceled, cancel the term algebraically and then substitute into the resulting expression. If no factors can be canceled, it could be that the limit does not exist at that point due to asymptotes.

Example A

Which of the following limits can you determine using direct substitution? Find that limit.

$\lim_{x \to 2}\frac{x^2-4}{x-2},\ \lim_{x \to 3}\frac{x^2-4}{x-2}$

Solution: The limit on the right can be evaluated using direct substitution because the hole exists at  $x=2$ not $x=3$ .

$\lim_{x \to 3}\frac{x^2-4}{x-2}=\frac{3^2-4}{3-2}=\frac{9-4}{1}=5$

Example B

Evaluate the following limit by canceling first and then using substitution.

$\lim \limits_{x \to 2}\frac{x^2-4}{x-2}$

Solution:

$\lim_{x \to 2}\frac{x^2-4}{x-2}&=\lim_{x \to 2}\frac{\left(x-2\right)\left(x+2\right)}{\left(x-2\right)} \\&=\lim_{x \to 2}(x+2) \\&=2+2 \\&=4$

Example C

Evaluate the following limit analytically:  $\lim_{x \to 4}\frac{x^2-x-12}{x-4}$

Solution:

$\lim_{x \to 4}\frac{x^2-x-12}{x-4}&=\lim_{x \to 4} \frac{\left(x-4\right)\left(x+3\right)}{\left(x-4\right)} \\&=\lim_{x \to 4}(x+3) \\&=4+3 \\&=7$

Concept Problem Revisited

In order to decide whether substitution is an appropriate first step you can always just try it. You’ll know it won’t work if you end up trying to evaluate an expression with a denominator equal to zero.  If this happens, go back and try to factor and cancel, and then try substituting again.

#### Vocabulary

Substitution is a method of determining limits where the value that  $x$ is approaching is substituted into the function and the result is evaluated. This is one way to evaluate a limit analytically.

#### Guided Practice

1. Evaluate the following limit analytically.

$\lim \limits_{x \to 3}\frac{x^2-9}{x-3}$

2. Evaluate the following limit analytically.

$\lim \limits_{t \to 4} \sqrt{t+32}$

3. Evaluate the following limit analytically.

$\lim \limits_{y \to 4}\frac{3|y-1|}{y+4}$

1. $\lim_{x \to 3}\frac{x^2-9}{x-3}=\lim_{x \to 3}\frac{\left(x-3\right)\left(x+3\right)}{\left(t-3\right)}=\lim_{x \to 3}(x+3)=6$

2. $\lim_{t \to 4}\sqrt{t+32}=\sqrt{4+32}=\sqrt{36}=6$

3. $\lim_{y \to 4}\frac{3|y-1|}{y+4}=\frac{3|4-1|}{4+4}=\frac{3 \cdot 3}{8}=\frac{9}{8}$

#### Practice

Evaluate the following limits analytically.

1. $\lim_{x \to 5}\frac{x^2-25}{x-5}$

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

3. $\lim_{x \to 5}\sqrt{5x}-12$

4. $\lim_{x \to 0}\frac{x^3+3x^2-x}{5x}$

5. $\lim_{x \to 1}\frac{3x|x-4|}{x+1}$

6. $\lim_{x \to 2}\frac{x^2+5x-14}{x-2}$

7. $\lim_{x \to 1}\frac{x^2-8x+7}{x-1}$

8. $\lim_{x \to 0}\frac{5x-1}{2x^2+3}$

9. $\lim_{x \to 1} 4x^2-2x+5$

10. $\lim_{x \to 0}\frac{x^2+5x}{x}$

11. $\lim_{x \to -3}\frac{x^2-9}{x+3}$

12. $\lim_{x \to 0}\frac{5x+1}{x}$

13. $\lim_{x \to 1}\frac{5x+1}{x}$

14. $\lim_{x \to 5}\frac{x^2-25}{x^3-125}$

15. $\lim_{x \to -1}\frac{x-2}{x+1}$