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GRE Subject Test: Math : Defining Derivatives with Limits

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Example Questions

Example Question #13 : Finding Derivatives

Evaluate: $\lim _{x \to 2} \frac {x^2-4}{x^3-8}$ $\displaystyle \lim _{x \to 2} \frac {x^2-4}{x^3-8}$

Possible Answers:

$\frac {1}{3}$ $\displaystyle \frac {1}{3}$

$\displaystyle -3$

Does Not Exist

$-\frac {1}{3}$ $\displaystyle -\frac {1}{3}$

Correct answer:

$\frac {1}{3}$ $\displaystyle \frac {1}{3}$

Explanation:

Step 1: Try plugging in x=2 $\displaystyle x=2$ into the denominator of the function. We want to make sure that the bottom does not become $\displaystyle 0$...

$\displaystyle 2^3-8=8-8=0$.. We got zero, and we cannot have zero in the denominator. So, we must try and factor the function (numerator and denominator):

Step 2: Factor:

$\frac {(x+2)(x-2)}{(x-2)(x^2+2x+4)}$ $\displaystyle \frac {(x+2)(x-2)}{(x-2)(x^2+2x+4)}$

Step 3: Reduce:

$\frac {x+2}{x^2+2x+4}$ $\displaystyle \frac {x+2}{x^2+2x+4}$

Step 4: Now that we got rid of the factor that made the denominator zero, we know that this function has a limit.

Step 5: Plug in x=2 $\displaystyle x=2$ into the reduced factor form:

Simplify as much as possible...

$\frac {2+2}{2^2+2(2)+4}=\frac {4}{12}=\frac {1}{3}$ $\displaystyle \frac {2+2}{2^2+2(2)+4}=\frac {4}{12}=\frac {1}{3}$

The limit of this function as x approaches $\displaystyle 2$ is $\frac {1}{3}$ $\displaystyle \frac {1}{3}$

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GRE Subject Test: Math : Defining Derivatives with Limits

Study concepts, example questions & explanations for GRE Subject Test: Math

All GRE Subject Test: Math Resources

Example Questions

Example Question #13 : Finding Derivatives

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