Evaluate the limit, if it exists. \lim \frac{t^2-64}{2t^2+17t+8} \left(t-8\right) . Show...

Question:

Evaluate the limit, if it exists.

{eq}\lim \frac{t^2-64}{2t^2+17t+8} \left(t-8\right) {/eq}.

Show steps!

Limit:


We have to evaluate the limit.

Here, in our problem, we use the factorization method to evaluate the value of a limit.

In the numerator, we apply the mathematical identity and in the denominator, we use the factorization method. Then cancel the common factor and finally substitute the value of the limit.

Answer and Explanation:


Given

{eq}\lim_{t \rightarrow -8} \dfrac{t^2-64}{2t^2+17t+8} {/eq}.


We have to evaluate the limit.


{eq}\begin{align} \lim_{t \rightarrow -8} \dfrac{t^2-64}{2t^2+17t+8} &=\lim_{t \rightarrow -8} \dfrac{t^2-8^2}{2t^2+16t+t+8}\\ &=\lim_{t \rightarrow -8} \dfrac{(t+8)(t-8)}{2t(t+8)+1(t+8)}\ & \left [ \because a^2-b^2=(a+b)(a-b) \right ]\\ &=\lim_{t \rightarrow -8} \dfrac{(t+8)(t-8)}{(2t+1)(t+8)}\\ &=\lim_{t \rightarrow -8} \dfrac{t-8}{2t+1}\\ &=\dfrac{-8-8}{2(-8)+1}\\ &=\dfrac{-16}{-16+1}\\ &=\dfrac{-16}{-15}\\ &=\dfrac{16}{15} \end{align} {/eq}


{eq}\color{blue}{\lim_{t \rightarrow -8} \dfrac{t^2-64}{2t^2+17t+8}=\dfrac{16}{15}} {/eq}


Learn more about this topic:

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How to Determine the Limits of Functions

from Math 104: Calculus

Chapter 6 / Lesson 4
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