# 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.

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} 