Question

use lhôpitals rule to evaluate $lim_{t \to 0}\frac{- 5sin(5t^{8})}{5t}$. $lim_{t \to 0}\frac{- 5sin(5t^{8})}{5t}=square$ (type an exact answer.)

Explanation:

Step1: Check the form

As $t
ightarrow0$, $\sin(5t^{8})
ightarrow0$. So, $\lim_{t
ightarrow0}\frac{- 5\sin(5t^{8})}{5t}$ is in the $\frac{0}{0}$ - form.

Step2: Apply L'Hopital's Rule

Differentiate the numerator and denominator. The derivative of $-5\sin(5t^{8})$ using the chain - rule: Let $u = 5t^{8}$, then $\frac{d}{dt}(-5\sin(u))=-5\cos(u)\cdot\frac{du}{dt}=-5\cos(5t^{8})\cdot40t^{7}=-200t^{7}\cos(5t^{8})$. The derivative of $5t$ is $5$. So, $\lim_{t
ightarrow0}\frac{-5\sin(5t^{8})}{5t}=\lim_{t
ightarrow0}\frac{-200t^{7}\cos(5t^{8})}{5}$.

Step3: Evaluate the limit

As $t
ightarrow0$, $\cos(5t^{8})
ightarrow\cos(0) = 1$. Then $\lim_{t
ightarrow0}\frac{-200t^{7}\cos(5t^{8})}{5}=\frac{-200\cdot0^{7}\cdot1}{5}=0$.

Answer:

$0$