Question

use lhôpitals rule to find the limit of $lim_{\theta
ightarrow0}\frac{5^{sin\theta}-1}{4\theta}$. $lim_{\theta
ightarrow0}\frac{5^{sin\theta}-1}{4\theta}=square$

Explanation:

Step1: Check indeterminate form

When $\theta
ightarrow0$, we have $\frac{5^{\sin\theta}-1}{4\theta}=\frac{5^{\sin(0)}-1}{4\times0}=\frac{5^{0}-1}{0}=\frac{1 - 1}{0}=\frac{0}{0}$, so L'Hopital's rule can be applied.

Step2: Differentiate numerator and denominator

The derivative of the numerator $y = 5^{\sin\theta}-1$ using the chain - rule. Let $u=\sin\theta$, then $y = 5^{u}-1$. $\frac{dy}{du}=5^{u}\ln5$ and $\frac{du}{d\theta}=\cos\theta$, so $\frac{d}{d\theta}(5^{\sin\theta}-1)=5^{\sin\theta}\ln5\cos\theta$. The derivative of the denominator $\frac{d}{d\theta}(4\theta)=4$.

Step3: Find the new limit

$\lim_{\theta
ightarrow0}\frac{5^{\sin\theta}\ln5\cos\theta}{4}$. Substitute $\theta = 0$ into the new function: $\frac{5^{\sin(0)}\ln5\cos(0)}{4}=\frac{5^{0}\ln5\times1}{4}=\frac{\ln5}{4}$.

Answer:

$\frac{\ln5}{4}$