Question

use lhôpitals rule to find the limit of $limlimits_{\theta\to0}\frac{2^{sin\theta}-1}{4\theta}$
$limlimits_{\theta\to0}\frac{2^{sin\theta}-1}{4\theta}=square$

Explanation:

Step1: Check the form of the limit

When \(\theta
ightarrow0\), \(2^{\sin\theta}-1
ightarrow2^{0}-1 = 0\) and \(4\theta
ightarrow0\). So, it is in the \(\frac{0}{0}\) form, and l'Hôpital's rule can be applied.

Step2: Differentiate the numerator and denominator

The derivative of \(y = 2^{\sin\theta}-1\) using the chain rule:
Let \(u=\sin\theta\), then \(y = 2^{u}-1\). \(\frac{dy}{du}=2^{u}\ln2\) and \(\frac{du}{d\theta}=\cos\theta\). So, \(\frac{d}{d\theta}(2^{\sin\theta}-1)=2^{\sin\theta}\ln2\cos\theta\).
The derivative of \(y = 4\theta\) is \(\frac{d}{d\theta}(4\theta)=4\).

Step3: Apply l'Hôpital's rule

\(\lim_{\theta
ightarrow0}\frac{2^{\sin\theta}-1}{4\theta}=\lim_{\theta
ightarrow0}\frac{2^{\sin\theta}\ln2\cos\theta}{4}\).

Step4: Evaluate the new limit

Substitute \(\theta = 0\) into \(\frac{2^{\sin\theta}\ln2\cos\theta}{4}\).
When \(\theta = 0\), \(2^{\sin0}\ln2\cos0=2^{0}\ln2\times1=\ln2\).
So, \(\lim_{\theta
ightarrow0}\frac{2^{\sin\theta}\ln2\cos\theta}{4}=\frac{\ln2}{4}\).

Answer:

\(\frac{\ln2}{4}\)