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

Explanation:

Step1: Check the form of the limit

When (\theta\rightarrow0), (2^{\sin\theta}-1\rightarrow2^{0}-1 = 0) and (4\theta\rightarrow0). 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\rightarrow0}\frac{2^{\sin\theta}-1}{4\theta}=\lim_{\theta\rightarrow0}\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\rightarrow0}\frac{2^{\sin\theta}\ln2\cos\theta}{4}=\frac{\ln2}{4}).

Answer:

(\frac{\ln2}{4})