Question

evaluate the following limit. use ihôpitals rule when it is convenient and applicable.
lim _ { \theta
ightarrow 0 } \frac { 2 ^ { sin \theta } - 1 } { 5 \theta }
how should the given limit be evaluated? select the correct choice below and, if necessary, fill in the answer box to comple
(use parentheses to clearly denote the argument of each function.)
a. use ihôpitals rule exactly once to rewrite the limit as ( lim _ { \theta
ightarrow 0 } )
b. use direct substitution.
c. use ihôpitals rule more than once to rewrite the limit in its final form as ( lim _ { \theta
ightarrow 0 } )
d. multiply the expression by a unit fraction to obtain ( lim _ { \theta
ightarrow 0 } )

Explanation:

Step1: Check the form of the limit

When \(\theta
ightarrow0\), \(2^{\sin\theta}-1
ightarrow2^{0}-1 = 0\) and \(5\theta
ightarrow0\). So, the limit \(\lim_{\theta
ightarrow0}\frac{2^{\sin\theta}-1}{5\theta}\) is in the \(\frac{0}{0}\) form.

Step2: Apply L'Hôpital's Rule

Differentiate the numerator and the 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 \(y = 5\theta\) is \(\frac{d}{d\theta}(5\theta)=5\)
By L'Hôpital's Rule, \(\lim_{\theta
ightarrow0}\frac{2^{\sin\theta}-1}{5\theta}=\lim_{\theta
ightarrow0}\frac{2^{\sin\theta}\ln2\cos\theta}{5}\)

Answer:

A. Use l'Hôpital's Rule exactly once to rewrite the limit as \(\lim_{\theta
ightarrow0}\frac{2^{\sin\theta}\ln2\cos\theta}{5}\)