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# Find the limit. Use l'Hospital's Rule where appropriate. If there is a more elementary method, consider using it. If l'Hospital's Rule doesn't apply, explain why.$\displaystyle \lim_{x\to 0^+} (1 + \sin 3x)^{1/x}$

## $$\begin{array}{l}y=(1+\sin 3 x)^{1 / x} \Rightarrow \ln y=\frac{1}{x} \ln (1+\sin 3 x) \Rightarrow \\\lim _{x \rightarrow 0^{+}} \ln y=\lim _{x \rightarrow 0^{+}} \frac{\ln (1+\sin 3 x)}{x}=\lim _{x \rightarrow 0^{+}} \frac{[1 /(1+\sin 3 x)] \cdot 3 \cos 3 x}{1}=\lim _{x \rightarrow 0^{+}} \frac{3 \cos 3 x}{1+\sin 3 x}=\frac{3 \cdot 1}{1+0}=3 \Rightarrow \\\lim _{x \rightarrow 0^{+}}(1+\sin 3 x)^{1 / x}=\lim _{x \rightarrow 0^{+}} e^{\ln y}=e^{3}\end{array}$$

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So for this problem, what we have is using little towels rule to evaluate X. As it approaches here from the right of one plus sign three X. The one X. One over X. So um first thing we're going to do is take the natural log of both sides to put it in a simpler form. And that form is going to be the natural log of one class reacts writing by X. Now, when we evaluate zero from the right we see that we still get the indeterminate form. Um so we're going to use local towels rule that makes things much simpler because now we have three co sign X over one plus sign three acts over one. Now, with this we can discover that when we evaluate this at X equals zero, we end up getting 3/1. As the results, we get three. Remember that? That's the natural log of Y equals three. So the final answer will be Y equals me to the third.

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