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Problem 37 Hard Difficulty

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.

$ \lim_{x\to 0^+} \frac{\arctan(2x)}{\ln x} $

Answer

This limit has the form $\frac{0}{\infty},$ so 1 'Hospital's Rule doesn't apply. As $x \rightarrow 0^{+}, \arctan (2 x) \rightarrow 0$ and $\ln x \rightarrow-\infty,$ so
$\lim _{x \rightarrow 0^{+}} \frac{\arctan (2 x)}{\ln x}=0$

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Video Transcript

here. We're going to be finding the limit of the function through the hope Charles rule. So the limit is the arc or the functions are tangent. Uh two X. So divided by natural log of X. So we see that when we plug in zero directly, we end up getting the indeterminate farm. So we want to take um we want to rewrite this expression. Um and if we rewrite the expression as a multiplication, we can take the limit of them separately. So we have the limit as the of the arc tangent of this times the limit of one over the natural log of acts. So that's gonna end up giving us This times one over natural log of axe. This is going to be zero times zero. We don't need to use locals role in this case. It doesn't apply. So we'll get zero when we apply the limit here and we'll get zero. When we apply the limit here, giving us zero as our final answer.