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Numerade Educator



Problem 49 Easy 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.

$ \displaystyle \lim_{x\to 1^+} \ln x \tan(\pi x/2) $


This limit has the form $0 \cdot(-\infty)$
\lim _{x \rightarrow 1^{+}} \ln x \tan (\pi x / 2)=\lim _{x \rightarrow 1^{+}} \frac{\ln x}{\cot (\pi x / 2)} \stackrel{\mathrm{H}}{=} \lim _{x \rightarrow 1^{+}} \frac{1 / x}{(-\pi / 2) \csc ^{2}(\pi x / 2)}=\frac{1}{(-\pi / 2)(1)^{2}}=-\frac{2}{\pi}


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Video Transcript

Yeah, this problem, my hair, we're going to be using what cows rule to find. Uh the limit of the function. Natural log of X. Get that uh times the tangent. So we're going to do over the co tangent of X. High over to. So that way when we plug in one plus we end up getting a negative chew over pie. So in this case we don't need to use locales role as long as we convert it to this form right here, once we plug in the one here we'll get pi over two. Um And then this will be uh Approaches one from the right, we'll get Natural Log of one. So this is through taking the derivative here. Actually we do have to take the derivative because we get 0/0. So that's one over X. And then this is going to be negative coast second squared uh exp I over to you Time high over two. Then when we plug in one we would get two divided by a negative high