Can L'Hopital's Rule Solve tan(pi*x/2)ln(x) as x Approaches 1?

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SUMMARY

The limit of tan(pi*x/2)ln(x) as x approaches 1 can be evaluated using L'Hôpital's Rule. The expression can be rewritten as ln(x)/cot(pi*x/2), which presents a 0/0 indeterminate form suitable for differentiation. By applying L'Hôpital's Rule, one differentiates the numerator and denominator to find the limit. The correct approach does not involve integration, as integrating is irrelevant to solving this limit problem.

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Homework Statement


Find the limit as x goes to 1 of tan(pi*x/2)lnx


Homework Equations


using l'hospital's, lim f'(x)/g'(x)=answer


The Attempt at a Solution


i tried integrating by making the tangent part squared so i can divide by the tangent part, but i keep getting stuck. i think the answer is either 1 or infinity.

help please? thanks
 
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I really don't know how you are trying to do the problem. l'Hopital's rule doesn't tell you to integrate anything. Write it as ln(x)/cot(pi*x/2). Now it's a 0/0 form. Now differentiate numerator and denominator like l'Hopital says.
 
woops i meant taking the derivative
 
Integrating has nothing to do with this problem. Try rewriting your limit as
\lim_{x \to 1}\frac{ln(x)}{cot(x*\pi/2)}

Now you have something you can use L'Hopital's Rule on.
 

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