`lim_(x->1^+) lnx tan((pix)/2)` 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.

Expert Answers

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You need to evaluate the limit, such that:

`lim_(x->1^+) ln x*tan((pi*x)/2) = 0*(-oo)`

You need to prepare the limit for l'Hospital's rule, by writing it as a quotient:

`lim_(x->1^+) (ln x)/(1/(tan((pi*x)/2))) = 0/0`

Use `cot ((pi*x)/2)` instead of `1/(tan((pi*x)/2)):`

`lim_(x->1^+) (ln x)/(cot ((pi*x)/2)) = 0/0`

You need to use l'Hospital's rule, such that:

`lim_(x->1^+) (1/x)/(-(pi/2)csc^2((pi*x)/2)) = -2/pi`

Hence, evaluating the limit, using l'Hospital's rule, yields `lim_(x->1^+) ln x*tan((pi*x)/2) = -2/pi.`

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