`lim_(x->(pi/2)^-) (tan(x))^(cos(x))` Evaluate the limit.

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Chapter 4, Review - Problem 14 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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gsarora17 | (Level 2) Associate Educator

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`lim_(x->(pi/2)^-)(tan(x))^cos(x)`

`=lim_(x->(pi/2)^-)e^(cos(x)lntan(x))`

applying the limit chain rule,

`lim_(x->(pi/2)^-)cos(x)lntan(x)`

`=lim_(x->(pi/2)^-)(lntan(x)/(1/cos(x)))`

Apply L'Hospital rule, Test L'Hospital condition:

`=lim_(x->(pi/2)^-)((lntan(x))')/((1/cos(x))')`

`=lim_(x->(pi/2)^-)((sec^2(x))/tan(x))/(sin(x)/(cos^2(x)))`

`=lim_(x->(pi/2)^-)(sec^2(x)cos^2(x))/(tan(x)sin(x))`

`=lim_(x->(pi/2)^-)1/(tan(x)sin(x))`

plug in the value to evaluate limit

`=1/(tan(pi/2)sin(pi/2))`  

`=1/(oo*1)=1/oo=0`

`:.lim_(x->(pi/2)^-)e^(cos(x)lntan(x))=lim_(x->(pi/2)^-)e^0=1`

 

 

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