# `lim_(x->1) (2 - x)^(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...

`lim_(x->1) (2 - x)^(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.

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You need to evaluate the limit, hence, you need to replace 1 for x:

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

You need to use the logarithm special technique but first you need to define the followings:

`f(x) = (2 - x)^(tan((pi*x)/2))`

Taking logarithms both sides yields:

`ln f(x) = ln (2 - x)^(tan((pi*x)/2))`

l`n f(x) = ((tan((pi*x)/2))) * ln (2 - x)`

Taking the limit:

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

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

You may use that `1/((tan((pi*x)/2))) = cot((pi*x)/2)`

`lim_(x->1) (ln (2 - x))/(cot ((pi*x)/2)) `

You may use l'Hospital's rule:

`lim_(x->1) (ln (2 - x))/(cot ((pi*x)/2)) = lim_(x->1) ((ln (2 - x))')/((cot ((pi*x)/2))')`

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

**Hence, evaluating the limit, yields `lim_(x->1) (2 - x)^(tan((pi*x)/2)) = e^(2/pi).` **