# Find `lim_(x->0^+)(cotx)^(sinx)` L' hôpital the answer is 1. I need steps

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Find `lim_(x->0^+)(cotx)^(sinx)`

`=lim_(x->0^+)(1/tanx)^(sinx)`

This is of the form `oo^0` , so rewrite as

`=e^(lim_(x->0^+)ln(1/tanx)^sinx)`

`=e^(lim_(x->0^+)sinxln(1/tanx))`

The limit is of the form `0*oo` so rewrite as:

`=e^(lim_(x->0^+)(ln(1/tanx)/(1/sinx)))` Now the limit is of the form `oo/oo` so we apply L'Hopital's rule:

`=e^(lim_(x->0^+)((sin^2x(tan^2x+1))/(cosxtanx)))` Now the limit is of the form `0/0` so we apply L'Hopital's rule again:

`=e^(lim_(x->0^+)(2tanx)/cos^3x)` The limit of a quotient is the quotient of limits

`=e^((lim_(x->0^+)2tanx)/(lim_(x->0^+)cos^3x))`

`=e^(2lim_(x->0^+)tanx)`

`=e^0=1`

**Thus the limit is 1.**

**Sources:**