Question #82d8c

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Answer 1 · 2017-05-15T03:35:08

#lim_(x->0^+) (tanx)^x = 1#

#color(blue)(lim_(x->0) (tanx)^x)#

Take the limit of the natural log of this, then at the end raise the new limit to an exponent with base #e#.

#lim_(x->0^+) ln((tanx)^x)#

#=lim_(x->0^+) x ln(tanx)#

Rewrite so that L'Hospital's rule will be applicable:
#=lim_(x->0^+) frac{ln(tanx)}{(1/x)#

After applying L'Hospital's rule because this is an indeterminate case of #frac{-oo}{oo}#

#= lim_(x->0^+) frac{ (frac{sec^2 x}{tanx}) }{((-1)/x^2)}#

#= lim_(x->0^+) frac{(frac{1}{sinxcosx})}{((-1)/x^2)}#

#= lim_(x->0^+) frac{-x^2}{sinxcosx}#

#= [lim_(x->0^+) frac{x}{sinx}]*[lim_(x->0^+) frac{-x}{cosx}]#

#=1*0#

#=0#

#lim_(x->0^+) ln((tanx)^x)=0#

#color(blue)(lim_(x->0^+) (tanx)^x = e^0 = 1)#