lim-x-0-cos-x-log-x-

Question Number 129510 by Adel last updated on 16/Jan/21

\lim_{x \to 0} {(\cos x)}^{\log x} = ?

Commented by mr W last updated on 16/Jan/21

\Rightarrow Q 129387

Answered by greg_ed last updated on 16/Jan/21

{(\cos x)}^{l n x} = \cos^{l n x} x = e^{l n x l n (\cos x)}

{let}^{'} s rewrite this u (x) = l n x l n (\cos x) .

u (x) = \frac{l n x}{\frac{1}{l n (\cos x)}}

1 . Applying L^{'} {Hopital}^{'} s rule, we get

u (x) = \frac{l n^{2} (\cos x)}{x t a n x}

2 . Applying L^{'} {Hopital}^{'} s rule, we get

u (x) = - \frac{2 l n (\cos x) t a n x}{t a n x + x s e c^{2} x}

3 . Applying L^{'} {Hopital}^{'} s rule, we get

u (x) = - \frac{- t a n^{2} x + \sec^{2} x l n (\cos x)}{\sec^{2} x (1 + x t a n x)}

Plugging x = 0 in this, we get

\lim_{x \to 0} u (x) = 0

So, finally

\lim_{u \to 0} e^{u} = 1 .

Answered by mathmax by abdo last updated on 16/Jan/21

let f (x) = {(cosx)}^{logx} \Rightarrow f (x) = e^{logxlog (cosx)} we have

\cos (x) \sim 1 - \frac{x^{2}}{2} \Rightarrow \log (cosx) \sim \log (1 - \frac{x^{2}}{2}) \sim - \frac{x^{2}}{2} \Rightarrow

logxlog (cosx) \sim - \frac{x^{2}}{2} logx \to 0 (x \to 0) \Rightarrow \lim_{x \to 0^{+}} f (x) = e^{0} = 1