Question

lim x→0 (sin x/x)^(1/x^2) by l hospital​

Answer 1

Answer:

$\frac{1}{ {e}^{6} }$

Step-by-step explanation:

First take log

$\lim_{x \to \: 0} {e}^{ log( { (\frac{ \sin(x) }{x} )}^{ \frac{1}{ {x}^{2} } } ) } \\ = \lim_{x \to \: 0} {e}^{ \frac{1}{ {x}^{2} } log{ (\frac{ \sin(x) }{x} )}}$

Now use L hopital

$\lim_{x \to \: 0} {e}^{ \frac{1}{ \frac {\sin(x) }{x} } \frac{x \cos(x) - \sin(x) }{ {x}^{2} } . \frac{1}{2x} }$

$\lim_{x \to \: 0} {e}^{ { (\frac{x \cot(x) - 1 }{2 {x}^{2} } )}}$

You can use series expansion of cot x to finish it now or use two times more l hospital rule choice is yours

$\cot(x) = \frac{1}{x} - \frac{x}{3} - \frac{ {x}^{3} }{45} ...$

$\lim_{x \to \: 0} {e}^{{ (\frac{ x( \frac{1}{x} - \frac{x}{3} + \frac{ {x}^{3} }{45} ..) - 1 }{2 {x}^{2} } )}}$

$\lim_{x \to \: 0} {e}^{{ (\frac{ 1 - \frac{ {x}^{2} }{3} + \frac{ {x}^{4} }{45} .. - 1 }{2 {x}^{2} } )}}$

$\lim_{x \to \: 0} {e}^{{ ( - \frac{1}{6} + \frac{ {x}^{3} }{90} ..) }{}} = {e}^{ - \frac{1}{6} } = \frac{1}{ {e}^{6} }$