How to find this type of limits ? Pls do tell me how to proceed further

Question

Question $\lim_{x->0} x^{sinx}$

Solution

put y= $\lim_{x->0} x^{sinx}$

log y = $\lim_{x->0} sinx log x $   ( 0. infinity form)

 

         = $\lim_{x->0} \frac{sin x}{\frac{1}{(log x)}}$ 


          =$\lim_{x->0} \frac{cos x}{\frac{-1}{(log x)^{2}}*\frac{1}{x}}$  ( using LH rule )

 

          =  how to proceed further ?

Comments

When we apply the limit it will be $\frac{0}{-\infty * \infty }$ form , right ?

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U r troubling urself make it simple

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Got it... Thanks :)
Perfect ans..

Answer 1

Let $y=\lim_{x\rightarrow 0} x^{\sin x}$

$\log y=\lim_{ x\rightarrow 0 }\sin x\times \log x$

$\log y=\lim_{ x\rightarrow 0 }\frac{\log x}{\mathrm{cosec }\; x}$

$\log y =\lim_{ x\rightarrow 0 }\frac{1}{-x\;\mathrm{cosec }\; x \cot x}$

$\log y=\lim_{ x\rightarrow 0 } (-\tan x)$

$\log y=0 ( \tan 0=0$)

so $y=e^{0}=1$

Comments

in ur step 4

after applying lh rule

$\frac{\frac{1}{x}}{-cosecx * cot x }$  = - tax  ??
How is this possible ?

how did you write

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when we apply limit it will be 0/0 form... is it not ?

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1/x cosec x cotx =sinx /x cotx

now  lim x->0sinx /x=1

and 1/cot x=tanx