Question

please ans limit class 12th​

Answer 1

Question:- To find the limit

$( = > 0) { \frac{sinx}{x} }^{ \frac{1}{1 - cosx} }$

Solution:-From the formula of limit we know,

$\frac{sinx}{x} = 1$

putting this value in given terms we get

${1}^{ \frac{1}{1 - cosx} }$

now simply put the value of x=0

we get

${1}^{ \infty }$

which is in the form of

${e}^{l}$

where

$l = (f(x) - 1) \times \: g(x)$

where

$g(x)$

is the terms in power

now, putting the value of f(x) and g(x) we get

$l = \frac{sinx - x}{x} \times \frac{1}{1 - cosx} \\ l = \frac{sinx - x}{x - xcosx}$

putting x=0 we get 0/0 which is indiscriminate form

so,

Differentiate it using L'Hopital rule

$\frac{dl}{dx} = \frac{cosx - 1}{1 + xsinx - cosx}$

now putting x=0,again we are getting as 0/0 form

so differentiate it again

$\frac{dl}{dx} = \frac{ - sinx}{xcosx + sinx + sinx}$

putting x=0 again we get 0/0 as indiscriminate form

this time rather than diffrentiating it we will try to use the formula

divide the numerator and denominator by x

$\frac{dl}{dx} = \frac{ \frac{ - sinx}{x} }{ \frac{xcosx}{x} + \frac{2sinx}{x} }$

now we know that,

$\frac{sinx}{x} = 1$

putting this value,we get

L=$\frac{ - 1}{1 + 2} = \frac{ - 1}{3}$

so, putting this of l value in ${e}^{l}$

,we get

required limit=${e}^{ \frac{ - 1}{3} }$

{hope it helps you}