Question

please solve limit class 12th​

Answer 1

Step-by-step explanation:

y = lim x->0 (sinx/x)^(1/1-cosx)

=> lgy = limx->0 log(sinx/x) / (1-cosx)

=> logy = limx->0 (x/sinx)*(xcosx-sinx)/x² / (sinx)

=> logy = limx->0 (1/sinx/x) * (xcosx-sinx) /(x³sinx/x)

=> logy = limx->0 (1) * (xcosx-sinx)/x³*1

=> logy = limx->0 (cosx - xsinx - cosx) /3x²

=> logy = limx->0 -(xcosx + sinx) /6x

=> logy = limx->0 -(cosx - xsinx + cosx) /6

=> logy = -(1 - 0+1)/6

=> logy = -2/6

=> logy = -1/3

=> y = e^-1/3

Answer 2

Using the formula of limit we can say,

$\sf \dfrac{sin \ x}{x} = 1$

Now,

The limit is

$\lim_{x \longrightarrow} \dfrac{sinx}{x})^{\dfrac{1}{1-cos x\ x}}$

Now,

Putting x=0 in this ,we get

$\sf \lim_{x \longrightarrow 0} \bigg(\dfrac{sin x}{x}\bigg)^{\frac{1}{(1-cosx)}} \\ \\ =1^{\infty}$

The limit of this form $\sf 1^{\infty}$ is e^L

where

L=[f(x)-1].g(x) , where g(x) is the power

$\sf L=\bigg(\dfrac{sin}{x}-1\bigg) \times \dfrac{1}{1-cosx} \\ \\ = \dfrac{sinx-x}{x-xcosx}$

Putting x=0,in this we get

$\sf \dfrac{sin0-0}{0-xcos0 \\ \\ = \dfrac{0}{0}$

This is in indiscriminate form

Use L'hopital rule

$\sf \dfrac{dL}{dx} = \dfrac{cosx-1}{1+xsinx-cosx}$

Putting again x=0 we get 0/0,so diffrentiate it again

$\sf \dfrac{dL'}{dx'} = \\dfrac{-sinx}{0+sinx+xcosx+sinx} \\ \\ = - \dfrac{sinx}{2sinx+xcosx}$

if we put 0 again,then we will again find 0/0 form

But this time we will try to apply formula

divide the numerator and denominator by x

$= \dfrac{{\dfrac{-sinx}{x}}{\dfrac{\dfrac{2sinx}{x+xcosx}}{x}}}$

we know,that

$\sf \dfrac{sinx}{x} = 1$

putting this value we get

$\sf L= \dfrac{-1}{2+cosx}$

now putting x=0,we get

$\sf L=-\dfrac{1}{2+1} \\ \\ =-\dfrac{1}{3}$

$\sf Limit =e^L=e^{\dfrac{-1}{3}}$