limits

Question

the value of $lim_{x-> 0}(1-sinx.cosx)^{cosec2x}$ is:

A $e^{2}$

B $e^{-2}$

C$e^{1/2}$

D $e^{-1/2}$

Comments

please give easy explanation

Answer 1

Let ,

$y=\lim_{{x->0}}(1-\sin x\cos x)^{cosec2x}$

$\log_{e}y=\lim_{{x->0}}\log_{e}(1-\sin x\cos x)^{cosec2x}$

$\log_{e}y=\lim_{{x->0}}cosec2x\log_{e}(1-\sin x\cos x)$

$\log_{e}y=\lim_{{x->0}}\frac{\log_{e}(1-\sin x\cos x)}{sin2x}$

 

$\log_{e}y=\lim_{{x->0}}\frac{\log_{e}(1-\sin x\cos x)}{2sinxcosx}$

$\log_{e}y=\frac{1}{2}\lim_{{x->0}}\frac{\log_{e}(1-\sin x\cos x)}{sinxcosx}$

Let , $z=\sin x\cos x$ ,

    now as , $x\rightarrow 0$ then $z\rightarrow 0$ .

So,$\log_{e}y=\frac{1}{2}\lim_{z\rightarrow 0}\frac{\log_{e}(1-z)}{z}$

Now, $\lim_{z\rightarrow 0}\frac{\log_{e}(1-z)}{z}=-1$  [ Its easy so not calculating]

So,$\log_{e}y=\frac{1}{2}(-1)=-\frac{1}{2}$

So, $\large y=e^{-\frac{1}{2}}$  ………………….………….. (D) is the answer ..

Answer 2

Drop a comment  for correction ….