Question

Its difficult for me ​

Answer 1

41. Given,

$\displaystyle\longrightarrow L=\lim_{x\to\infty}\left(\dfrac{1}{x}\right)^{\frac{1}{x}}$

Taking log,

$\displaystyle\longrightarrow\log L=\lim_{x\to\infty}\dfrac{1}{x}\log\left(\dfrac{1}{x}\right)$

$\displaystyle\longrightarrow\log L=\lim_{x\to\infty}\dfrac{\log\left(\dfrac{1}{x}\right)}{x}$

$\displaystyle\longrightarrow\log L=\lim_{x\to\infty}\dfrac{\log\left(x^{-1}\right)}{x}$

$\displaystyle\longrightarrow\log L=-\lim_{x\to\infty}\dfrac{\log x}{x}$

Applying L'hospital's Rule,

$\displaystyle\longrightarrow\log L=-\lim_{x\to\infty}\dfrac{\left(\dfrac{1}{x}\right)}{1}$

$\displaystyle\longrightarrow\log L=-\lim_{x\to\infty}\dfrac{1}{x}$

$\displaystyle\longrightarrow\log L=-\dfrac{1}{\infty}$

$\displaystyle\longrightarrow\log L=0$

$\displaystyle\longrightarrow\underline{\underline{L=1}}$

42. Given,

$\displaystyle\longrightarrow L=\lim_{x\to0}(\sin x)^{\tan x}$

Taking log,

$\displaystyle\longrightarrow\log L=\lim_{x\to0}\tan x\cdot\log(\sin x)$

$\displaystyle\longrightarrow\log L=\lim_{x\to0}\dfrac{\log(\sin x)}{\cot x}$

Applying L'hospital's Rule,

$\displaystyle\longrightarrow\log L=\lim_{x\to0}\dfrac{\left(\dfrac{\cos x}{\sin x}\right)}{-\csc^2x}$

$\displaystyle\longrightarrow\log L=-\lim_{x\to0}\dfrac{\left(\dfrac{\sin x\cos x}{\sin^2x}\right)}{\left(\dfrac{1}{\sin^2x}\right)}$

$\displaystyle\longrightarrow\log L=-\lim_{x\to0}\sin x\cos x$

$\displaystyle\longrightarrow\log L=-\dfrac{1}{2}\lim_{x\to0}\sin(2x)$

$\displaystyle\longrightarrow\log L=-\dfrac{1}{2}\sin(2\cdot0)$

$\displaystyle\longrightarrow\log L=0$

$\displaystyle\longrightarrow\underline{\underline{L=1}}$

Answer 2

Cᴀʟᴄᴜʟᴀᴛɪᴏɴ of 41 :

$\displaystyle\implies \rm \: L=\lim_{x\to\infty}\left(\dfrac{1}{x}\right)^{\frac{1}{x}}$

Taking log, on both sides, we get

$\displaystyle\implies \rm logL=\lim_{x\to\infty}log \: \left(\dfrac{1}{x}\right)^{\frac{1}{x}}$

$\displaystyle\implies \rm logL=\lim_{x\to\infty} \: \dfrac{1}{x} log \: \left(\dfrac{1}{x}\right)$

$\displaystyle\implies \rm logL=\lim_{x\to\infty} \: \dfrac{ log( {x}^{ - 1} ) }{x}$

$\displaystyle\implies \rm logL=\lim_{x\to\infty}\dfrac{ - log(x) }{x}$

Since, it an indeterminant form,

so

Using L - Hospital Rule, we get

$\displaystyle\implies \rm logL=\lim_{x\to\infty}\dfrac{ - 1}{x}$

$\rm :\implies\: log(L) = 0$

$\rm :\implies\:L = {e}^{0} = 1$

$\boxed{ \pink{ \bf \: \displaystyle\implies \rm \lim_{x\to\infty}\: \left(\dfrac{1}{x}\right)^{\frac{1}{x}} \: = \tt \:1 }}$

Cᴀʟᴄᴜʟᴀᴛɪᴏɴ of 42 :

$\displaystyle\implies \rm L=\lim_{x\to0}\: \left(sinx\right)^{tanx}$

Taking log, on both sides, we get

$\displaystyle\implies \rm log \: L=\lim_{x\to0}log \: \left(sinx\right)^{tanx}$

$\displaystyle\implies \rm logL=\lim_{x\to0}tanx \: log \: \left(sinx\right)$

$\displaystyle\implies \rm logL=\lim_{x\to0} \: \dfrac{ log(sinx) }{cotx}$

Since, its an indeterminant form,

so,

Using L - Hospital Rule, we get

$\displaystyle\implies \rm logL=\lim_{x\to0} \: \dfrac{\dfrac{1}{sinx} \times cosx}{ - {cosec}^{2} x}$

$\displaystyle\implies \rm logL=\lim_{x\to0} \: \dfrac{ - {sin}^{2} x \times cosx }{sinx}$

$\displaystyle\implies \rm logL=\lim_{x\to0}( - sinx \: cosx)$

$\rm :\implies\: log(L) = 0$

$\rm :\implies\:L = {e}^{0} = 1$

$\: \boxed{ \pink{ \bf \: \displaystyle\implies \rm \lim_{x\to0}log \: \left(sinx\right)^{tanx}\: = \tt \: 1}}$