BITSAT 2014
Exam BITSAT
Year 2014
Subject Mathematics
Chapter Limits BITSAT 2014 · Mathematics · Limits
_ x arrow 0 (cosec x)^ 1 / x is equal to :
lim x → 0 ( cosec x ) 1 / log x \lim _{x \rightarrow 0}(\operatorname{cosec} x)^{1 / \log x} lim x → 0 ( cosec x ) 1/ l o g x is equal to :
Options A 0 B 1 C 1 e \frac{1}{e} e 1 D None of these Correct answer C. 1 e \frac{1}{e} e 1
Step-by-step solution
Let
y = lim x → 0 ( cosec x ) 1 / log x y=\lim _{x \rightarrow 0}(\operatorname{cosec} x)^{1 / \log x} y = lim x → 0 ( cosec x ) 1/ l o g x Taking
log \log log on both sides, we get
log y = lim x → 0 log cosec x log x [ ∞ ∞ \log y=\lim _{x \rightarrow 0} \frac{\log \operatorname{cosec} x}{\log x}\left[\frac{\infty}{\infty}\right. log y = lim x → 0 l o g x l o g cosec x [ ∞ ∞ form
] = lim x → 0 − cot x 1 / x ]=\lim _{x \rightarrow 0} \frac{-\cot x}{1 / x} \quad ] = lim x → 0 1/ x − c o t x (ByL' Hospital rule)
= − lim x → 0 x tan x ( ∵ cot x = 1 tan x ) ⇒ log y = − 1 ⇒ y = e − 1 = 1 e =-\lim _{x \rightarrow 0} \frac{x}{\tan x} \quad\left(\because \cot x=\frac{1}{\tan x}\right)\Rightarrow \log y=-1\Rightarrow y=e^{-1}=\frac{1}{e} = − lim x → 0 t a n x x ( ∵ cot x = t a n x 1 ) ⇒ log y = − 1 ⇒ y = e − 1 = e 1 Hence, required limit
= 1 e =\frac{1}{e} = e 1
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