Given a real valued function ' f ' such that f(x)=\ array cc ^2\ x\ x^2-[x]^2 & for x>0 1 & for x=0 \ x\ \ x\ & for x < 0 array . then

Correct answer: D. _ x arrow 0 f(x) does not exists _ x arrow 0^ - f(x)= _ h arrow 0 \ -h\ \ -h\ = _ h arrow 0 (1-h) (1-h) = 1 _ x arrow 0^ + f(x)= _ h arrow 0 ^2\ h\ h^2-[h]^2 = _ h arrow 0 ( ^2 h)/(h^2)=1 _ x arrow 0 f(x) does not exist.

BITSAT 2024Paper 2

Exam: BITSAT
Year: 2024
Paper: Paper 2
Subject: Mathematics
Chapter: Limits

BITSAT 2024 · Paper 2 · Mathematics · Limits

Given a real valued function ' f ' such that f(x)=\ array cc ^2\ x\...

Given a real valued function '

f

' such that

f(x)=\left\{

\begin{array}{cc}\frac{\tan ^2\{x\}}{x^2-[x]^2} & \text{ for } x>0 \\ 1 & \text{ for } x=0 \\ \sqrt{\{x\} \cot \{x\}} & \text{ for } x < 0\end{array}

\right.

then

Options

A $\mathrm{LHL}=1$
B $\mathrm{RHL}=\sqrt{\cot 1}$
C $\lim _{x \rightarrow 0} f(x)$ exist
D $\lim _{x \rightarrow 0} f(x)$ does not exists

Correct answer

D. $\lim _{x \rightarrow 0} f(x)$ does not exists

Step-by-step solution

\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0} \sqrt{\{-h\} \cot \{-h\}}=\lim _{h \rightarrow 0} \sqrt{(1-h) \cot (1-h)}=\sqrt{\cot 1}\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} \frac{\tan ^2\{h\}}{h^2-[h]^2}=\lim _{h \rightarrow 0} \frac{\tan ^2 h}{h^2}=1\therefore \lim _{x \rightarrow 0} f(x)

does not exist.

Practice Limits on Quantrex Academy →

More from Limits

Let f(x)= x, g(x)= x, h(x)=x^2 then _ x arrow 1 (f(g(h(x)))-f(g(h(1))))/(x-1)= BITSAT 2024 If f(x)=\ array l x^2-1,0 < x < 2 2 x+3,2 ≤ x < 3 array . , then the quadratic equation… BITSAT 2023 _ x arrow 0 (| x|)/(x) is equal to BITSAT 2023 The value of _ x arrow 0 (1- (1- x))/(x^4) is BITSAT 2022 The value of _ x arrow 0 (1+x)^ (1)/(x) -e+(1)/(2) e x x^2 is BITSAT 2022 Value of _ x arrow 0 a^ x -1 x is BITSAT 2021 _ x arrow 0 2 ^ 2 3 x x^ 2 is equal to BITSAT 2021 _ x arrow 0 x- x x+ ^ 2 x is equal to BITSAT 2021 The value of _ x → (3π)/(4) 4 ^ 2 xcosx - x + x x + x is BITSAT 2019 The value of a , so that _ x → 0 (1)/(x2)(e^ ax - e^ x - x) = (3)/(2) ,& is BITSAT 2018 _ n → ∞ sin [π n^ 2 + 1 ] is equal to BITSAT 2017 If _ n → ∞ (an - 1 + n^ 2 1 + n ) = b , where a is finite number, then BITSAT 2017 _ n arrow ∞ a^ n +b^ n a^ n -b^ n , where a > b > 1, is equal to BITSAT 2016 _ x arrow ∞ ∫_ 0 ^ 2 x x e^ x^ 2 d x e^ 4 x^ 2 equals BITSAT 2015 If _ x arrow x ((1)/(x))=A and _ x arrow 0 x ((1)/(x))=B , then which one of the… BITSAT 2015 _ x arrow 0 (cosec x)^ 1 / x is equal to : BITSAT 2014 Full Limits list Limits JEE Maths PYQs JEE Mathematics PYQs All BITSAT PYQs