EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 21 - Limits>JEE Advanced Paper 2 - 2017>Q 3

HARD

JEE Advanced

IMPORTANT

Earn 100

Let $f (x) = \frac{1 - x (1 + |1 - x|)}{| 1 - x |} \cos (\frac{1}{1 - x}) for x \neq 1,$ then

35% studentsanswered this correctly

Important Questions on Limits

HARD

JEE Advanced

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 21 - Limits>JEE Advanced Paper 2 - 2016>Q 4

Let

f : R \to (0, \infty)

and

g : R \to R

be twice differentiable functions such that

f^{″}

and

g^{″}

are continuous functions on

R

. Suppose

f^{'} (2) = g (2) = 0, f^{″} (2) \neq 0

and

g^{'} (2) \neq 0

,

\lim_{x \to 2} \frac{f (x) g (x)}{f^{'} (x) g^{'} (x)} = 1

, then

HARD

JEE Advanced

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 21 - Limits>JEE Advanced Paper 1 - 2020>Q 6

Let

e

denote the base of the natural logarithm. The value of the real number

a

for which the right-hand limit

\lim_{x \to 0^{+}} \frac{{(1 - x)}^{\frac{1}{x}} - e^{- 1}}{x^{a}}

is equal to a non-zero real number, is