Mathematics Crash Course JEE Main>Chapter 16 - Limits>Level 2>Q 20

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Let $f (x)$ and $g (x)$ be differentiable functions on $(- \infty, \infty)$ and let $f^{'} (x)$ and $g^{'} (x)$ denote derivatives of $f (x)$ and $g (x)$ , respectively. If $f (0) = \frac{1}{2}, g (0) = \frac{1}{3}, f^{'} (0) = 1$ and $g^{'} (0) = 2,$ then the value of $\lim_{x \to 0} [\frac{2 f (2 x^{2} + 3 x) - 1}{3 g (x) - 1}]$ is

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Important Questions on Limits

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JEE Main

IMPORTANT

Mathematics Crash Course JEE Main>Chapter 16 - Limits>Level 2>Q 21

If for some real number

a, \lim_{x \to 0} \frac{\sin 2 x + a \sin x}{x^{3}}

exists, then the limit is equal to

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Mathematics Crash Course JEE Main>Chapter 16 - Limits>Level 3>Q 1

If

g (x)

is a polynomial satisfying

g (x) g (y) = g (x) + g (y) + g (x y) - 2

for all real

x

and

y

and

g (2) = 5

, then

\lim_{x \to 3} g (x)

is

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Mathematics Crash Course JEE Main>Chapter 16 - Limits>Level 3>Q 2

If

x_{n + 1} = \sqrt{\frac{1 + x_{n}}{2}}

and

|x_{0}| < 1, n \geq 0,

then

\lim_{n \to \infty} (\frac{\sqrt{1 - {x^{2}}_{0}}}{x_{1} x_{2} x_{3} . . . . x_{n + 1}})

is equal to:

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Mathematics Crash Course JEE Main>Chapter 16 - Limits>Level 3>Q 3

A function

f

is said to be a rational function, if

f (x) = \frac{g (x)}{h (x)}

, where

g (x)

and

h (x)

are polynomials such that

h (x) \neq 0

, then

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Mathematics Crash Course JEE Main>Chapter 16 - Limits>Level 3>Q 4

The value of

\lim_{x \to 0^{+}} \frac{\tan x^{\frac{1}{5}}}{{(\tan^{- 1} \sqrt{x})}^{2}} \frac{\ln (1 + 5 x)}{(e^{3 \sqrt[5]{x}} - 1)}

is

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IMPORTANT

Mathematics Crash Course JEE Main>Chapter 16 - Limits>Level 3>Q 5

The value of

\lim_{x \to 0^{+}} {(\frac{1 - \cos (\sin^{2} x)}{x^{2}})}^{\frac{\log_{e} (1 - 2 x^{2})}{\sin^{2} x}}

is

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Mathematics Crash Course JEE Main>Chapter 16 - Limits>Level 3>Q 6

If

\lim_{x \to 0} [\frac{α x e^{x} - β \log_{e} (1 + x) + γ x^{2} e^{- x}}{x \sin^{2} x}] = 10, α, β, γ \in R

, then the value of

α + β + γ

is __________.

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Mathematics Crash Course JEE Main>Chapter 16 - Limits>Level 3>Q 7

If

\underset{x \to 0}{l i m} \frac{f (x)}{x^{2}} = a

and

\underset{x \to 0}{l i m} \frac{f (1 - c o s x)}{g (x) {s i n}^{2} x} = b

(where

b \neq 0

), then

\underset{x \to 0}{l i m} \frac{g (1 - c o s 2 x)}{x^{4}}

is