JEE Advanced Maths PYQ – Limits and Derivatives | Previous Year Questions with Solutions

Q 1 :

Let $α (a)$ and $β (a)$ be the roots of the equation $(\sqrt[3]{1 + a} - 1) x^{2} + (\sqrt{1 + a} - 1) x + (\sqrt[6]{1 + a} - 1) = 0,$

where $a > - 1$ . Then $\lim_{a \to 0^{+}} α (a) and \lim_{a \to 0^{+}} β (a)$ are [2012]

$- \frac{5}{2} and 1$
$- \frac{1}{2} and - 1$
$- \frac{7}{2} and 2$
$- \frac{9}{2} and 3$

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(2)

$(\sqrt[3]{1 + a} - 1) x^{2} + (\sqrt{1 + a} - 1) x + (\sqrt[6]{1 + a} - 1) = 0$

Let $a + 1 = y,$ then equation reduces to

$(y^{1 / 3} - 1) x^{2} + (y^{1 / 2} - 1) x + (y^{1 / 6} - 1) = 0$

On dividing both sides by $y - 1$ , we get

$(\frac{y^{1 / 3} - 1}{y - 1}) x^{2} + (\frac{y^{1 / 2} - 1}{y - 1}) x + (\frac{y^{1 / 6} - 1}{y - 1}) = 0$

On taking limit as $y \to 1 i.e. a \to 0$ on both sides, we get

$\frac{1}{3} x^{2} + \frac{1}{2} x + \frac{1}{6} = 0$ $\Rightarrow 2 x^{2} + 3 x + 1 = 0$

$\Rightarrow x = - 1, - \frac{1}{2}$ (roots of the equation)

$∴$ $\lim_{a \to 0^{+}} α (a) = - 1, \lim_{a \to 0^{+}} β (a) = - \frac{1}{2}$

Q 2 :

Let $x_{0}$ be the real number such that $e^{x_{0}} + x_{0} = 0 .$ For a given real number $α$ , define $g (x) = \frac{3 x e^{x} + 3 x - α e^{x} - α x}{3 (e^{x} + 1)}$ for all real numbers $x$ . Then which one of the following statements is TRUE? [2025]

For $α = 2$ , $\lim_{x \to x_{0}}$ $| \frac{g (x) + e^{x_{0}}}{x - x_{0}} |$ $= 0$
For $α = 2$ , $\lim_{x \to x_{0}}$ $| \frac{g (x) + e^{x_{0}}}{x - x_{0}} |$ $= 1$
For $α = 3$ , $\lim_{x \to x_{0}}$ $| \frac{g (x) + e^{x_{0}}}{x - x_{0}} |$ $= 0$
For $α = 3$ , $\lim_{x \to x_{0}}$ $| \frac{g (x) + e^{x_{0}}}{x - x_{0}} |$ $= \frac{2}{3}$

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(3)

$Given that e^{x_{0}} + x_{0} = 0$

$g (x) = \frac{3 x (e^{x} + 1) - α (e^{x} + x)}{3 (e^{x} + 1)} = x - \frac{α (e^{x} + x)}{3 (e^{x} + 1)}$ $\lim_{x \to x_{0}} \frac{g (x) + e^{x_{0}}}{x - x_{0}}$

$= \lim_{h \to 0} \frac{x_{0} + h - \frac{α (e^{x_{0} + h} + x_{0} + h)}{3 (e^{x_{0} + h} + 1)} + e^{x_{0}}}{h}$ $[∵ e^{x_{0}} + x_{0} = 0]$

$= \lim_{h \to 0} 1 - \frac{α (e^{x_{0} + h} - e^{x_{0}} + h)}{3 h (e^{x_{0} + h} + 1)}$

$= \lim_{h \to 0} 1 - \frac{α}{3 (e^{x_{0} + h} + 1)} {\frac{e^{x_{0}} (e^{h} - 1)}{h} + 1}$ $[∵ \lim_{h \to 0} \frac{e^{h} - 1}{h} = 1]$

$= 1 - \frac{α}{3 (e^{x_{0}} + 1)} (e^{x_{0}} + 1) = 1 - \frac{α}{3}$

$So for α = 2, required limit is 1 - \frac{2}{3} = \frac{1}{3}$

$for α = 3, required limit is 1 - \frac{3}{3} = 0$

Q 3 :

If $\lim_{x \to \infty} (\frac{x^{2} + x + 1}{x + 1} - a x - b) = 4,$ then [2012]

$a = 1, b = 4$
$a = 1, b = - 4$
$a = 2, b = - 3$
$a = 2, b = 3$

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(2)

$Given: \lim_{x \to \infty} (\frac{x^{2} + x + 1}{x + 1} - a x - b) = 4$

$\Rightarrow \lim_{x \to \infty} \frac{x^{2} + x + 1 - a x^{2} - a x - b x - b}{x + 1} = 4$

$\Rightarrow \lim_{x \to \infty} \frac{(1 - a) x^{2} + (1 - a - b) x + (1 - b)}{x + 1} = 4$

$For this limit to be finite, 1 - a = 0 \Rightarrow a = 1$

$Then given limit reduces to:$

$\lim_{x \to \infty} \frac{- b x + (1 - b)}{x + 1} = 4$ $\Rightarrow \lim_{x \to \infty} \frac{- b + \frac{(1 - b)}{x}}{1 + \frac{1}{x}} = 4$

$\Rightarrow - b = 4$ or $b = - 4$ $∴$ $a = 1, b = - 4$

Q 4 :

If $\lim_{x \to 0} \frac{((a - n) n x - \tan x) \sin n x}{x^{2}} = 0,$ where $n$ is a nonzero real number, then $a$ is equal to [2003]

$0$
$\frac{n + 1}{n}$
$n$
$n + \frac{1}{n}$

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(4)

$\lim_{x \to 0} \frac{[(a - n) n x - \tan x] \sin n x}{x^{2}} = 0$

$\Rightarrow \lim_{x \to 0} n \cdot \frac{\sin n x}{n x} [{(a - n) n - \frac{\tan x}{x}}] = 0$

$\Rightarrow n \cdot 1 \cdot [(a - n) n - 1] = 0$ $\Rightarrow (a - n) n - 1 = 0$

$\Rightarrow a = \frac{1}{n} + n$ $[∵ n is non zero real number]$

Q 5 :

$\lim_{x \to 0} \frac{\sin (π \cos^{2} x)}{x^{2}}$ equals [2001]

$- π$
$π$
$\frac{π}{2}$
$1$

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(2)

$\lim_{x \to 0} \frac{\sin (π \cos^{2} x)}{x^{2}} = \lim_{x \to 0} \frac{\sin (π - π \sin^{2} x)}{x^{2}}$

$= \lim_{x \to 0} \frac{\sin (π \sin^{2} x)}{π \sin^{2} x} \times \frac{π \sin^{2} x}{x^{2}} = π$

Q 6 :

The value of the limit $\lim_{x \to \frac{π}{2}} \frac{4 \sqrt{2} (\sin 3 x + \sin x)}{(2 \sin 2 x \sin \frac{3 x}{2} + \cos \frac{5 x}{2}) - (\sqrt{2} + \sqrt{2} \cos 2 x + \cos \frac{3 x}{2})}$ is ____________ [2020]

(8)

$\lim_{x \to \frac{π}{2}} \frac{4 \sqrt{2} \cdot 2 \sin 2 x \cos x}{2 \sin 2 x \sin \frac{3 x}{2} + (\cos \frac{5 x}{2} - \cos \frac{3 x}{2}) - \sqrt{2} (1 + \cos 2 x)}$

$= \lim_{x \to \frac{π}{2}} \frac{8 \sqrt{2} \cdot 2 \sin x \cos x \cos x}{2 \sin 2 x \sin \frac{3 x}{2} - 2 \sin 2 x \sin \frac{x}{2} - 2 \sqrt{2} \cos^{2} x}$

$= \lim_{x \to \frac{π}{2}} \frac{16 \sqrt{2} \sin x \cos^{2} x}{2 \sin 2 x (\sin \frac{3 x}{2} - \sin \frac{x}{2}) - 2 \sqrt{2} \cos^{2} x}$

$= \lim_{x \to \frac{π}{2}} \frac{16 \sqrt{2} \sin x \cos^{2} x}{4 \sin x \cos x (2 \cos x . \sin \frac{x}{2}) - 2 \sqrt{2} \cos^{2} x}$

$= \frac{16 \sqrt{2} \sin x \cos^{2} x}{2 \cos^{2} x (4 \sin x \sin \frac{x}{2} - \sqrt{2})}$

$= \lim_{x \to \frac{π}{2}} \frac{8 \sqrt{2} \sin x}{4 \sin x . \sin \frac{x}{2} - \sqrt{2}} = 8$

Q 7 :

Let $m$ and $n$ be two positive integers greater than 1. If $\lim_{α \to 0} (\frac{e^{\cos (α^{n})} - e}{α^{m}}) = - (\frac{e}{2}),$ then the value of $\frac{m}{n}$ is [2015]

(2)

$\lim_{α \to 0} \frac{e^{\cos α^{n}} - e}{α^{m}} = - \frac{e}{2}$

$\Rightarrow \lim_{α \to 0} \frac{e [e^{\cos α^{n} - 1} - 1]}{\cos α^{n} - 1} \times \frac{\cos α^{n} - 1}{α^{m}} = - \frac{e}{2}$

$\Rightarrow e \lim_{α \to 0} \frac{- 2 \sin^{2} \frac{α^{n}}{2}}{{(\frac{α^{n}}{2})}^{2}} \times \frac{{(\frac{α^{n}}{2})}^{2}}{α^{m}} = - \frac{e}{2}$

$\Rightarrow \frac{- e}{2} α^{2 n - m} = \frac{- e}{2}$

$\Rightarrow \frac{m}{n} = 2$

Q 8 :

The largest value of non-negative integer $a$ for which $\lim_{x \to 1} {\frac{- a x + \sin (x - 1) + a}{x + \sin (x - 1) - 1}}^{\frac{1 - x}{1 - \sqrt{x}}} = \frac{1}{4}$ is [2014]

(2)

$\lim_{x \to 1} {\frac{- a x + \sin (x - 1) + a}{x + \sin (x - 1) - 1}}^{\frac{1 - x}{1 - \sqrt{x}}} = \frac{1}{4}$

$\Rightarrow \lim_{x \to 1} {\frac{a (1 - x) + \sin (x - 1)}{(x - 1) + \sin (x - 1)}}^{1 + \sqrt{x}}$

$\Rightarrow \lim_{x \to 1} {\frac{- a + \frac{\sin (x - 1)}{x - 1}}{1 + \frac{\sin (x - 1)}{x - 1}}}^{1 + \sqrt{x}}$

$\Rightarrow {(\frac{- a + 1}{2})}^{2} = \frac{1}{4}$

$\Rightarrow a = 0 or 2$

$∴$ $Largest value of a is 2 .$

Q 9 :

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 ________. [2020]

(1)

$\lim_{x \to 0^{+}} \frac{{(1 - x)}^{1 / x} - e^{- 1}}{x^{a}} = 1$ $= \lim_{x \to 0^{+}} \frac{e^{(\frac{l n (1 - x)}{x})} - \frac{1}{e}}{x^{a}}$ $[∵ {(1 - x)}^{1 / x} = e^{\frac{1}{x} \ln (1 - x)}]$

$= \lim_{x \to 0^{+}} \frac{1}{e} \cdot \frac{e^{(1 + \frac{\ln (1 - x)}{x})} - 1}{x^{a}}$ $= \frac{1}{e} \lim_{x \to 0^{+}} \frac{l n (1 - x) + x}{x^{a + 1}}$

$= \frac{1}{e} \lim_{x \to 0^{+}} \frac{(- x - \frac{x^{2}}{2} - \frac{x^{3}}{3} - \dots) + x}{x^{a + 1}}$

$∴$ $a = 1$

Q 10 :

Let $f (x) = \frac{1 - x (1 + | 1 - x |)}{| 1 - x |} \cos (\frac{1}{1 - x}), for x \neq 1 .$ Then [2017]

${lim}_{x \to 1^{-}} f (x) = 0$
${lim}_{x \to 1^{-}} f (x)$ does not exist
${lim}_{x \to 1^{+}} f (x) = 0$
${lim}_{x \to 1^{+}} f (x)$ does not exist

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Select one or more options

(1, 4)

$Given: f (x) = \frac{1 - x (1 + | 1 - x |)}{| 1 - x |} \cos (\frac{1}{1 - x}), x \neq 1$

$\lim_{x \to 1^{-}} f (x) = \lim_{h \to 0} \frac{1 - (1 - h) (1 + h)}{h} \cos (\frac{1}{h})$

$= \lim_{h \to 0} \frac{1 - 1 + h^{2}}{h} \cos (\frac{1}{h}) = \lim_{h \to 0} h \cos (\frac{1}{h}) = 0$

$\lim_{x \to 1^{+}} f (x) = \lim_{h \to 0} \frac{1 - (1 + h) (1 + h)}{h} \cos (\frac{1}{h})$

$= \lim_{h \to 0} \frac{- 2 h - h^{2}}{h} \cos (\frac{1}{h}) = \lim_{h \to 0} (- 2 - h) \cos (\frac{1}{h})$

$= - 2 \times (some value oscillating between - 1 and 1)$

$∴$ $\lim_{x \to 1^{+}} f (x) does not exist.$

Q 11 :

For $a \in ℝ$ (the set of all real numbers), $a \neq - 1$ , $\lim_{n \to \infty} \frac{(1^{a} + 2^{a} + \dots + n^{a})}{{(n + 1)}^{a - 1} [(n a + 1) + (n a + 2) + \dots + (n a + n)]} = \frac{1}{60} .$

Then $a =$ [2013]

$5$
$7$
$\frac{- 15}{2}$
$\frac{- 17}{2}$

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Select one or more options

(2, 4)

$Given:$

$\lim_{n \to \infty} \frac{1^{a} + 2^{a} + \dots + n^{a}}{{(n + 1)}^{a - 1} [(n a + 1) + (n a + 2) + \dots + (n a + n)]} = \frac{1}{60}$

$\Rightarrow \lim_{n \to \infty} \frac{n^{a} [{(\frac{1}{n})}^{a} + {(\frac{2}{n})}^{a} + \dots + {(\frac{n}{n})}^{a}]}{{(n + 1)}^{a - 1} [n^{2} a + \frac{n (n + 1)}{2}]} = \frac{1}{60}$

$\Rightarrow \lim_{n \to \infty} \frac{n^{a - 1}}{{(n + 1)}^{a - 1}} \frac{\frac{1}{n} \sum_{r = 1}^{n} {(\frac{r}{n})}^{a}}{[a + \frac{1}{2} (1 + \frac{1}{n})]} = \frac{1}{60}$

$\Rightarrow \lim_{n \to \infty} {(\frac{1}{1 + \frac{1}{n}})}^{a - 1} \frac{\frac{1}{n} \sum_{r = 1}^{n} {(\frac{r}{n})}^{a}}{a + \frac{1}{2} (1 + \frac{1}{n})} = \frac{1}{60}$

$∵$ $\frac{1}{n} \sum_{r = 1}^{n} {(\frac{r}{n})}^{a} = \int_{0}^{1} x^{a} d x$ as $\frac{1}{n} = d x$ and $\frac{r}{n} = x$

$when r = 1, n \to \infty \Rightarrow x \to 0$

$when r = n \Rightarrow x \to 1$

$\Rightarrow \frac{\int_{0}^{1} x^{a} d x}{a + \frac{1}{2}} = \frac{1}{60}$ $\Rightarrow \frac{{[x^{a + 1}]}_{0}^{1}}{(a + 1) (a + \frac{1}{2})} = \frac{1}{60}$

$\Rightarrow \frac{1}{(a + 1) (a + \frac{1}{2})} = \frac{1}{60}$

$\Rightarrow 2 a^{2} + 3 a - 119 = 0$ $\Rightarrow (a - 7) (2 a - 17) = 0$

$∴$ $a = 7 or - \frac{17}{2}$

Q 12 :

Let $f : R \to R$ be a function. We say that $f$ has

PROPERTY 1: If $\lim_{h \to 0} \frac{f (h) - f (0)}{\sqrt{| h |}}$ exists and is finite, and

PROPERTY 2: If $\lim_{h \to 0} \frac{f (h) - f (0)}{h^{2}}$ exists and is finite.

Then which of the following options is/are correct? [2019]

$f (x) = x^{2 / 3}$ has PROPERTY 1
$f (x) = \sin x$ has PROPERTY 2
$f (x) = | x |$ has PROPERTY 1
$f (x) = x | x |$ has PROPERTY 2

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Select one or more options

(1, 3)

Property 1: $\lim_{h \to 0} \frac{f (h) - f (0)}{\sqrt{| h |}}$ exists and is finite

Property 2: $\lim_{h \to 0} \frac{f (h) - f (0)}{h^{2}}$

(1) $f (x) = x^{2 / 3}$ for Property 1

$\lim_{h \to 0} \frac{h^{2 / 3} - 0}{\sqrt{| h |}} = \lim_{h \to 0} \frac{| h |^{2 / 3}}{| h |^{1 / 2}} = \lim_{h \to 0} | h |^{1 / 6} = 0$

$∴$ $option (1) is correct.$

(2) $f (x) = \sin x$ for Property 2

$\lim_{h \to 0} \frac{\sin h - \sin 0}{h^{2}} = \lim_{h \to 0} \frac{\sin h}{h} \times \frac{1}{h}$ which does not exist.

$∴$ $(2) is incorrect option.$

(3) $f (x) = | x |$ for Property 1

$\lim_{h \to 0} \frac{| h | - 0}{\sqrt{| h |}} = \lim_{h \to 0} \sqrt{| h |} = 0$

$∴$ $option (3) is correct.$

(4) $f (x) = x | x |$ for Property 2

$\lim_{h \to 0} \frac{h | h | - 0}{h^{2}} = \lim_{h \to 0} \frac{| h |}{h}$

LHL $= - 1$ and RHL $= 1$

$∴$ $\lim_{h \to 0} \frac{| h |}{h} does not exist$

$∴$ $option (4) is incorrect.$

Topic Question Set

Q 1 :

Let $α (a)$ and $β (a)$ be the roots of the equation $(\sqrt[3]{1 + a} - 1) x^{2} + (\sqrt{1 + a} - 1) x + (\sqrt[6]{1 + a} - 1) = 0,$

where $a > - 1$ . Then $\lim_{a \to 0^{+}} α (a) and \lim_{a \to 0^{+}} β (a)$ are [2012]

Q 2 :

Let $x_{0}$ be the real number such that $e^{x_{0}} + x_{0} = 0 .$ For a given real number $α$ , define $g (x) = \frac{3 x e^{x} + 3 x - α e^{x} - α x}{3 (e^{x} + 1)}$ for all real numbers $x$ . Then which one of the following statements is TRUE? [2025]

Q 3 :

If $\lim_{x \to \infty} (\frac{x^{2} + x + 1}{x + 1} - a x - b) = 4,$ then [2012]

Q 4 :

If $\lim_{x \to 0} \frac{((a - n) n x - \tan x) \sin n x}{x^{2}} = 0,$ where $n$ is a nonzero real number, then $a$ is equal to [2003]

Q 5 :

$\lim_{x \to 0} \frac{\sin (π \cos^{2} x)}{x^{2}}$ equals [2001]

Q 6 :

The value of the limit $\lim_{x \to \frac{π}{2}} \frac{4 \sqrt{2} (\sin 3 x + \sin x)}{(2 \sin 2 x \sin \frac{3 x}{2} + \cos \frac{5 x}{2}) - (\sqrt{2} + \sqrt{2} \cos 2 x + \cos \frac{3 x}{2})}$ is ____________ [2020]

Q 7 :

Let $m$ and $n$ be two positive integers greater than 1. If $\lim_{α \to 0} (\frac{e^{\cos (α^{n})} - e}{α^{m}}) = - (\frac{e}{2}),$ then the value of $\frac{m}{n}$ is [2015]

Q 8 :

The largest value of non-negative integer $a$ for which $\lim_{x \to 1} {\frac{- a x + \sin (x - 1) + a}{x + \sin (x - 1) - 1}}^{\frac{1 - x}{1 - \sqrt{x}}} = \frac{1}{4}$ is [2014]

Q 9 :

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 ________. [2020]

Q 10 :

Let $f (x) = \frac{1 - x (1 + | 1 - x |)}{| 1 - x |} \cos (\frac{1}{1 - x}), for x \neq 1 .$ Then [2017]

Q 11 :

For $a \in ℝ$ (the set of all real numbers), $a \neq - 1$ , $\lim_{n \to \infty} \frac{(1^{a} + 2^{a} + \dots + n^{a})}{{(n + 1)}^{a - 1} [(n a + 1) + (n a + 2) + \dots + (n a + n)]} = \frac{1}{60} .$

Then $a =$ [2013]

Q 12 :

Let $f : R \to R$ be a function. We say that $f$ has

PROPERTY 1: If $\lim_{h \to 0} \frac{f (h) - f (0)}{\sqrt{| h |}}$ exists and is finite, and

PROPERTY 2: If $\lim_{h \to 0} \frac{f (h) - f (0)}{h^{2}}$ exists and is finite.

Then which of the following options is/are correct? [2019]

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Topic Question Set

Q 1 : Let α(a) and β(a) be the roots of the equation (1+a3-1)x2+(1+a-1)x+(1+a6-1)=0, where a>-1. Then lima→0+α(a) and lima→0+β(a) are [2012]

Q 2 : Let x0 be the real number such that ex0+x0=0. For a given real number α, define g(x)=3xex+3x-αex-αx3(ex+1) for all real numbers x. Then which one of the following statements is TRUE? [2025]

Q 3 : If limx→∞(x2+x+1x+1-ax-b)=4, then [2012]

Q 4 : If limx→0((a-n)nx-tanx)sinnxx2=0, where n is a nonzero real number, then a is equal to [2003]

Q 5 : limx→0sin(πcos2x)x2 equals [2001]

Q 6 : The value of the limit limx→π242(sin3x+sinx)(2sin2xsin3x2+cos5x2)-(2+2cos2x+cos3x2) is ____________ [2020]

Q 7 : Let m and n be two positive integers greater than 1. If limα→0(ecos(αn)-eαm)=-(e2), then the value of mn is [2015]

Q 8 : The largest value of non-negative integer a for which limx→1{-ax+sin(x-1)+ax+sin(x-1)-1}1-x1-x=14 is [2014]

Q 9 : Let e denote the base of the natural logarithm. The value of the real number a for which the right hand limit limx→0+(1-x)1x-e-1xa is equal to a non-zero real number, is ________. [2020]

Q 10 : Let f(x)=1-x(1+|1-x|)|1-x|cos(11-x), for x≠1. Then [2017]

Q 11 : For a∈ℝ (the set of all real numbers), a≠-1, limn→∞(1a+2a+⋯+na)(n+1)a-1[(na+1)+(na+2)+⋯+(na+n)]=160. Then a= [2013]

Q 12 : Let f:R→R be a function. We say that f has PROPERTY 1: If limh→0f(h)-f(0)|h| exists and is finite, and PROPERTY 2: If limh→0f(h)-f(0)h2 exists and is finite. Then which of the following options is/are correct? [2019]

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Q 1 :

Let $α (a)$ and $β (a)$ be the roots of the equation $(\sqrt[3]{1 + a} - 1) x^{2} + (\sqrt{1 + a} - 1) x + (\sqrt[6]{1 + a} - 1) = 0,$

where $a > - 1$ . Then $\lim_{a \to 0^{+}} α (a) and \lim_{a \to 0^{+}} β (a)$ are [2012]

Q 2 :

Let $x_{0}$ be the real number such that $e^{x_{0}} + x_{0} = 0 .$ For a given real number $α$ , define $g (x) = \frac{3 x e^{x} + 3 x - α e^{x} - α x}{3 (e^{x} + 1)}$ for all real numbers $x$ . Then which one of the following statements is TRUE? [2025]

Q 3 :

If $\lim_{x \to \infty} (\frac{x^{2} + x + 1}{x + 1} - a x - b) = 4,$ then [2012]

Q 4 :

If $\lim_{x \to 0} \frac{((a - n) n x - \tan x) \sin n x}{x^{2}} = 0,$ where $n$ is a nonzero real number, then $a$ is equal to [2003]

Q 5 :

$\lim_{x \to 0} \frac{\sin (π \cos^{2} x)}{x^{2}}$ equals [2001]

Q 6 :

The value of the limit $\lim_{x \to \frac{π}{2}} \frac{4 \sqrt{2} (\sin 3 x + \sin x)}{(2 \sin 2 x \sin \frac{3 x}{2} + \cos \frac{5 x}{2}) - (\sqrt{2} + \sqrt{2} \cos 2 x + \cos \frac{3 x}{2})}$ is ____________ [2020]

Q 7 :

Let $m$ and $n$ be two positive integers greater than 1. If $\lim_{α \to 0} (\frac{e^{\cos (α^{n})} - e}{α^{m}}) = - (\frac{e}{2}),$ then the value of $\frac{m}{n}$ is [2015]

Q 8 :

The largest value of non-negative integer $a$ for which $\lim_{x \to 1} {\frac{- a x + \sin (x - 1) + a}{x + \sin (x - 1) - 1}}^{\frac{1 - x}{1 - \sqrt{x}}} = \frac{1}{4}$ is [2014]

Q 9 :

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 ________. [2020]

Q 10 :

Let $f (x) = \frac{1 - x (1 + | 1 - x |)}{| 1 - x |} \cos (\frac{1}{1 - x}), for x \neq 1 .$ Then [2017]

Q 11 :

For $a \in ℝ$ (the set of all real numbers), $a \neq - 1$ , $\lim_{n \to \infty} \frac{(1^{a} + 2^{a} + \dots + n^{a})}{{(n + 1)}^{a - 1} [(n a + 1) + (n a + 2) + \dots + (n a + n)]} = \frac{1}{60} .$

Then $a =$ [2013]

Q 12 :

Let $f : R \to R$ be a function. We say that $f$ has

PROPERTY 1: If $\lim_{h \to 0} \frac{f (h) - f (0)}{\sqrt{| h |}}$ exists and is finite, and

PROPERTY 2: If $\lim_{h \to 0} \frac{f (h) - f (0)}{h^{2}}$ exists and is finite.

Then which of the following options is/are correct? [2019]