Jamia Millia Islamia Limit Previous Year Question Paper and Solution with PDF

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Jamia Millia Islamia Previous Year Questions (PYQs)

Jamia Millia Islamia Limit PYQ

Qus : 1 Jamia Millia Islamia PYQ

Let $f(t) = \begin{vmatrix} \cos t & 1 & 1 \\ 2\sin t & 2t & 1 \\ \sin t & t & t \end{vmatrix}$, then $\displaystyle \lim_{t \to 0}\dfrac{f(t)}{t^2}$ is equal to:

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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MILLIA ISLAMIA MCA 2021 PYQ

Solution

Expand the determinant using first row: $f(t) = \cos t \begin{vmatrix} 2t & 1 \\ t & t \end{vmatrix} - 1 \begin{vmatrix} 2\sin t & 1 \\ \sin t & t \end{vmatrix} + 1 \begin{vmatrix} 2\sin t & 2t \\ \sin t & t \end{vmatrix}$ Simplify and expand around $t \to 0$ using $\sin t \approx t$ and $\cos t \approx 1 - t^2/2$. After simplification, $\dfrac{f(t)}{t^2} \to 3$. $\boxed{\text{Answer: (D) 3}}$

Qus : 2 Jamia Millia Islamia PYQ

For $x \in \mathbb{R}$, find $\displaystyle \lim_{x \to \infty} \left(\dfrac{x - 3}{x + 2}\right)^x = ?$

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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MILLIA ISLAMIA MCA 2023 PYQ

Solution

Let $L = \left(\dfrac{x - 3}{x + 2}\right)^x = \left(1 - \dfrac{5}{x + 2}\right)^x.$ Take $\ln L = x \ln\!\left(1 - \dfrac{5}{x + 2}\right).$ For large $x$, use $\ln(1 - y) \approx -y$. So, $\ln L \approx x \left(-\dfrac{5}{x + 2}\right) \to -5.$ Hence, $L = e^{-5}.$

Qus : 3 Jamia Millia Islamia PYQ

$\displaystyle \lim_{x\to0}\frac{1-\cos x}{x^{2}}$ is equal to …

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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MCA 2017 PYQ

Solution

Standard limit: $\frac{1-\cos x}{x^{2}}\to \frac12$.

Qus : 4 Jamia Millia Islamia PYQ

$\displaystyle \lim_{x\to\infty}\Big(x-\sqrt{x^{2}+x}\,\Big)$ is equal to …

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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MCA 2017 PYQ

Solution

$\sqrt{x^{2}+x}=x\sqrt{1+1/x}=x\left(1+\tfrac{1}{2x}+o(1/x)\right)=x+\tfrac12+o(1)$. Hence limit $=x-(x+\tfrac12)= -\tfrac12$.

Qus : 5 Jamia Millia Islamia PYQ

What will be the limiting value of $f(x)=|x|-5$ when $x \to 5$?

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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MILLIA ISLAMIA MCA 2020 PYQ

Solution

$\lim_{x \to 5} (|x|-5) = |5|-5 = 5-5 = 0.$

Qus : 6 Jamia Millia Islamia PYQ

$\displaystyle \lim_{x \to \frac{\pi}{4}} \frac{\sin x - \cos x}{x - \frac{\pi}{4}}$ is equal to …

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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MCA 2016 PYQ

Solution

his is the derivative of $\sin x-\cos x$ at $x=\frac{\pi}{4}$. $\,(\sin x-\cos x)'=\cos x+\sin x \Rightarrow \cos\frac{\pi}{4}+\sin\frac{\pi}{4} =\tfrac{\sqrt2}{2}+\tfrac{\sqrt2}{2}=\sqrt2.$

Qus : 7 Jamia Millia Islamia PYQ

$\displaystyle \lim_{h \to 0} \frac{\sqrt{x + h} - \sqrt{x}}{h}$ is equal to …

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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MCA 2016 PYQ

Solution

$\sqrt{x}$ at $x$: $\dfrac{d}{dx}\sqrt{x}=\dfrac{1}{2\sqrt{x}}.$

Qus : 8 Jamia Millia Islamia PYQ

If $\displaystyle \lim_{x \to a} \frac{x^{\alpha} - x^{a}}{x - a} = -1$, then $\alpha$ is equal to …

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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MCA 2016 PYQ

Solution

Limit $\Rightarrow \dfrac{d}{dx}x^{\alpha}=\alpha x^{\alpha-1}$. At $x=a$, $\alpha a^{\alpha-1}=-1$. If $a=1$, $\alpha=-1$.

Qus : 9 Jamia Millia Islamia PYQ

$\displaystyle \lim_{x\to a}\frac{x^{10}-a^{10}}{x^{2}-a^{2}}$ is equal to …

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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MCA 2016 PYQ

Solution

Using L’Hôpital’s rule or factorization, $\dfrac{10a^{9}}{2a}=5a^{8}$.

Qus : 10 Jamia Millia Islamia PYQ

$\displaystyle \lim_{x\to1}\frac{x+x^{2}+\ldots+x^{10}-10}{5x-5}$ is equal to …

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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MCA 2016 PYQ

Solution

Let $f(x)=x+x^{2}+\ldots+x^{10}-10$. Then $f'(x)=1+2x+3x^{2}+\ldots+10x^{9}$. At $x=1$, $f'(1)=1+2+3+\ldots+10=55$. Hence limit $=\dfrac{f'(1)}{5}=\dfrac{55}{5}=11$.

Qus : 11 Jamia Millia Islamia PYQ

What will the following evaluate to? $\displaystyle \lim_{x \to 4} \left(\frac{4x+3}{x-2}\right)$

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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MILLIA ISLAMIA MCA 2020 PYQ

Solution

Direct substitution (function is continuous at $x=4$): $\displaystyle \frac{4(4)+3}{4-2} = \frac{16+3}{2} = \frac{19}{2}$

Qus : 12 Jamia Millia Islamia PYQ

What is $\displaystyle \lim_{x\to 0}\left(\frac{x\tan x}{\cot x}\right)$?

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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MILLIA ISLAMIA MCA 2024 PYQ

Solution

$\tan x\sim x,\ \cot x\sim \frac1x \Rightarrow \dfrac{x\tan x}{\cot x}\sim \dfrac{x\cdot x}{1/x}=x^3\to0.$

Qus : 13 Jamia Millia Islamia PYQ

Evaluate $\displaystyle \lim_{x\to 0}\left\lfloor \frac{\sin x}{x}\right\rfloor$, where $[;]$ denotes the greatest-integer function.

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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MILLIA ISLAMIA MCA 2019 PYQ

Solution

$\dfrac{\sin x}{x}=1-\dfrac{x^2}{6}+O(x^4)<1$ for $x\ne0$ near $0$, and $>0$. So $\left\lfloor \dfrac{\sin x}{x}\right\rfloor=0$ in a punctured neighborhood of $0$.

Qus : 14 Jamia Millia Islamia PYQ

Evaluate $\displaystyle \lim_{x\to 0}\frac{\sqrt{1-\cos 2x}}{x}$.

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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MILLIA ISLAMIA MCA 2019 PYQ

Solution

$1-\cos 2x=2\sin^2 x \Rightarrow \dfrac{\sqrt{1-\cos 2x}}{x}=\sqrt{2},\dfrac{|\sin x|}{x}$. As $x\to0^+$, this $\to\sqrt{2}$; as $x\to0^-$, it $\to -\sqrt{2}$. Thus two-sided limit does not exist; the right-hand limit is $\sqrt{2}$.

Qus : 15 Jamia Millia Islamia PYQ

If $\displaystyle \lim_{x \to \infty} \left(1 + \frac{a}{x} + \frac{b}{x^2}\right)^{2x} = e^2$, then values of $a$ and $b$ are:

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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MILLIA ISLAMIA MCA 2019 PYQ

Solution

Let $\ln L = 2x \ln\left(1 + \frac{a}{x} + \frac{b}{x^2}\right)$ Using expansion $\ln(1+z) = z - \frac{z^2}{2} + \dots$ $\ln L = 2x\left(\frac{a}{x} + \frac{b}{x^2} - \frac{a^2}{2x^2}\right)$ $= 2a + \frac{2(b - a^2/2)}{x} + \dots$ As $x \to \infty$, $\ln L \to 2a$ Given $\ln L = 2 \Rightarrow 2a = 2 \Rightarrow a = 1$. Hence, $b$ can be any real number.

Qus : 16 Jamia Millia Islamia PYQ

Given that limit exists, find $\displaystyle \lim_{(x,y,z)\to(-2,-2,-2)} \frac{\sin((x+2)(y+5)(z+1))}{(x+2)(y+7)}$

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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MILLIA ISLAMIA MCA 2018 PYQ

Solution

Using $\sin t / t \to 1$, limit $=1$.

Qus : 17 Jamia Millia Islamia PYQ

Which of the following is not an indeterminate form?

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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MILLIA ISLAMIA MCA 2019 PYQ

Solution

Indeterminate forms: $0^0, \infty^0, 1^\infty$. $1^0$ is determinate (equals 1).

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