Jamia Millia Islamia MCA Limit Previous Year Questions (PYQs) – Page 1 of 2
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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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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, $\dfrac{f(t)}{t^2} \to 3$.
$\boxed{\text{Answer: (D) 3}}$
For $x \in \mathbb{R}$, find $\displaystyle \lim_{x \to \infty} \left(\dfrac{x - 3}{x + 2}\right)^x = ?$
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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, $L = e^{-5}.$
$\displaystyle \lim_{x\to0}\frac{1-\cos x}{x^{2}}$ is equal to …
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Solution
Standard limit: $\frac{1-\cos x}{x^{2}}\to \frac12$.
$\displaystyle \lim_{x\to\infty}\Big(x-\sqrt{x^{2}+x}\,\Big)$ is equal to …
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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 $=x-(x+\tfrac12)= -\tfrac12$.
What will be the limiting value of $f(x)=|x|-5$ when $x \to 5$?
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Solution
$\lim_{x \to 5} (|x|-5) = |5|-5 = 5-5 = 0.$
$\displaystyle \lim_{x \to \frac{\pi}{4}} \frac{\sin x - \cos x}{x - \frac{\pi}{4}}$ is equal to …
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Solution
his is the derivative of $\sin x-\cos x$ at $x=\frac{\pi}{4}$.
$\,(\sin x-\cos x)'=\cos x+\sin x \Rightarrow \cos\frac{\pi}{4}+\sin\frac{\pi}{4}
=\tfrac{\sqrt2}{2}+\tfrac{\sqrt2}{2}=\sqrt2.$
$\displaystyle \lim_{h \to 0} \frac{\sqrt{x + h} - \sqrt{x}}{h}$ is equal to …
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Solution
$\sqrt{x}$ at $x$:
$\dfrac{d}{dx}\sqrt{x}=\dfrac{1}{2\sqrt{x}}.$
If $\displaystyle \lim_{x \to a} \frac{x^{\alpha} - x^{a}}{x - a} = -1$, then $\alpha$ is equal to …
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Solution
Limit $\Rightarrow \dfrac{d}{dx}x^{\alpha}=\alpha x^{\alpha-1}$.
At $x=a$, $\alpha a^{\alpha-1}=-1$.
If $a=1$, $\alpha=-1$.
$\displaystyle \lim_{x\to a}\frac{x^{10}-a^{10}}{x^{2}-a^{2}}$ is equal to …
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Solution
Using L’Hôpital’s rule or factorization,
$\dfrac{10a^{9}}{2a}=5a^{8}$.
$\displaystyle \lim_{x\to1}\frac{x+x^{2}+\ldots+x^{10}-10}{5x-5}$ is equal to …
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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 $=\dfrac{f'(1)}{5}=\dfrac{55}{5}=11$.
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