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AMU MCA Previous Year Questions (PYQs)

AMU MCA Limit PYQ

AMU MCA PYQ
$\lim_{(x,y)\to(0,0)} f(x,y)$ where $f(x,y)=\tan^{-1}\left(\frac{|x|+|y|}{x^2+y^2}\right)$ is






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AMU MCA Previous Year PYQ AMU MCA AMU MCA 2020 PYQ

Solution

As $(x,y)\to(0,0)$, $|x|+|y|\sim r$ and $x^2+y^2=r^2$ so $\frac{|x|+|y|}{x^2+y^2}\sim\frac{1}{r}\to\infty$

Hence $\tan^{-1}(\cdot)\to\frac{\pi}{2}$

AMU MCA PYQ
$\lim_{x \to 0} \dfrac{\cos(\sin x) - 1}{x^2}$ equals





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AMU MCA Previous Year PYQ AMU MCA AMU MCA 2019 PYQ

Solution

AMU MCA PYQ
$\lim_{x\to0}\dfrac{\sin x - x + \frac{x^3}{6}}{x^5}$ is equal to





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AMU MCA Previous Year PYQ AMU MCA AMU MCA 2022 PYQ

Solution

AMU MCA PYQ
If $\lim_{x\to0}|x|^{[\cos x]}=l$, where $[.]$ denotes the greatest integer function, then the value of $l$ is





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AMU MCA Previous Year PYQ AMU MCA AMU MCA 2018 PYQ

Solution

AMU MCA PYQ
Let $f(x)$ be a function defined by $f(x)=\begin{cases}4x-5,\ x\le2 \ x-\lambda,\ x>2 \end{cases}$ If $\lim_{x\to2} f(x)$ exists, then the value of $\lambda$ is





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AMU MCA Previous Year PYQ AMU MCA AMU MCA 2017 PYQ

Solution

For limit to exist,

$ \lim_{x\to2^-} f(x)=\lim_{x\to2^+} f(x) $

$ \lim_{x\to2^-} f(x)=4(2)-5=3 $

$ \lim_{x\to2^+} f(x)=2-\lambda $

Equating,

$ 3=2-\lambda $

$ \Rightarrow \lambda=-1 $

AMU MCA PYQ
$ \lim_{x→0} (cosec x)^{1/log x} $





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AMU MCA Previous Year PYQ AMU MCA AMU MCA 2025 PYQ

Solution

As $x \to 0$, $\csc x \sim \frac{1}{x}$. So expression becomes $\left(\frac{1}{x}\right)^{1/\log x}$. Using standard limit result, the value is $\frac{1}{e}$.
AMU MCA PYQ
If $\lim_{x\to 0} \frac{\log(3+x) - \log(3-x)}{x} = K$, then $K$ is





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AMU MCA Previous Year PYQ AMU MCA AMU MCA 2016 PYQ

Solution

Use derivative: $K = \left[\frac{d}{dx}(\log(3+x)-\log(3-x))\right]_{x=0}$ $= \frac{1}{3} + \frac{1}{3} = \frac{2}{3}$

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