NIMCET Limit Previous Year Question Paper and Solution with PDF

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NIMCET Previous Year Questions (PYQs)

NIMCET Limit PYQ

Qus : 1 NIMCET PYQ

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

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NIMCET Mathematics PYQ NIMCET Limit Continuity Differentiability (Disha JEE) PYQ

Solution

Qus : 2 NIMCET PYQ

$\lim _{{x}\rightarrow1}\frac{{x}^4-1}{x-1}=\lim _{{x}\rightarrow k}\frac{{x}^3-{k}^2}{{x}^2-{k}^2}=$, then find k

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NIMCET Previous Year PYQ NIMCET NIMCET 2023 PYQ

Solution

Quick DL Method Solution

Given:

\[ \lim_{x \to 1} \frac{x^4 - 1}{x - 1} = \lim_{x \to k} \frac{x^3 - k^2}{x^2 - k^2} \]

LHS using derivative:

\[ \lim_{x \to 1} \frac{x^4 - 1}{x - 1} = \left.\frac{d}{dx}(x^4)\right|_{x=1} = 4x^3|_{x=1} = 4 \]

RHS using DL logic:

\[ \lim_{x \to k} \frac{x^3 - k^2}{x^2 - k^2} \approx \frac{3k^2(x - k)}{2k(x - k)} = \frac{3k}{2} \]

Equating both sides:

\[ \frac{3k}{2} = 4 \Rightarrow k = \frac{8}{3} \]

\[ \boxed{k = \frac{8}{3}} \]

Qus : 3 NIMCET PYQ

Let $f(x)=\frac{x^2-1}{|x|-1}$. Then the value of $lim_{x\to-1} f(x)$ is

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NIMCET Previous Year PYQ NIMCET NIMCET 2023 PYQ

Solution

Given $f(x)=\dfrac{x^2-1}{|x|-1}$

$x=-2$

So $f(x)=\dfrac{(x-1)(x+1)}{-(x+1)}$ Cancel $(x+1)$: $f(x)=-(x-1)=1-x$

Now take the limit:

$\lim_{x\to -1}(1-x)$

$=1-(-1)=2$

Qus : 4 NIMCET PYQ

The value of $\displaystyle \lim_{n\to\infty} \frac{\pi}{n}\left[\sin\frac{\pi}{n}+\sin\frac{2\pi}{n}+\cdots+\sin\frac{(n-1)\pi}{n}\right]$ is:

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NIMCET Previous Year PYQ NIMCET NIMCET 2012 PYQ

Solution

This is a Riemann sum.

Rewrite the expression:

$\displaystyle \frac{\pi}{n}\sum_{k=1}^{n-1}\sin\left(\frac{k\pi}{n}\right)$

Let $x_k = \frac{k\pi}{n}$

Then spacing $\Delta x = \frac{\pi}{n}$

As $n\to\infty$, this becomes:

$\displaystyle \int_{0}^{\pi} \sin x , dx$

Now evaluate: $\displaystyle \int_{0}^{\pi} \sin x dx = [-\cos x]_{0}^{\pi}$

$= (-\cos\pi) - (-\cos 0)$

$= -(-1) - (-1)$

$= 1 + 1 = 2$

$\therefore$ the value of the limit is 2.

Qus : 5 NIMCET PYQ

$ \displaystyle \lim_{x\to 0} \frac{x+\sin x}{\sqrt{x}-\cos x} $

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NIMCET Previous Year PYQ NIMCET NIMCET 2011 PYQ

Solution

Use expansions: $\sin x = x - \frac{x^{3}}{6} + \cdots$ $\cos x = 1 - \frac{x^{2}}{2} + \cdots$ $\sqrt{x}$ near $0$ goes to $0$ Numerator: $ x + (x - \frac{x^{3}}{6}) = 2x + O(x^{3}) $ Denominator: $ \sqrt{x} - \cos x = \sqrt{x} - 1 + \frac{x^{2}}{2} + \cdots $ As $x\to 0$, denominator → $-1$ So limit = $0$.

Qus : 6 NIMCET PYQ

$\lim_{x\to3} \dfrac{\sqrt{3x}-3}{\sqrt{2x-4}-\sqrt{2}}$ is equal to

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NIMCET Previous Year PYQ NIMCET NIMCET 2019 PYQ

Solution

Rationalize numerator and denominator:

\[ \frac{\sqrt{3x}-3}{\sqrt{2x-4}-\sqrt{2}} =\frac{\dfrac{3(x-3)}{\sqrt{3x}+3}}{\dfrac{2(x-3)}{\sqrt{2x-4}+\sqrt{2}}} =\frac{3}{\sqrt{3x}+3}\cdot\frac{\sqrt{2x-4}+\sqrt{2}}{2}. \]

Now let $x\to 3$: $\sqrt{3x}\to 3$ and $\sqrt{2x-4}\to \sqrt{2}$. Hence

\[ \lim_{x\to 3}=\frac{3}{3+3}\cdot\frac{\sqrt{2}+\sqrt{2}}{2} =\frac{1}{2}\cdot\sqrt{2}=\boxed{\frac{1}{2\sqrt{2}}}. \]

Qus : 7 NIMCET PYQ

The value of the limit $$\lim _{{x}\rightarrow0}\Bigg{(}\frac{{1}^x+{2}^x+{3}^x+{4}^x}{4}{\Bigg{)}}^{1/x}$$ is

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NIMCET Previous Year PYQ NIMCET NIMCET 2024 PYQ

Solution

We need to find: $\displaystyle \lim_{x\to 0}\left( \frac{1^x + 2^x + 3^x + 4^x}{4} \right)^{1/x}$

Rewrite each term: $k^x = e^{x\ln k}$

So: $1^x + 2^x + 3^x + 4^x = e^{0} + e^{x\ln 2} + e^{x\ln 3} + e^{x\ln 4}$

As $x \to 0$ use expansion: $e^{x\ln k} = 1 + x\ln k + O(x^2)$

So: $1^x + 2^x + 3^x + 4^x$

$= 1 + (1 + x\ln 2) + (1 + x\ln 3) + (1 + x\ln 4)$ $= 4 + x(\ln 2 + \ln 3 + \ln 4) + O(x^2)$

Thus: $\frac{1^x + 2^x + 3^x + 4^x}{4} = 1 + \frac{x(\ln 2 + \ln 3 + \ln 4)}{4} + O(x^2)$

Now apply the standard limit:

$\lim_{x\to 0} (1 + kx)^{1/x} = e^{k}$

Here: $k = \frac{\ln 2 + \ln 3 + \ln 4}{4}$

Hence the limit is: $e^{(\ln 2 + \ln 3 + \ln 4)/4}$

Combine logs: $\ln 2 + \ln 3 + \ln 4 = \ln(2\cdot 3 \cdot 4) = \ln 24$

Final answer: $\boxed{24^{1/4}}$

Qus : 8 NIMCET PYQ

$lim_{x\to0}\left [ \frac{tanx-x}{x^{2}tanx} \right ]$ is equal to

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NIMCET Previous Year PYQ NIMCET NIMCET 2013 PYQ

Solution

Qus : 9 NIMCET PYQ

Which of the following is NOT true?

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NIMCET Previous Year PYQ NIMCET NIMCET 2022 PYQ

Solution

Qus : 10 NIMCET PYQ

For $a\in R$ (the set of al real numbers), $a \ne 1$, $\lim _{{n}\rightarrow\infty}\frac{({1}^a+{2}^a+{\ldots+{n}^a})}{{(n+1)}^{a-1}\lbrack(na+1)+(na+2)+\ldots+(na+n)\rbrack}=\frac{1}{60}$ . Then one of the value of $a$ is

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NIMCET Previous Year PYQ NIMCET NIMCET 2022 PYQ

Solution

Qus : 11 NIMCET PYQ

The value of ${{Lt}}_{x\rightarrow0}\frac{{e}^x-{e}^{-x}-2x}{1-\cos x}$ is equal to

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NIMCET Previous Year PYQ NIMCET NIMCET 2024 PYQ

Solution

Evaluate: $$\lim_{x \to 0} \frac{e^x - e^{-x} - 2x}{1 - \cos x}$$

Step 1: Apply L'Hôpital's Rule (since it's 0/0):

First derivative: $$\frac{e^x + e^{-x} - 2}{\sin x}$$

Still 0/0 → Apply L'Hôpital's Rule again: $$\frac{e^x - e^{-x}}{\cos x}$$

Now, $$\lim_{x \to 0} \frac{1 - 1}{1} = 0$$

Final Answer: $$\boxed{0}$$

Qus : 12 NIMCET PYQ

The integer $n$ for which $\displaystyle \lim_{x\to0}\frac{(\cos x-1)(\cos x-e^x)}{x^n}$ is finite and non-zero

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NIMCET Previous Year PYQ NIMCET NIMCET 2008 PYQ

Solution

$\cos x-1 \sim -\frac{x^2}{2}$ $\cos x-e^x \sim -x$ Numerator $\sim x^3$ Hence $n=3$

Qus : 13 NIMCET PYQ

$\lim_{x\to \infty} (\frac{x+7}{x+2})^{x+5}$ equal to

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NIMCET Previous Year PYQ NIMCET NIMCET 2021 PYQ

Solution

Qus : 14 NIMCET PYQ

Find the value of

$\lim_{x\to\infty}\left(\frac{\sqrt{x}}{\sqrt{x+\sqrt{x+\sqrt{x}}}}\right)$

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NIMCET Previous Year PYQ NIMCET NIMCET 2026 PYQ

Solution

Given limit is

$\lim_{x\to\infty}\left(\frac{\sqrt{x}}{\sqrt{x+\sqrt{x+\sqrt{x}}}}\right)$

Divide numerator and denominator inside the radical by $x$.

$\frac{\sqrt{x}}{\sqrt{x+\sqrt{x+\sqrt{x}}}}=\sqrt{\frac{x}{x+\sqrt{x+\sqrt{x}}}}$

$=\sqrt{\frac{1}{1+\frac{\sqrt{x+\sqrt{x}}}{x}}}$

Now,

$\frac{\sqrt{x+\sqrt{x}}}{x}\to 0$ as $x\to\infty$

Therefore,

$\lim_{x\to\infty}\sqrt{\frac{1}{1+\frac{\sqrt{x+\sqrt{x}}}{x}}}$

$=\sqrt{\frac{1}{1+0}}$

$=1$

Qus : 15 NIMCET PYQ

The value of the limit:

$\lim_{x\to 0}\frac{|x|\log_e(1+|\sin 2x|)}{x^2(|x|+3)}$

is:

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NIMCET Previous Year PYQ NIMCET NIMCET 2026 PYQ

Solution

We have,

$\lim_{x\to 0}\frac{|x|\log_e(1+|\sin 2x|)}{x^2(|x|+3)}$

As $x\to 0$,

$|\sin 2x|\sim 2|x|$

Also,

$\log_e(1+|\sin 2x|)\sim |\sin 2x|$

So,

$\log_e(1+|\sin 2x|)\sim 2|x|$

Now the numerator becomes approximately:

$|x|\cdot 2|x|=2x^2$

The denominator becomes approximately:

$x^2(|x|+3)\to 3x^2$

Therefore,

$\lim_{x\to 0}\frac{|x|\log_e(1+|\sin 2x|)}{x^2(|x|+3)}=\frac{2x^2}{3x^2}$

$=\frac{2}{3}$

Hence, the limit exists and is equal to $\frac{2}{3}$.

Qus : 16 NIMCET PYQ

Let f(x) be a polynomial of degree four, having extreme value at x = 1 and x = 2. If $\lim _{{x}\rightarrow0}[1+\frac{f(x)}{{x}^2}]=3$, then f(2) is

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NIMCET Previous Year PYQ NIMCET NIMCET 2017 PYQ

Solution

Given it has extremum values at x=1 and x=2
⇒f′(1)=0 and f′(2)=0
Given f(x) is a fourth degree polynomial
Let $f(x)=a{x}^4+b{x}^3+c{x}^2+dx+e$

Given

$\lim _{{x}\rightarrow0}[1+\frac{f(x)}{{x}^2}]=3$

$\lim _{{x}\rightarrow0}\lbrack1+\frac{a{x}^4+b{x}^3+c{x}^2+\mathrm{d}x+e}{{x}^2}\rbrack=3$

$\lim _{{x}\rightarrow0}\lbrack1+a{x}^2+bx+c+\frac{d}{x}+\frac{e}{{x}^2}\rbrack=3$

For limit to have finite value, value of 'd' and 'e' must be 0

⇒d=0 & e=0

Substituting x=0 in limit

⇒ c+1=3

⇒ c=2

$f^{\prime}(x)=4a{x}^3+3b{x}^2+2cx+d$

$x=1$ and $x=2$ are extreme values,

⇒$f^{\prime}(1)=0$ and $f^{\prime}(2)=0

⇒ $4a+3b+4=0$ and $32a+12b+8=0$

By solving these equations

we get, $a=\frac{1}{2}$ and $b=-2$

So,

$f(x)=\frac{x^{4}}{2}-2x^{3}+2x^{2}$

⇒$f(x)=x^{2}(\frac{x^{2}}{2}-2x+2)$

⇒$f(2)=2^{2}(2-4+2)$

⇒$f(2)=0$

Qus : 17 NIMCET PYQ

The value of $\lim_{x\to a} \frac{\sqrt{a+2x}-\sqrt{3x}}{\sqrt{3a+x}-2\sqrt{x}}$

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NIMCET Previous Year PYQ NIMCET NIMCET 2015 PYQ

Solution

Qus : 18 NIMCET PYQ

What is the value of $\lim _{{x}\rightarrow\infty}-(x+1)\Bigg{(}{e}^{\frac{1}{x+1}}-1\Bigg{)}$?

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NIMCET Previous Year PYQ NIMCET NIMCET 2025 PYQ

Solution

We want to evaluate $ \displaystyle \lim_{x \to \infty} -(x+1)\left(e^{\frac{1}{x+1}} - 1\right). $

Rewrite it as a product: $ -(x+1)\left(e^{\frac{1}{x+1}} - 1\right) = -\dfrac{e^{\frac{1}{x+1}} - 1}{\frac{1}{x+1}}. $

Now let $ t = \frac{1}{x+1} \Rightarrow t \to 0^+ $ as $ x \to \infty $.

The expression becomes: $ -\dfrac{e^{t} - 1}{t}. $

Now apply L'Hospital’s Rule to the limit: $ \displaystyle \lim_{t \to 0} \dfrac{e^{t} - 1}{t} = \lim_{t \to 0} \dfrac{e^{t}}{1} = 1. $

So the original limit is: $ -1. $

Final Answer: $ -1 $.

Qus : 19 NIMCET PYQ

$\lim_{x\to0} \dfrac{x \tan x}{(1-\cos x)}$

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NIMCET Previous Year PYQ NIMCET NIMCET 2017 PYQ

Solution

Given limit is indeterminate of type $\frac{0}{0}$.

(DL Rule) Differentiate numerator and denominator:

$\displaystyle \lim_{x\to0}\frac{x\tan x}{1-\cos x}=\lim_{x\to0}\frac{\tan x+x\sec^2 x}{\sin x}$

As $x\to0$, $\tan x\approx x$, $\sin x\approx x$, $\sec^2 x\approx1$

$\Rightarrow \dfrac{x+x}{x}=2$

Answer: $\boxed{2}$ ✅

Qus : 20 NIMCET PYQ

If $A = \begin{bmatrix} \cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha \end{bmatrix},$ then for any positive integer $n$, $A^n$ is

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NIMCET Previous Year PYQ NIMCET NIMCET 2020 PYQ

Solution

Qus : 21 NIMCET PYQ

Find

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NIMCET Previous Year PYQ NIMCET NIMCET 2020 PYQ

Solution

Qus : 22 NIMCET PYQ

If $f(x)=\lim _{{x}\rightarrow0}\, \frac{{6}^x-{3}^x-{2}^x+1}{\log _e9(1-\cos x)}$ is a real number then $\lim _{{x}\rightarrow0}\, f(x)$

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NIMCET Previous Year PYQ NIMCET NIMCET 2023 PYQ

Solution

Qus : 23 NIMCET PYQ

$\lim_{x\to 0}\frac{x^2+2\cos x-2}{x\sin^3 x}$

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NIMCET Previous Year PYQ NIMCET NIMCET 2026 PYQ

Solution

$\lim_{x\to 0}\frac{x^2+2\cos x-2}{x\sin^3x}$

Using expansion:

$\cos x=1-\frac{x^2}{2}+\frac{x^4}{24}$

$x^2+2\cos x-2=x^2+2\left(1-\frac{x^2}{2}+\frac{x^4}{24}\right)-2$

$=x^2+2-x^2+\frac{x^4}{12}-2$

$=\frac{x^4}{12}$

Also,

$\sin x\approx x$

So, $x\sin^3x\approx x\cdot x^3=x^4$

Therefore, $\lim_{x\to 0}\frac{x^2+2\cos x-2}{x\sin^3x}$

$=\frac{x^4/12}{x^4}$ $=\frac{1}{12}$

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