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nimcet Previous Year Questions (PYQs)

nimcet Limit PYQ

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 PYQnimcet 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:

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

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

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 PYQnimcet NIMCET 2023 PYQ

Solution

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





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nimcet Previous Year PYQnimcet 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

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

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 PYQnimcet NIMCET 2024 PYQ

Solution

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





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nimcet Previous Year PYQnimcet NIMCET 2013 PYQ

Solution

nimcet PYQ
Which of the following is NOT true?





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nimcet Previous Year PYQnimcet NIMCET 2022 PYQ

Solution

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





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nimcet Previous Year PYQnimcet NIMCET 2022 PYQ

Solution

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 PYQnimcet 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: $$\boxed{0}$$

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





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nimcet Previous Year PYQnimcet NIMCET 2021 PYQ

Solution

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 PYQnimcet 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,
$f(x)=\frac{x^{4}}{2}-2x^{3}+2x^{2}$
⇒$f(x)=x^{2}(\frac{x^{2}}{2}-2x+2)$
⇒$f(2)=2^{2}(2-4+2)$
⇒$f(2)=0$

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 PYQnimcet NIMCET 2015 PYQ

Solution

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 PYQnimcet NIMCET 2025 PYQ

Solution

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





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nimcet Previous Year PYQnimcet 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: $\boxed{2}$ ✅

nimcet PYQ
Find 





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nimcet Previous Year PYQnimcet NIMCET 2020 PYQ

Solution

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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Solution


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