Limits and Derivatives
59 JEE Maths previous year questions on Limits and Derivatives — options free on every question; 6 include the answer & explanation free, the rest unlock with PYQ Pass.
The value of limit: \(\lim _{x\to 0}(\csc x)\left(\sqrt{2{\cos }^{2}x+3\cos x}-\sqrt{{\cos }^{2}x+\sin x+4}\right)\text{ is }\)
\(\frac{-1}{2\sqrt{5}}\)
\(\lim _{x\to 0}\csc x\left(\sqrt{2{\cos }^{2}x+3\cos x}-\sqrt{{\cos }^{2}x+\sin x+4}\right)\\ =\lim _{x\to 0}\frac{1}{\sin x}\left[\frac{2{\cos }^{2}x+3\cos x-{\cos }^{2}x-\sin x-4}{\sqrt{2{\cos }^{2}x+3\cos x}+\sqrt{{\cos }^{2}x+\sin x+4}}\right]\\ =\lim _{x\to 0}\frac{1}{2\sqrt{5}}(\frac{{\cos }^{2}x+3\cos x-\sin x-4}{\sin x})\\ \mathrm{using}\mathrm{L}'\mathrm{hospital}\mathrm{rule}-\\ =\frac{1}{2\sqrt{5}}\lim _{x\to 0}\left(-\frac{2\cos x\sin x-3\sin x-\cos x}{\cos x}\right)\\ =\frac{-1}{2\sqrt{5}}\\ \\\)
The value of \(\lim _{n\to \infty }\left(\sum _{k=1}^{n}\frac{{k}^{3}+6{k}^{2}+11k+5}{(k+3)!}\right)\) is
[JEE Main 2025, 29 Jan (Shift 1)]
\(\frac{5}{3}\)
\({k}^{3}+6{k}^{2}+11k+5\\ ={k}^{3}+6{k}^{2}+11k+6-1\\ =(k+1)(k+2)(k+3)-1\)
\(\lim _{n\to \infty }\sum _{k=1}^{n}\frac{{k}^{3}+6{k}^{2}+11k+5}{(k+3)!}\\ =\lim _{n\to \infty }\sum _{k=1}^{n}\frac{1}{k!}-\frac{1}{(k+3)!}\\ =\left(1+\frac{1}{2!}+\frac{1}{3!}+........\right)\\ -\left(\frac{1}{4!}+\frac{1}{5!}+\frac{1}{6!}+.........\right)\\ =1+\frac{1}{2}+\frac{1}{6}=\frac{5}{3}\)
\(f(x)-6 f\left(\frac{1}{x}\right)=\frac{35}{3 x}-\frac{5}{2} . \lim _{\text {If }}\left(\frac{1}{\alpha x}+f(x)\right)=\beta . \quad \text { find }(\alpha+2 \beta)\). (24 Jan, Shift I, Memory based)
4
\( f(x)-6 f\left(\frac{1}{x}\right)=\frac{35}{3 x}-\frac{5}{2}\)
\( f\left(\frac{1}{x}\right)-6 f(x)=\left(\frac{35 x}{3}-\frac{5}{2}\right)\)
\( 6 f\left(\frac{1}{x}\right)-36 f(x)=\left(\frac{35 x}{3}-\frac{5}{2}\right) \times 6 \)
\( -35 f(x)=\frac{35}{3 x}-\frac{5}{2}+70 x-15 \)
\( -35 f(x)=70 x+\frac{35}{3 x}-\frac{35}{2} \)
\( f(x)=\frac{1}{2}-2 x-\frac{1}{3 x}\)
\( {lim}_{x \rightarrow 0} \frac{1}{\alpha x}+\frac{1}{2}-2 x-\frac{1}{3 x}=\beta \)
\(={lim}_{x \rightarrow 0}\left(\frac{1}{\alpha}-\frac{1}{3}\right)\frac{1}{x}+\frac{1}{2}-2 x=\beta\)
\( \alpha=3 \)
\( \beta=\frac{1}{2}\)
\(\lim _{x \rightarrow \infty}\left(\left(\frac{e}{1-e}\right)\left(\frac{1}{e}-\frac{x}{1+x}\right)\right)^x=\alpha, \text { find } \frac{\ln \alpha}{1+\ln \alpha}\) (22 Jan, Shift II, Memory Based)
e
\(\begin{aligned}& \lim _{x \rightarrow \infty}\left(\left(\frac{e}{e-1}\right)\left(\frac{1}{e}-\frac{x}{1+x}\right)\right)^x \\& \lim _{x \rightarrow \infty}\left(\left(\frac{1}{e-1}-\frac{e x}{(e-1)(x+1)}\right)^x\right) \\& \alpha=e^{\lim _{x \rightarrow \infty}\left(\frac{1}{e-1}-\frac{e x}{(e-1)(x+1)}-1\right)x} \\& \ln (\alpha)=\lim _{x \rightarrow \infty}\left(\frac{e }{(1-e)}-\frac{ex}{(1-e)(1+x)}\right) x \\& \lim _{x \rightarrow \infty} \frac{ex}{(1-e)(1+x)}=\frac{e}{1-e} \\& \text { Now, } \frac{\ln \alpha}{1+\ln \alpha}=\frac{\frac{e}{1-e}}{1+\frac{e}{1-e}}=e\end{aligned}\)
\(\text { Evaluate } \lim _{x \rightarrow \infty}\left(\frac{2 x^2-3 x+10}{3 x^2+4 x-2}\right) \frac{(3 x-1)^{\frac{x}{2}}}{(\sqrt{3 x+2})^x}=\)
\(\frac{2}{3\sqrt{e}}\)
Ans. (3)
Sol.
\(\begin{aligned}& \lim _{x \rightarrow \infty} \frac{x^2\left(2-\frac{3}{x}+\frac{10}{x^2}\right)}{x^2\left(3+\frac{4}{x}-\frac{2}{x^2}\right)} \cdot\left(\frac{3 x-1}{3 x+2}\right)^{x / 2} \\& \lim _{x \rightarrow \infty} \frac{2}{3} \times e^{\left(\frac{-3}{3 x+2}\right)\left(\frac{x}{2}\right)} \\& =\frac{2}{3} e^{-\frac{1}{2}}=\frac{2}{3 \sqrt{e}}\end{aligned}\)
If \(\lim _{x\to 0}\frac{{e}^{(a−1)x}+2\cos bx+(c−2){e}^{−x}}{x\cos x−{\log }_{e}(1+x)}=2,\) then \({a}^{2}+{b}^{2}+{c}^{2}\) is equal to:
[JEE Main 2026, 22 Jan (Shift 2)]
\(7\)
Given: \(\lim _{x\to 0}\frac{{e}^{(a−1)x}+2\cos bx+(c−2){e}^{−x}}{x\cos x−{\log }_{e}(1+x)}=2,\)
\({\text{⇒lim}}_{x\to 0}\frac{\left(1+\left(a−1\right)x+\frac{{(a−1)}^{2}{x}^{2}}{2!}\right)+2\left(1−\frac{{b}^{2}{x}^{2}}{2!}\right)+\left(c−2\right)\left(1−x+\frac{{x}^{2}}{2!}\right)}{x\left(1−\frac{{x}^{2}}{2!}\right)−\left(x−\frac{{x}^{2}}{2}\ldots \right)}=2\)
\({\text{⇒lim}}_{x\to 0}\frac{\left(1+2+c−2\right)+x\left(a−1−c+2\right)+{x}^{2}\left(\frac{{(a−1)}^{2}}{2}−{b}^{2}+\left(\frac{c−2}{2}\right)\right)}{\frac{{x}^{2}}{2}−\frac{{x}^{3}}{2!}+\ldots }=2\)
For which
\(∵c+1=0\Rightarrow c=−1\)
\(∵a−c=−1\Rightarrow a=−2\)
\(∵\frac{{(\text{a}−1)}^{2}}{2}−{\text{b}}^{2}+\left(\frac{\text{c}−2}{2}\right)=1\)
So, \(\frac{9}{2}−{\text{b}}^{2}−\frac{3}{2}=1\Rightarrow {\text{ b}}^{2}=2\)
Hence, \({a}^{2}+{b}^{2}+{c}^{2}=4+2+1=7\)
Let \(f\) be a differentiable function on \(R\) such that \(f(2)=1\), \({f}^{'}(2)=4\). Let \(\lim _{x \rightarrow 0}(f(2+x))^{\frac{3}{x}}=e^{\alpha}\). Then the number of times the curve \(y=4{x}^{3}-4{x}^{2}-4(\alpha -7)x-\alpha\) meets x -axis is :-
[JEE Main 2025, 4 Apr (Shift 2)]
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If \(\lim _{x\to 0}\frac{{e}^{(a−1)x}+2\cos bx+(c−2){e}^{−x}}{x\cos x−{\log }_{e}(1+x)}=2,\) then \({a}^{2}+{b}^{2}+{c}^{2}\) is equal to:
[JEE Main 2026, 22 Jan (Shift 2)]
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The value of \(\lim _{x\to 0}\left(\frac{{x}^{2}{\sin }^{2}x}{{x}^{2}-{\sin }^{2}x}\right)\) is:
[JEE Main 2026, 6 Apr (Shift 1)]
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\(\begin{aligned}&\text { Evaluate }\\&\lim _{x \rightarrow 0} \operatorname{cosec} x .\left(\sqrt{2 \cos ^2 x+3 \cos x}-\sqrt{\cos ^2 x+\sin x+4}\right)\end{aligned}\) (24 Jan, Shift I, Memory Based)
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Let f be a differentiable function on \(R\) such that \(f(2)=1\), \({f}^{'}(2)=4\). Let \(\lim _{\mathrm{x}\to 0}(\mathrm{f}(2+\mathrm{x}){)}^{3/\mathrm{x}}={\mathrm{e}}^{\alpha }\). Then the number of times the curve \(\mathrm{y}=4{\mathrm{x}}^{3}-4{\mathrm{x}}^{2}-4(\alpha -7)\mathrm{x}-\alpha\) meets x -axis is :-
[JEE Main 2025, 4 Apr (Shift 2)]
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Let \(k\in \mathrm{ℝ}\). If \(\lim _{x\to {0}^{+}}(\sin {(\sin kx)+\cos x+x)}^{2/x}={e}^{6}\), then the value of k is
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The value of \(\lim _{n\to \infty }\left(\sum _{k=1}^{n}\frac{{k}^{3}+6{k}^{2}+11k+5}{(k+3)!}\right)\) is
[JEE Main 2025, 29 Jan (Shift 1)]
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\(f(x)-6 f\left(\frac{1}{x}\right)=\frac{35}{3 x}-\frac{5}{2} . \lim _{\text {If }}\left(\frac{1}{\alpha x}+f(x)\right)=\beta . \quad \text { find }(\alpha+2 \beta)\). (24 Jan, Shift I, Memory based)
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If the function \(f\left(x\right)=\frac{{e}^{x}\left({e}^{\tan x−x}−1\right)+{\log }_{e}(\sec x+\tan x)−x}{\tan x−x}\) is continuous at \(x =0\), then the value of \(f (0)\) is equal to
[JEE Main 2026, 24 Jan (Shift 1)]
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\(\text { Evaluate } \lim _{x \rightarrow \infty}\left(\frac{2 x^2-3 x+10}{3 x^2+4 x-2}\right) \frac{(3 x-1)^{\frac{x}{2}}}{(\sqrt{3 x+2})^x}=\)
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Let f be a differentiable function on \(R\) such that \(f(2)=1\), \({f}^{'}(2)=4\). Let \(\lim _{\mathrm{x}\to 0}(\mathrm{f}(2+\mathrm{x}){)}^{3/\mathrm{x}}={\mathrm{e}}^{\alpha }\). Then the number of times the curve \(\mathrm{y}=4{\mathrm{x}}^{3}-4{\mathrm{x}}^{2}-4(\alpha -7)\mathrm{x}-\alpha\) meets x -axis is :-
[JEE Main 2025, 4 Apr (Shift 2)]
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The value of limit: \(\lim _{x\to 0}(\csc x)\left(\sqrt{2{\cos }^{2}x+3\cos x}-\sqrt{{\cos }^{2}x+\sin x+4}\right)\text{ is }\)
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\(\begin{aligned}&\text { Evaluate }\\&\lim _{x \rightarrow 0} \operatorname{cosec} x .\left(\sqrt{2 \cos ^2 x+3 \cos x}-\sqrt{\cos ^2 x+\sin x+4}\right)\end{aligned}\) (24 Jan, Shift I, Memory Based)
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\(\lim _{x \rightarrow \infty}\left(\left(\frac{e}{1-e}\right)\left(\frac{1}{e}-\frac{x}{1+x}\right)\right)^x=\alpha, \text { find } \frac{\ln \alpha}{1+\ln \alpha}\) (22 Jan, Shift II, Memory Based)
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Let \(\lim _{x\to 2}\frac{(\tan (x-2))\left(r{x}^{2}+(p-2)x-2p\right)}{(x-2{)}^{2}}=5\) for some \(r, p \in R\). If the set of all possible values of \(q\), such that the roots of the equation \(r{x}^{2}-px+q=0\) lie in \((0,2)\), be the interval \((\alpha ,\beta ]\), then \(4(\alpha +\beta )\) equals:
[JEE Main 2026, 6 Apr (Shift 2)]
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Value of \(\lim _{x \rightarrow 0} \frac{a^{\sin x}-1}{\sin x}\) is
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Among:
(S1): \(\lim _{n\to \infty }\frac{1}{{n}^{2}}(2+4+6+\ldots \ldots \ldots +2n)=1\)
(S2) : \(\lim _{n\to \infty }\frac{1}{{n}^{16}}\left({1}^{15}+{2}^{15}+{3}^{15}+\ldots \ldots \ldots .+{n}^{15}\right)=\frac{1}{16}\)
[JEE Main 2023, 13 Apr (Shift 1)]
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\(\begin{matrix}\lim _{x\to \infty }\frac{(\sqrt{3x+1}+\sqrt{3x-1}{)}^{6}+(\sqrt{3x+1}-\sqrt{3x-1}{)}^{6}}{{\left(x+\sqrt{{x}^{2}-1}\right)}^{6}+{\left(x-\sqrt{{x}^{2}-1}\right)}^{6}}{x}^{3} \\ \text{ }\end{matrix}\)
[JEE Main 2023, 31 Jan (Shift 2)]
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\(\lim _{x \rightarrow 0} \frac{|\sin x|}{x}\) is equal to
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The value of \(\lim _{n \rightarrow \infty} \frac{[\mathrm{r}]+[2 \mathrm{r}]+\ldots .+[\mathrm{nr}]}{n^2}\), where \(r\) is a non-zero real number and\([r]\) denotes the greatest integer less than equal to '\(r\) ', is equal to:
[JEE Main 2021, 17 Mar (Shift 2)]
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The value of \(\lim _{x\to 0}\left(\frac{x}{\sqrt[8]{1-\sin x}-\sqrt[8]{1+\sin x}}\right)\) is equal to :
[JEE Main 2021, 27 Jul (Shift 2)]
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\(\ \lim _{x \rightarrow 0} \frac{2 \sin ^2 3 x}{x^2} \) is equal to :
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\(\lim _{x\to 0}\left(\left(\frac{1-{\cos }^{2}(3x)}{{\cos }^{3}(4x)}\right)\left(\frac{{\sin }^{3}(4x)}{{\left({\log }_{e}(2x+1)\right)}^{5}}\right)\right)\) is equal to
[JEE Main 2023, 8 Apr (Shift 1)]
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The value of limit \( \lim _{\theta \rightarrow 0} \frac{\tan \left(\pi \cos ^{2} \theta\right)}{\sin \left(2 \pi \sin ^{2} \theta\right)} \) is equal to:
[JEE Main 2021, 17 Mar (Shift 2)]
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If \( \alpha, \beta \) are the distinct roots of \( x^{2}+b x+c=0 \), then
\( \lim _{x \rightarrow \beta} \frac{e^{2\left(x^{2}+b x+c\right)}-1-2\left(x^{2}+b x+c\right)}{(x-\beta)^{2}} \) is equal to
[JEE Main 2021, 27 Aug (Shift 1)]
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Let \(f:R\to R\) be a function such that\(f(2)=4\) and \({f}^{'}(2)=1\). Then the value of \(\lim _{x\to 2}\frac{{x}^{2}f(2)-4f(x)}{x-2}\) is equal to:
[JEE Main 2021, 27 Jul (Shift 1)]
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Let \(f:R\to R\) be a function such that\(f(2)=4\) and \({f}^{'}(2)=\) 1. Then the value of \(\lim _{x\to 2}\frac{{x}^{2}f(2)-4f(x)}{x-2}\) is equal to:
[JEE Main 2021, 27 Jul (Shift 1)]
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Let \( p=\lim _{x \rightarrow 0^{+}}\left(1+\tan ^{2} \sqrt{x}\right)^{1 / 2 x} \), then \( \log p \) is equal to
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If \(\alpha >\beta >0\) are the roots of the equation \(a{x}^{2}+bx+1=0\), and \(\lim _{x\to \frac{1}{\alpha }}{\left(\frac{1-\cos \left({x}^{2}+bx+a\right)}{2(1-\alpha x{)}^{2}}\right)}^{\frac{1}{2}}=\frac{1}{k}\left(\frac{1}{\beta }-\frac{1}{\alpha }\right)\), then \(k\) is equal to
[JEE Main 2023, 8 Apr (Shift 2)]
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Among
\((S1):\lim _{n\to \infty }\frac{1}{{n}^{2}}(2+4+6+\ldots +2n)=1\\ (S2):\lim _{n\to \infty }\frac{1}{{n}^{16}}\left({1}^{15}+{2}^{15}+{3}^{15}+\ldots ..+{n}^{15}\right)=\frac{1}{16}\)
[JEE Main 2023, 13 Apr (Shift 1)]
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The derivative of \(e^{x^3}\) with respect to \(\log x\) is
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Let \(f(x)=\frac{\sin x+\cos x-\sqrt{2}}{\sin x-\cos x},x\in [0,\pi ]-\left\{\frac{\pi }{4}\right\}\). Then \(f\left(\frac{7\pi }{12}\right){f}^{''}\left(\frac{7\pi }{12}\right)\) is equal to
[JEE Main 2023, 08 Apr (Shift 1)]
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\(\lim _{x\to 0}\frac{x\left({e}^{\frac{\sqrt{1+{x}^{2}+{x}^{4}}-1}{x}}-1\right)}{\sqrt{1+{x}^{2}+{x}^{4}}-1}\)
[JEE Main 2020, 5 Sep (Shift 2)]
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If \(\lim _{x\to 0}\frac{{e}^{ax}-\cos (bx)-\frac{cx{e}^{-cx}}{2}}{1-\cos (2x)}=17\), then \(5{a}^{2}+{b}^{2}\) is equal to
[JEE Main 2023, 13 Apr (Shift 2)]
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\(\lim _{x\to 2}\left(\sum _{n=1}^{9}\frac{x}{n(n+1){x}^{2}+2(2n+1)x+4}\right)\)is equal to:
[JEE Main 2021, 26 Aug (Shift 2)]
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\(\lim _{n\to \infty }\left\{\left({2}^{\frac{1}{2}}-{2}^{\frac{1}{3}}\right)\left({2}^{\frac{1}{2}}-{2}^{\frac{1}{5}}\right)⋯⋯\cdot \left({2}^{\frac{1}{2}}-{2}^{\frac{1}{2n+1}}\right)\right\}\) is equal to
[JEE Main 2023, 6 Apr (Shift 2)]
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\(\lim _{t\to 0}{\left({1}^{\frac{1}{{\sin }^{2}t}}+{2}^{\frac{1}{{\sin }^{2}t}}+\ldots +{n}^{\frac{1}{{\sin }^{2}t}}\right)}^{{\sin }^{2}t}\) is equal to
[JEE Main 2023, 24 Jan (Shift 1)]
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\(\ \lim _{x \rightarrow 0} \sqrt{\frac{x-\sin x}{x+\sin ^2 x}} \) is equal to
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If \(\alpha ,\beta\) are the distinct roots of \({x}^{2}+bx+c=0\), then \(\lim _{x\to \beta }\frac{{e}^{2\left({x}^{2}+bx+c\right)}-1-2\left({x}^{2}+bx+c\right)}{(x-\beta {)}^{2}}\) is equal to
[JEE Main 2021, 27 Aug (Shift 1)]
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If \(\lim _{x\to \infty }\left(\sqrt{{x}^{2}-x+1}-ax\right)=b\), then the ordered pair (a, b) is :
[JEE Main 2021, 27 Aug (Shift 2)]
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\( \lim _{x \rightarrow 0} \frac{\sin ^{2}\left(\pi \cos ^{4} x\right)}{x^{4}} \) is equal to:
[JEE Main 2021, 31 Aug (Shift 1)]
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If \(\alpha >\beta >0\) are the roots of the equation \(a{x}^{2}+bx+1=0\), and \(\lim _{x\to \frac{1}{a}}{\left(\frac{1-\cos \left({x}^{2}+bx+a\right)}{2(1-\alpha x{)}^{2}}\right)}^{\frac{1}{2}}=\frac{1}{k}\left(\frac{1}{\beta }-\frac{1}{\alpha }\right)\), then k is equal to
[JEE Main 2023, 8 Apr (Shift 2)]
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Let \(x=2\) be a root of the equation \(x^2+p x+q=0\) and
\(f(x)=\left\{\begin{matrix}\frac{1-\cos \left({x}^{2}-4px+{q}^{2}+8q+16\right)}{(x-2p{)}^{4}}, & x\neq 2p \\ 0 & x=2p\end{matrix}\right.\)
Then \(\lim _{x \rightarrow 2 p^{+}}[f(x)]\) where [.] denotes greatest integer function, is
[JEE Main 2023, 29 Jan (Shift 1)]
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The value of \(\lim _{x \rightarrow 0}\left(\frac{x}{\sqrt[8]{1-\sin x}-\sqrt[8]{1+\sin x}}\right)\) is equal to :
[JEE Main 2021, 27 Jul (Shift 2)]
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\(\lim _{n\to \infty }{\left(1+\frac{1+\frac{1}{2}+\ldots \ldots \ldots +\frac{1}{n}}{{n}^{2}}\right)}^{n}\) is equal to:
[JEE Main 2021, 25 Feb (Shift 1)]
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If \( \lim _{x \rightarrow 0} \frac{e^{a x}-\cos (b x)-\frac{c x e^{-c x}}{2}}{1-\cos (2 x)}=17 \), then \( 5 a^{2}+b^{2} \) is equal to
[JEE Main 2023, 13 Apr (Shift 2)]
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If \(\alpha >\beta >0\) are the roots of the equation \(a{x}^{2}+bx+1=0\), and \(\lim _{x\to \frac{1}{a}}{\left(\frac{1-\cos \left({x}^{2}+bx+a\right)}{2(1-\alpha x{)}^{2}}\right)}^{\frac{1}{2}}=\frac{1}{k}\left(\frac{1}{\beta }-\frac{1}{\alpha }\right)\), then k is equal to
[JEE Main 2023, 08 Apr (Shift 2)]
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If \(\lim _{x\to 0}\frac{{e}^{ax}-\cos \left(bx\right)-\frac{cx{e}^{-cx}}{2}}{1-\cos \left(2x\right)}=17\), then \(5{a}^{2}+{b}^{2}\) is equal to
[JEE Main 2023, 13 Apr (Shift 2)]
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If \(xy+{y}^{2}=\tan x+y\), then find \(\frac{dy}{dx}\) is
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Which is \(\lim _{x\to 0}\frac{\sin x−\tan x}{x}\) equal to?
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The value of \(\lim _{x \rightarrow 0^{+}} \frac{\cos ^{-1}\left(x-[x]^2\right) \cdot \sin ^{-1}\left(x-[x]^2\right)}{x-x^3}\), where \([x]\) denotes the greatest integer \(\leq x\) is:
[JEE Main 2021, 17 Mar (Shift 1)]
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Among:
\(\text{S}1:\) \(\lim _{n\to \infty }\frac{1}{{n}^{2}}(2+4+6+\ldots \ldots \ldots +2n)=1\)
\(\text{S}2:\) \(\lim _{n\to \infty }\frac{1}{{n}^{16}}\left({1}^{15}+{2}^{15}+{3}^{15}+\ldots \ldots \ldots .+{n}^{15}\right)=\frac{1}{16}\)
[JEE Main 2023, 13 Apr (Shift 1)]
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If \(\lim _{x\to \infty }\left(\sqrt{{x}^{2}-x+1}-ax\right)=b\), then the ordered pair \((a, b)\) is :
[JEE Main 2021, 27 Aug (Shift 2)]
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