Limits JEE Advanced previous year questions with solutions

5 solved JEE Advanced questions on Limits, free to read — no sign-in needed. The full chapter has 31 questions; sign in to attempt the remaining 26 in the exam simulator.

1. Q1JEE Advanced Adv 2025 (Paper 1)
   Let $\alpha$ and $\beta$ be the real numbers such that $\lim _{x \rightarrow 0} \frac{1}{x^3}\left(\frac{\alpha}{2} \int_0^x \frac{1}{1-t^2} d t+\beta x \cos x\right)=2$. Then the value of $\alpha+\beta$ is _______
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   Answer: 2.4

   $\lim _{x \rightarrow 0} \frac{\frac{\alpha}{2} \int_0^x \frac{1}{1-t^2} d t+\beta x \cos x}{x^3}$ $\begin{aligned} & =\lim _{x \rightarrow 0} \frac{\frac{\alpha}{2}\left(\frac{1}{1-x^2}\right)+\beta \cos x-\beta x \sin x}{3 x^2} \\ & =\frac{\frac{\alpha}{2}\left(1-x^2\right)^{-1}+\beta\left(1-\frac{x^2}{2!}+\frac{x^4}{4!} \ldots\right)-\beta x\left(x-\frac{x^3}{3!}+\frac{x^5}{5!} \ldots\right)}{3 x^2} \\ & =\frac{\frac{\alpha}{2}\left(1+x^2+x^4 \ldots\right)+\beta\left(1-\frac{x^2}{2!}+\frac{x^4}{4!} \ldots\right)-\beta\left(x^2-\frac{x^4}{3!} \ldots\right)}{3 x^2}\end{aligned}$ $\begin{aligned} & =\frac{\left(\frac{\alpha}{2}+\beta\right)+\mathrm{x}^2\left(\frac{\alpha}{2}-\frac{\beta}{2}-\beta\right)+\mathrm{x}^4() \ldots}{3 \mathrm{x}^2}=2(\text { Given }) \\ & \therefore \frac{\alpha}{2}+\beta=0 \text { and } \frac{\alpha-3 \beta}{6}=2 \\ & \Rightarrow \alpha=-2 \beta \text { and } \alpha=12+3 \beta \\ & \Rightarrow \beta=-\frac{12}{5} \text { and } \alpha=\frac{24}{5}\end{aligned}$ $\therefore \alpha+\beta=\frac{12}{5}=2.40$
2. Q2JEE Advanced Adv 2024 (Paper 2)
   Let $S$ be the set of all $(\alpha, \beta) \in \mathbb{R} \times \mathbb{R}$ such that $\lim _{x \rightarrow \infty} \frac{\sin \left(x^2\right)\left(\log _e x\right)^\alpha \sin \left(\frac{1}{x^2}\right)}{x^{\alpha \beta}\left(\log _e(1+x)\right)^\beta}=0$. Then which of the following is (are) correct?
   1. A.$(-1,3) \in S$
   2. B.$(-1,1) \in S$
   3. C.$(1,-1) \in S$
   4. D.$(1,-2) \in S$
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   Answer: B,C

   $\begin{aligned} & \lim _{x \rightarrow \infty} \frac{\sin x^2 \cdot\left(\log _e x\right)^\alpha \cdot \sin \frac{1}{x^2}}{x^{\alpha \beta} \cdot\left(\log _e(1+x)\right)^\beta}=0 \\ & \lim _{x \rightarrow \infty} \frac{\left(\log _e x\right)^\alpha}{\left(\log _e(x+1)\right)^\beta \cdot x^{\alpha \beta+2}}=0\end{aligned}$ $\begin{aligned} & \lim _{x \rightarrow \infty}\left(\frac{\log _e x}{\log _e(x+1)}\right)^\beta \cdot \frac{\left(\log _e x\right)^{\alpha-\beta}}{x^{\alpha \beta+2}}=0 \\ & \lim _{x \rightarrow \infty} \frac{\left(\log _e x\right)^{\alpha-\beta}}{x^{\alpha \beta+2}}=0 \quad \text { Put } \log _e x=t \\ & \lim _{t \rightarrow \infty} \frac{t^{\alpha-\beta}}{\left(e^t\right)^{\alpha \beta+2}}=0\end{aligned}$ As we know $\lim _{x \rightarrow \infty} \frac{x}{e^x}=0$ $\alpha \beta+2\gt 0 \Rightarrow \alpha \beta\gt -2$
3. Q3JEE Advanced Adv 2024 (Paper 2)
   Let $k \in \mathbb{R}$. If $\lim _{x \rightarrow 0^+}(\sin (\sin k x)+\cos x+x)^{\frac{2}{x}}=e^6$, then the value of $k$ is
   1. A.$1$
   2. B.$2$
   3. C.$3$
   4. D.$4$
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   Answer: (B)

   $\begin{aligned} & l=\lim _{x \rightarrow 0^{+}}(\sin (\sin k x)+\cos x+x)^{\frac{2}{x}}=e^6 \\ & \Rightarrow \ln l=\lim _{x \rightarrow 0^{+}} \frac{2}{x}(\sin (\sin k x)+\cos x+x-1) \\ & \Rightarrow \ln l=\lim _{x \rightarrow 0^{+}} 2\left(\frac{\sin (\sin k x)}{\sin k x} \cdot \frac{\sin k x}{k x} \cdot \frac{k x}{x}+1-\frac{(1-\cos x)}{x^2} \cdot x\right) \\ & \Rightarrow \ln l=2(k+1) \\ & \Rightarrow l=e^{2(k+1)}=e^6 \\ & k+1=3 \\ & \Rightarrow k=2\end{aligned}$
4. Q4JEE Advanced Adv 2024 (Paper 1)
   Let $f(x)$ be a continuously differentiable function on the interval $(0, \infty)$ such that $f(1)=2$ and $\underset{t \rightarrow x}{\lim} \frac{t^{10} f(x)-x^{10} f(t)}{t^9-x^9}=1$ for each $x\gt 0$. Then, for all $x\gt 0, f(x)$ is equal to
   1. A.$\frac{31}{11 x}-\frac{9}{11} x^{10}$
   2. B.$\frac{9}{11 x}+\frac{13}{11} x^{10}$
   3. C.$\frac{-9}{11 x}+\frac{31}{11} x^{10}$
   4. D.$\frac{13}{11 x}+\frac{9}{11} x^{10}$
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   Answer: (B)

   $\begin{aligned} & \lim _{t \rightarrow x} \frac{t^{10} f(x)-x^{10} f(t)}{t^9-x^9}=1 \\ & \Rightarrow \lim _{t \rightarrow x} \frac{10 t^9 f(x)-x^{10} f^{\prime}(t)}{9 t^8}=1\end{aligned}$ $\begin{aligned} & \Rightarrow 10 x f(x)-x^2 f^{\prime}(x)=9 \\ & \Rightarrow x^2 f^{\prime}(x)=10 x f(x)-9 \\ & \Rightarrow f^{\prime}(x)=\frac{10 f(x)}{x}-\frac{9}{x^2} \\ & \Rightarrow \frac{d y}{d x}-\frac{10}{x} y=-\frac{9}{x^2} \\ & \Rightarrow y \cdot \frac{1}{x^{10}}=\int-\frac{9}{x^2} \cdot \frac{1}{x^{10}} d x \\ & \Rightarrow \frac{y}{x^{10}}=\frac{9}{11 x^{11}}+c...(1)\end{aligned}$ $\begin{aligned} & \because \mathrm{f}(1)=2 \Rightarrow \frac{2}{1}=\frac{9}{11}+\mathrm{c} \Rightarrow \mathrm{c}=\frac{13}{11} \\ & \therefore \mathrm{f}(\mathrm{x})=\frac{9}{11 \mathrm{x}}+\frac{13}{11} \mathrm{x}^{10}\end{aligned}$ $\Rightarrow$ Option (2) is correct.
5. Q5JEE Advanced Adv 2023 (Paper 1)
   Let $f:\left(0,1\right)\rightarrow ℝ$ be the function defined as $f\left(x\right)=\sqrt{n}$ if $x\in [\dfrac{1}{n+1},\dfrac{1}{n})$ where $n\in ℕ$. Let $g:\left(0,1\right)\rightarrow ℝ$ be a function such that ${\int }_{{x}^{2}}^{x}\sqrt{\dfrac{1-t}{t}}dt\lt g\left(x\right)\lt 2\sqrt{x}$ for all $x\in \left(0,1\right)$. Then $\lim _{x\rightarrow 0}f\left(x\right)g\left(x\right)$
   1. A.Does NOT exist
   2. B.is equal to $1$
   3. C.is equal to $2$
   4. D.is equal to $3$
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   Answer: (C)

   Given, $f:\left(0,1\right)\rightarrow ℝ$ be the function defined as $f\left(x\right)=\sqrt{n}$ if $x\in [\dfrac{1}{n+1},\dfrac{1}{n})$ where $n\in ℕ$ And $g:\left(0,1\right)\rightarrow ℝ$ be a function such that ${\int }_{{x}^{2}}^{x}\sqrt{\dfrac{1-t}{t}}dt\lt g\left(x\right)\lt 2\sqrt{x}$ for all $x\in \left(0,1\right)$. Now we need to solve $1$ sided limit here to get some answer, as $\lim _{x\rightarrow {0}^{-}}$ doesn’t exist here (not in domain) As $\dfrac{1}{n+1}\leq x\lt \dfrac{1}{n}$ $\Rightarrow n+1\geq \dfrac{1}{x}\gt n$ $\Rightarrow n\geq \dfrac{1}{x}-1\Rightarrow \sqrt{n}\geq \sqrt{\dfrac{1}{x}-1}$ So, let $f\left(x\right)=\sqrt{\left(\dfrac{1}{x}\right)-1}$ where $\left(\cdot \right)=$ least integer function Now multiplying $f\left(x\right)$ in ${\int }_{{x}^{2}}^{x}\sqrt{\dfrac{1-t}{t}}dt\lt g\left(x\right)\lt 2\sqrt{x}$ and taking limit we get, $\lim _{x\rightarrow {0}^{+}}{\int }_{{x}^{2}}^{x}\sqrt{\dfrac{1-t}{t}}dt\cdot \sqrt{\left(\dfrac{1}{x}\right)-1}\leq \lim _{x\rightarrow {0}^{+}}f\left(x\right)\cdot g\left(x\right)\leq \lim _{x\rightarrow {0}^{+}}\sqrt{\left(\dfrac{1}{x}\right)-1}\times 2\sqrt{x}$ Now $\lim _{x\rightarrow {0}^{+}}\sqrt{\left(\dfrac{1}{x}\right)-1}\times 2\sqrt{x}=\lim _{x\rightarrow {0}^{+}}2\sqrt{x}\sqrt{\left[\dfrac{1}{x}\right]}\left(\dfrac{1}{x}\notin Z\right)$ $=\lim _{x\rightarrow {0}^{+}}2\sqrt{x\left(\dfrac{1}{x}-\left\{\dfrac{1}{x}\right\}\right)}=2$ $=\lim _{x\rightarrow {0}^{+}}2\sqrt{1-x\left(\dfrac{1}{x}\right)}=2;\left(\dfrac{1}{x}\notin Z\right)$ So, $\lim _{x\rightarrow {0}^{+}}{\int }_{{x}^{2}}^{x}\sqrt{\dfrac{1-t}{t}}dt\cdot \sqrt{\dfrac{1}{x}-\left\{\dfrac{1}{x}\right\}}=\dfrac{{\int }_{{x}^{2}}^{x}\sqrt{\dfrac{1-t}{t}}dt\cdot \sqrt{1-x\left\{\dfrac{1}{x}\right\}}}{\sqrt{x}}$ And $\lim _{x\rightarrow {0}^{+}}\dfrac{{\int }_{{x}^{2}}^{x}\sqrt{\dfrac{1-t}{t}}dt}{\sqrt{x}}=\lim _{x\rightarrow {0}^{+}}\dfrac{\sqrt{\dfrac{1-x}{x}}-2x\sqrt{\dfrac{1-{x}^{2}}{{x}^{2}}}}{\dfrac{1}{2\sqrt{x}}}$ {Using L-hospital rule} $=\lim _{x\rightarrow {0}^{+}}2\sqrt{1-x}-4\sqrt{x}\cdot \sqrt{1-{x}^{2}}=2$ Similarly for $\dfrac{1}{x}\in Z$ is equal to $2$.