Limits JEE Main previous year questions with solutions

5 solved JEE Main questions on Limits, free to read — no sign-in needed. The full chapter has 145 questions; sign in to attempt the remaining 140 in the exam simulator.

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1. Q1JEE Main 2026 (02 Apr, Shift 1)Existance of Limit
   If $\displaystyle\lim_{x \to 2} \dfrac{\sin(x^3 - 5x^2 + ax + b)}{(\sqrt{x-1} - 1)\log_e(x-1)} = m$, then $a + b + m$ is equal to :
   1. A.$5$
   2. B.$6$
   3. C.$8$
   4. D.$10$
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   Answer: (B)

   Let $x - 2 = t$. As $x \to 2$, $t \to 0$. The given limit can be written as: $\lim_{t \to 0} \dfrac{\sin((t+2)^3 - 5(t+2)^2 + a(t+2) + b)}{(\sqrt{t+1} - 1)\log_e(t+1)} = m$ The denominator can be approximated for small $t$ as: $\lim_{t \to 0} \dfrac{\sqrt{t+1} - 1}{t} \cdot \dfrac{\log_e(t+1)}{t} \cdot t^2 = \dfrac{1}{2} \cdot 1 \cdot t^2 = \dfrac{t^2}{2}$ For the limit to be finite, the argument of the sine function in the numerator must have $t^2$ as its lowest degree term in its expansion. Let $P(t) = (t+2)^3 - 5(t+2)^2 + a(t+2) + b$ $P(t) = t^3 + 6t^2 + 12t + 8 - 5(t^2 + 4t + 4) + at + 2a + b$ $P(t) = t^3 + t^2 + (a - 8)t + (2a + b - 12)$ For the limit to exist, the constant and linear terms must be zero: $a - 8 = 0 \Rightarrow a = 8$ $2a + b - 12 = 0 \Rightarrow 16 + b - 12 = 0 \Rightarrow b = -4$ Substituting $a$ and $b$ back into $P(t)$: $P(t) = t^3 + t^2$ The limit becomes: $m = \lim_{t \to 0} \dfrac{\sin(t^3 + t^2)}{t^2 / 2}$ $m = \lim_{t \to 0} \dfrac{\sin(t^2(t+1))}{t^2(t+1)} \cdot \dfrac{t^2(t+1)}{t^2 / 2} = 1 \cdot 2 = 2$ Therefore, $a + b + m = 8 - 4 + 2 = 6$. Answer: $6$
2. Q2JEE Main 2025 (02 Apr, Shift 1)Algebraic and Rational Limits
   For $\alpha, \beta, \gamma, \in \mathbf{R}$, if $\lim _{x \rightarrow 0} \frac{x^2 \sin \alpha x+(\gamma-1) e^{x^2}}{\sin 2 x-\beta x}=3$, then $\beta+\gamma-\alpha$ is equal to:
   1. A.7
   2. B.4
   3. C.6
   4. D.$-1$
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   Answer: (A)

   $\begin{aligned} & \lim _{x \rightarrow 10} \frac{x^2(\alpha x)+(\gamma-1)\left(1+\frac{x^2}{1}\right)}{2 x-\frac{8 x^3}{6}-\beta x}=3 \\ & \lim _{x \rightarrow 0} \frac{(\gamma-1)+(\gamma-1) x^2+\alpha x^3}{(2-\beta) x-\frac{4}{3} x^3}=3 \\ & \gamma-1, \beta=2, \frac{-3 \alpha}{4}=+3 \Rightarrow \alpha=-4 \\ & \beta+\gamma-\alpha=7\end{aligned}$
3. Q3JEE Main 2024 (09 Apr, Shift 2)Exponential and logarithmic limits
   $\lim _{x \rightarrow 0} \frac{e-(1+2 x)^{\frac{1}{2 x}}}{x}$ is equal to
   1. A.0
   2. B.$\frac{-2}{e}$
   3. C.e
   4. D.$e-e^2$
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   Answer: (C)

   $\begin{aligned} & \operatorname{Lim}_{x \rightarrow 0} \frac{e-e^{\frac{1}{2 x} \ln (1+2 x)}}{x} \\ & =\operatorname{Lim}_{x \rightarrow 0}(-e) \frac{\left(e^{\frac{\ln (1+2 x)}{2 x}-1}-1\right)}{x} \\ & =\operatorname{Lim}_{x \rightarrow 0}(-e) \frac{\ln (1+2 x)-2 x}{2 x^2} \\ & =(-e) \times(-1) \frac{4}{2 \times 2}=e\end{aligned}$
4. Q4JEE Main 2023 (06 Apr, Shift 1)Arithmetic Progression
   Let ${a}_{1},{a}_{2},{a}_{3},....,{a}_{n}$ be $n$ positive consecutive terms of an arithmetic progression. If $d\gt 0$ is its common difference, then $\lim _{n\rightarrow \infty }\sqrt{\dfrac{d}{n}}\left(\dfrac{1}{\sqrt{{a}_{1}}+\sqrt{{a}_{2}}}+\dfrac{1}{\sqrt{{a}_{2}}+\sqrt{{a}_{3}}}+\ldots +\dfrac{1}{\sqrt{{a}_{n-1}}+\sqrt{{a}_{n}}}\right)$ is
   1. A.$\dfrac{1}{\sqrt{d}}$
   2. B.$\sqrt{d}$
   3. C.$1$
   4. D.$2$
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   Answer: (C)

   Given, ${a}_{1},{a}_{2},{a}_{3},....,{a}_{n}$ are terms of an $A.P$, So common difference will be, $d={a}_{2}-{a}_{1}={a}_{3}-{a}_{2}=........={a}_{n}-{a}_{n-1}$ Now solving, $\lim _{n\rightarrow \infty }\sqrt{\dfrac{d}{n}}\left(\dfrac{1}{\sqrt{{a}_{1}}+\sqrt{{a}_{2}}}+\dfrac{1}{\sqrt{{a}_{2}}+\sqrt{{a}_{3}}}+\ldots +\dfrac{1}{\sqrt{{a}_{n-1}}+\sqrt{{a}_{n}}}\right)$ $=\lim _{n\rightarrow \infty }\sqrt{\dfrac{d}{n}}\left(\dfrac{\sqrt{{a}_{2}}-\sqrt{{a}_{1}}}{{a}_{2}-{a}_{1}}+\dfrac{\sqrt{{a}_{3}}-\sqrt{{a}_{2}}}{{a}_{3}-{a}_{2}}+\ldots +\dfrac{\sqrt{{a}_{n}}-\sqrt{{a}_{n-1}}}{{a}_{n}-{a}_{n-1}}\right)$ $=\lim _{n\rightarrow \infty }\sqrt{\dfrac{d}{n}}\times \dfrac{1}{d}\left(\sqrt{{a}_{n}}-\sqrt{{a}_{1}}\right)$ Now using the formula ${a}_{n}={a}_{1}+\left(n-1\right)d$ we get, $=\lim _{n\rightarrow \infty }\dfrac{1}{\sqrt{d}}\left(\dfrac{\sqrt{{a}_{1}+(n-1)d}-\sqrt{{a}_{1}}}{\sqrt{n}}\right)$ $=\lim _{n\rightarrow \infty }\dfrac{1}{\sqrt{d}}\left(\dfrac{\sqrt{n}\left(\sqrt{\dfrac{{a}_{1}}{n}+d-\dfrac{d}{n}}-\sqrt{\dfrac{{a}_{1}}{n}}\right)}{\sqrt{n}}\right)$ $=\dfrac{1}{\sqrt{d}}\left(\left(\sqrt{0+d-0}-\sqrt{0}\right)\right)$ $=\dfrac{1}{\sqrt{d}}\times \sqrt{d}=1$
5. Q5JEE Main 2022 (29 Jul, Shift 1)Miscellaneous
   <p>If $\lim _{x\rightarrow 0}\dfrac{\alpha {e}^{x}+\beta {e}^{-x}+\gamma \sin x}{x{\sin }^{2}x}=\dfrac{2}{3}$, where $\alpha ,\beta ,\gamma \in R$, then which of the following is NOT correct?</p>
   1. A.<p>${\alpha }^{2}+{\beta }^{2}+{\gamma }^{2}=6$</p>
   2. B.$\alpha \beta +\beta \gamma +\gamma \alpha +1=0$
   3. C.$\alpha {\beta }^{2}+\beta {\gamma }^{2}+\gamma {\alpha }^{2}+3=0$
   4. D.<p>${\alpha }^{2}-{\beta }^{2}+{\gamma }^{2}=4$</p>
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   Answer: (C)

   <p>Given, $\lim _{x\rightarrow 0}\dfrac{\alpha {e}^{x}+\beta {e}^{-x}+\gamma \sin x}{x{\sin }^{2}x}=\dfrac{2}{3}$ $\Rightarrow \lim _{x\rightarrow 0}\dfrac{\alpha {e}^{x}+\beta {e}^{-x}+\gamma \sin x}{{x}^{3}\dfrac{{\sin }^{2}x}{{x}^{2}}}=\dfrac{2}{3}$ $\Rightarrow \lim _{x\rightarrow 0}\dfrac{\alpha {e}^{x}+\beta {e}^{-x}+\gamma \sin x}{{x}^{3}}=\dfrac{2}{3}$ Now using expansion of given function in terms of $x$ we get, $=\lim _{x\rightarrow 0}\dfrac{\alpha \left(1+x+\dfrac{{x}^{2}}{2!}+\dfrac{{x}^{3}}{3!}+\ldots \right)+\beta \left(1-x+\dfrac{{x}^{2}}{2!}-\dfrac{{x}^{3}}{3!}+\ldots \right)+\gamma \left(x-\dfrac{{x}^{3}}{3!}+\ldots \right)}{{x}^{3}}$ Now for limit to exist constant terms should be zero $\Rightarrow a+\beta =0....\left(1\right)$ Also coefficient of $x$ should be zero $\Rightarrow \alpha -\beta +\gamma =0.......\left(2\right)$ Also coefficient of ${x}^{2}$ should be zero $\Rightarrow \dfrac{\alpha }{2}+\dfrac{\beta }{2}=0.......\left(3\right)$ So limit becomes, $\lim _{x\rightarrow 0}\dfrac{{x}^{3}\left(\dfrac{\alpha }{3!}-\dfrac{\beta }{3!}-\dfrac{\gamma }{3!}\right)+{x}^{4}\left(\dfrac{\alpha }{3!}-\dfrac{\beta }{3!}-\dfrac{\gamma }{3!}\right)}{{x}^{3}}=\dfrac{2}{3}$ Now putting the value of limit we get, $\dfrac{\alpha }{6}-\dfrac{\beta }{6}-\dfrac{\gamma }{6}=\dfrac{2}{3}........\left(4\right)$ Now on solving equation 1,2,3,4 we get, $\Rightarrow \alpha =1,\beta =-1,\gamma =-2$ Now putting these in all option we get, $\alpha {\beta }^{2}+\beta {\gamma }^{2}+\gamma {\alpha }^{2}+3=0$</p>