Limits JEE Main previous year questions with solutions

4 solved JEE Main questions on Limits, free to read — no sign-in needed. The full chapter has 147 questions; sign in to attempt the remaining 143 in the exam simulator.

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.$10$
   Show answer & solution

   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.$-1$
   Show answer & solution

   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.$e-e^2$
   Show answer & solution

   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 2022 (28 Jul, Shift 1)Special Forms
   $\lim _{x\rightarrow 0}{\left(\dfrac{{\left(x+2\cos x\right)}^{3}+2{\left(x+2\cos x\right)}^{2}+3\sin \left(x+2\cos x\right)}{{\left(x+2\right)}^{3}+2{\left(x+2\right)}^{2}+3\sin \left(x+2\right)}\right)}^{\dfrac{100}{x}}$ is equal to
   Show answer & solution

   Answer: 1

   Given,$lim_{x\rightarrow 0}{\left(\dfrac{{\left(x+2cosx\right)}^{3}+2{\left(x+2cosx\right)}^{2}+3sin\left(x+2cosx\right)}{{\left(x+2\right)}^{3}+2{\left(x+2\right)}^{2}+3sin\left(x+2\right)}\right)}^{\dfrac{100}{x}}$ Now by putting the value of limit we can see it is in form ${1}^{\infty }$ Now we know that, $\lim _{x\rightarrow 0}{\left(f\left(x\right)\right)}^{g\left(x\right)}$ when is in form of ${1}^{\infty }$, we can write this as ${e}^{\lim _{x\rightarrow 0}\left(f\left(x\right)-1\right)g\left(x\right)}$ Now using the above rule we get, $={e}^{\lim _{x\rightarrow 0}\left[\left(\dfrac{{\left(x+2\cos x\right)}^{3}+2{\left(x+2\cos x\right)}^{2}+3\sin \left(x+2\cos x\right)}{{\left(x+2\right)}^{3}+2{\left(x+2\right)}^{2}+3\sin \left(x+2\right)}\right)-1\right]\times \dfrac{100}{x}}$ $={e}^{\lim _{x\rightarrow 0}\left[\dfrac{100}{x}\left[\dfrac{{\left(x+2\cos x\right)}^{3}+2{\left(x+2\cos x\right)}^{2}+3\sin \left(x+2\cos x\right)-\left({\left(x+2\right)}^{3}+2{\left(x+2\right)}^{2}+3\sin \left(x+2\right)\right)}{{\left(x+2\right)}^{3}+2{\left(x+2\right)}^{2}+3\sin \left(x+2\right)}\right]\right]}$ $={e}^{\lim _{x\rightarrow 0}\dfrac{100}{x}\left[\left(\dfrac{{\left(x+2\cos x\right)}^{3}-{\left(x+2\right)}^{3}+2{\left(x+2\cos x\right)}^{2}-2{\left(x+2\right)}^{2}+3\sin \left(x+2\cos x\right)-3\sin \left(x+2\right)}{8+8+3\sin 2}\right)\right]}$ Using L-hospital method we get, $={e}^{\dfrac{100}{16+3\sin 2}\lim _{x\rightarrow 0}\dfrac{3{\left(x+2\cos x\right)}^{2}\times \left(1+2\sin x\right)-3{\left(x+2\right)}^{2}+4\left(x+2\cos x\right)\times \left(1-2\sin x\right)-4\left(x+2\right)+3\cos \left(x+2\cos x\right)\times \left(1-2\sin x\right)-3\cos \left(x+2\right)}{1}}$ $={e}^{\dfrac{100}{16+3\sin 2}\left(\dfrac{12-3\left(4\right)+8\times 1-8+3\cos 2-3\cos 2}{1}\right)}$ $={e}^{\dfrac{100}{16+3\sin 2}\times 0}$ $={e}^{0}=1$