Home > Limits > JEE Main 2021 > Limits-Q8
Limits
Taylor Series Expansion
JEE Main 2021
Grade 11

Question:

$\lim_{x \to 0} \frac{\sin x + \ln(1 - x)}{x^2}$
(1) $-\frac{1}{2}$
(2) $0$
(3) $\frac{1}{2}$
(4) $-1$

Step-by-Step Solution

Key Concept: Expand: $\sin x = x - x^2/2 + \dots$, $\ln(1 - x) = -x - x^2/2 - \dots$. Sum $= (x - x^2/2) + (-x - x^2/2) + O(x^3) = -x^2 + O(x^3)$. Dividing by $x^2$: limit $= -1$.
The detailed step-by-step mathematical proof is available inside the Mathbee app workspace.
Correct Answer: (1)

Master Limits with Mathbee

Practice this topic under real exam conditions with strict timers, or ask our AI Mentor to explain the concepts step-by-step.

Start Practicing for Free