TS EAMCET · Maths · Limits
\(\lim _{x \rightarrow 0} \frac{\left(1-e^x\right) \sin x}{x^2+x^3}\) is equal to
- A \(-1\)
- B \(1\)
- C \(0\)
- D \(2\)
Answer & Solution
Correct Answer
(A) \(-1\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \lim _{x \rightarrow 0} \frac{\left(1-e^x\right) \sin x}{\left(x+x^2\right) x} \\ & =\lim _{x \rightarrow 0} \frac{\left(-x-\frac{x^2}{2}-\frac{x^3}{3}-\ldots\right)}{x(1+x)} \times \lim _{x \rightarrow 0} \frac{\sin x}{x} \\ & =-1 \times 1=-1\end{aligned}\)
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