KCET · Maths · Limits
\( \lim _{x \rightarrow 0} \frac{x e^{x}-\sin x}{x} \) is equal to
- A \( 13 \)
- B \( 1 \)
- C \( 00 \)
- D \( 12 \)
Answer & Solution
Correct Answer
(C) \( 00 \)
Step-by-step Solution
Detailed explanation
Given that
\[
\begin{array}{l}
\lim _{x \rightarrow 0} \frac{x e^{x}-\sin x}{x}=\lim _{x \rightarrow 0}\left(\frac{x e^{x}}{x}-\frac{\sin x}{x}\right) \\
=\lim _{x \rightarrow 0} \frac{x e^{x}}{x}-\lim _{x \rightarrow 0} \frac{\sin x}{x} \\
=1-1=0
\end{array}
\]
\[
\begin{array}{l}
\lim _{x \rightarrow 0} \frac{x e^{x}-\sin x}{x}=\lim _{x \rightarrow 0}\left(\frac{x e^{x}}{x}-\frac{\sin x}{x}\right) \\
=\lim _{x \rightarrow 0} \frac{x e^{x}}{x}-\lim _{x \rightarrow 0} \frac{\sin x}{x} \\
=1-1=0
\end{array}
\]
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