KCET · Maths · Limits
\(\lim _{x \rightarrow 0} \frac{x 2^{x}-x}{1-\cos x}\) is equal to
- A \(2 \log 2\)
- B \(\log 2\)
- C \(\frac{1}{2} \log 2\)
- D \(\frac{1}{2}\)
Answer & Solution
Correct Answer
(B) \(\log 2\)
Step-by-step Solution
Detailed explanation
\[
\begin{aligned}
\lim _{x \rightarrow 0} \frac{x 2^{x}-x}{1-\cos x} \\
&=\lim _{x \rightarrow 0} \frac{x 2^{x} \log 2}{\sin x}
\end{aligned}
\]
\[
\left(\frac{0}{0} \text { form }\right)
\]
(L'Hopital's rule)
\[
=\log 2 \lim _{x \rightarrow 0} \frac{\left[x 2^{x} \log 2+2^{x} \times 1\right]}{\cos x}
\]
(L'Hopital's rule)
\[
=\log 2
\]
\begin{aligned}
\lim _{x \rightarrow 0} \frac{x 2^{x}-x}{1-\cos x} \\
&=\lim _{x \rightarrow 0} \frac{x 2^{x} \log 2}{\sin x}
\end{aligned}
\]
\[
\left(\frac{0}{0} \text { form }\right)
\]
(L'Hopital's rule)
\[
=\log 2 \lim _{x \rightarrow 0} \frac{\left[x 2^{x} \log 2+2^{x} \times 1\right]}{\cos x}
\]
(L'Hopital's rule)
\[
=\log 2
\]
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