WBJEE · Maths · Limits
\(\lim _{x \rightarrow 0} \frac{\pi^{x}-1}{\sqrt{1+x}-1}\)
- A does not exist
- B equals \(\log _{e}\left(\pi^{2}\right)\)
- C equals 1
- D lies between 10 and 11
Answer & Solution
Correct Answer
(B) equals \(\log _{e}\left(\pi^{2}\right)\)
Step-by-step Solution
Detailed explanation
\(\lim _{x \rightarrow 0} \frac{\pi^{x}-1}{\sqrt{1+x}-1}\) \(\frac{0}{0}\) form \(=\lim _{x \rightarrow 0} \frac{\pi^{x} \log _{e} \pi}{\frac{1}{2 \sqrt{1+x}}}\) \(=\lim _{x \rightarrow 0} 2 \sqrt{1+x}\left(\pi^{x} \log _{e} \pi\right)\)…
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