JEE Mains · Maths · STD 11 - 12. limits
\(\operatorname{Lim}_{x \rightarrow 0} \frac{e-(1+2 x)^{\frac{1}{2 x}}}{x}\) is equal to :
- A \(e\)
- B \(\frac{-2}{\mathrm{e}}\)
- C \(0\)
- D \(e-e^2\)
Answer & Solution
Correct Answer
(A) \(e\)
Step-by-step Solution
Detailed explanation
\( \operatorname{Lim}_{x \rightarrow 0} \frac{e-e^{\frac{1}{2 x} \ln (1+2 x)}}{x} \) \( =\operatorname{Lim}_{x \rightarrow 0}(-e) \frac{\left(e^{\frac{\ln (1+2 x)}{2 x}-1}-1\right)}{x} \) \( =\operatorname{Lim}_{x \rightarrow 0}(-e) \frac{\ln (1+2 x)-2 x}{2 x^2} \)…
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