Lim x → 0 e - (1 + 2x)^1/2x / x is equal to : (1) e (2) -2 / e (3) 0 (4) e–e^2

Question

\(\operatorname{Lim}_{x \rightarrow 0} \frac{e-(1+2 x)^{\frac{1}{2 x}}}{x}\) is equal to : 

(1) e

(2) \(\frac{-2}{\mathrm{e}}\)

(3) 0

(4) \( \mathrm{e}-\mathrm{e}^{2}\) 

Answer 1

Correct option is : (1) e 

\(\operatorname{Lim}_{x \rightarrow 0} \frac{e-e^{\frac{1}{2 x} \ln (1+2 x)}}{x}\)

\(=\operatorname{Lim}_{x \rightarrow 0}(-e) \frac{\left(e^{\frac{\ln (1+2 x)}{2 x}-1}-1\right)}{x}\)

\(=\operatorname{Lim}\limits_{x \rightarrow 0}(-e) \frac{\ln (1+2 x)-2 x}{2 x^{2}}\)

\(=(-\mathrm{e}) \times(-1) \frac{4}{2 \times 2}=\mathrm{e}\)