Question

The value of $\underset{x\to 0}{lim}\frac{{(1+x)}^{1/x}-e+\frac{ex}{2}}{{11x}^2}$ is equal to'

• A. $\frac{5}{24}e$
• B. $0$
• C. $\frac{11}{24}eq$
• D. $\frac{e}{24}$

Answer 1

The correct option is D $\frac{e}{24}$

$\begin{array}{c}{lim}_{x\to 0}\frac{{\left(1+x\right)}^{1/x}-e+\frac{e^x}{2}}{1+x^2}\\ {\left(1+x\right)}^{1/x}=e\left(\frac{1}{x}\log \left(1+x\right)\right)\\ =e\left(1-\frac{x}{2}+\frac{x^2}{8}......\right)\left(1+\frac{x^2}{3}\right)\\ =e\left(1-\frac{x}{2}+\frac{11}{24}{x}^2\right)\\ putthevalue\\ =\frac{1}{1+x^2}\left[e\left(1-\frac{x}{2}+\frac{11}{24}{x}^2\right)-\left(1-\frac{x}{2}\right)e\right]\\ =\frac{11}{24}\times \frac{x}{11x}e=\frac{e}{24}\end{array}$