Question

$\underset{x\to 0}{lim}\frac{(1+x{)}^{1/x}-e}{x}$

• A. $-\frac{1}{3}e$
• B. $-\frac{1}{2}e$
• C. $-\frac{1}{5}e$
• D. $-\frac{2}{3}e$

Answer 1

The correct option is B $-\frac{1}{2}e$

$\underset{x\to 0}{lim}\frac{{\left(1+x\right)}^{1/x}-e}{x}$

Using L-hospital rule,

$=\underset{x\to 0}{lim}\frac{d}{dx}{\left(1+x\right)}^{1/x}$

Let $y={\left(1+x\right)}^{1/x}$

$\log y=\frac{1}{x}\log (1+x)$

$\frac{1}{y}\frac{dy}{dx}=\frac{\left[\frac{x}{1+x}-\log (1+x)\right]}{x^2}$

$\frac{dy}{dx}={\left(1+x\right)}^{1/x}\frac{\left[\frac{x}{1+x}-\log (1+x)\right]}{x^2}$

Therefore,

$\underset{x\to 0}{lim}\frac{{\left(1+x\right)}^{1/x}-e}{x}=\underset{x\to 0}{lim}{\left(1+x\right)}^{1/e}\frac{\left[\frac{x}{1+x}-\log (1+x)\right]}{x^2}$

$=e\times \underset{x\to 0}{lim}\frac{\left[\frac{x}{1+x}-\log (1+x)\right]}{x^2}$

Again using L-hospital rule,

$=e\times \underset{x\to 0}{lim}\frac{\left[\frac{\left(1+x\right)-x}{{\left(1+x\right)}^2}-\frac{1}{(1+x)}\right]}{2x}$

$=e\times \underset{x\to 0}{lim}\frac{\left[\frac{1}{{\left(1+x\right)}^2}-\frac{1}{(1+x)}\right]}{2x}$

Again using L-hospital rule

$=e\times \underset{x\to 0}{lim}\frac{\left[\frac{-2}{{\left(1+x\right)}^3}+\frac{1}{{(1+x)}^2}\right]}{2}$

$=e\times \left(\frac{-2+1}{2}\right)=-\frac{1}{2}e$

Using L-hospital rule,

$=\underset{x\to 0}{lim}\frac{d}{dx}{\left(1+x\right)}^{1/x}$

Let $y={\left(1+x\right)}^{1/x}$

$\log y=\frac{1}{x}\log (1+x)$

$\frac{1}{y}\frac{dy}{dx}=\frac{\left[\frac{x}{1+x}-\log (1+x)\right]}{x^2}$

$\frac{dy}{dx}={\left(1+x\right)}^{1/x}\frac{\left[\frac{x}{1+x}-\log (1+x)\right]}{x^2}$

Therefore,

$\underset{x\to 0}{lim}\frac{{\left(1+x\right)}^{1/x}-e}{x}=\underset{x\to 0}{lim}{\left(1+x\right)}^{1/e}\frac{\left[\frac{x}{1+x}-\log (1+x)\right]}{x^2}$

$=e\times \underset{x\to 0}{lim}\frac{\left[\frac{x}{1+x}-\log (1+x)\right]}{x^2}$

Again using L-hospital rule,

$=e\times \underset{x\to 0}{lim}\frac{\left[\frac{\left(1+x\right)-x}{{\left(1+x\right)}^2}-\frac{1}{(1+x)}\right]}{2x}$

$=e\times \underset{x\to 0}{lim}\frac{\left[\frac{1}{{\left(1+x\right)}^2}-\frac{1}{(1+x)}\right]}{2x}$

Again using L-hospital rule

$=e\times \underset{x\to 0}{lim}\frac{\left[\frac{-2}{{\left(1+x\right)}^3}+\frac{1}{{(1+x)}^2}\right]}{2}$

$=e\times \left(\frac{-2+1}{2}\right)=-\frac{1}{2}e$