Question

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

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

We have,$\underset{x\to 0}{lim}\frac{(1+x{)}^{\frac{1}{x}}-e+\frac{1}{2}ex}{x^2}$ ........(1)

At x=0, the value of the given expression takes $\frac{0}{0}$ form

since,$\underset{x\to 0}{lim}(1+x)\frac{1}{x}=e$

$\frac{e-e+0}{0}=\frac{0}{0}$ form

$(1+x{)}^{\frac{1}{x}}=e^{\frac{1}{x}}lo{g}_e(1+x)$ $\{y=e^{lo{g}_{e^y}}\}$

We know expansion of $lo{g}_e(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}.......$

$(1+x{)}^{\frac{1}{x}}=e^{(1+x)\{x-\frac{x^2}{2}+\frac{x^3}{3}-.......\}}$

$=e^{\{1-\frac{x}{2}+\frac{x^2}{3}-.......\}}$

$=e.e^{-\frac{x}{2}+\frac{x^2}{3}-.......}$

Applying the expression of $e^x$

$(1+x{)}^{\frac{1}{x}}=e.[1+(-\frac{x}{2}+\frac{x^2}{3}-......)+\frac{1}{2!}{(-\frac{x}{2}+\frac{x^2}{3}-......)}^2+......]$

$\{e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+......\}$

$=e.\{1+(-\frac{x}{2}+\frac{x^2}{3}+.....)+\frac{1}{2!}(\frac{x^2}{4}+......termswithpowermorethan{x}^2\}$

$=e\{1-\frac{x}{2}+\frac{x^2}{3}+\frac{x^2}{8}+.....termswithpowermorethan{x}^2\}$

$(1+x{)}^{\frac{1}{x}}=e\{1-\frac{x}{2}+\frac{11}{24}{x}^2+.....termswithpowermorethan{x}^2\}$

Substituting the value of $(1+x{)}^{\frac{1}{x}}$ in equation 1

$\underset{x\to 0}{lim}\frac{e\{1-\frac{x}{2}+\frac{11}{24}{x}^2+.....termswithpowermorethan{x}^2\}-e+\frac{1}{2}ex}{x^2}$

$=\underset{x\to 0}{lim}\frac{\{e-\frac{e}{x}+\frac{11}{24}{x}^2+.....termswithpowermorethan{x}^2\}-e+\frac{1}{2}ex}{x^2}$

$=\underset{x\to 0}{lim}\frac{11}{24}e+.......$terms which has variable x

Putting$x=0$

$=\frac{11}{24}e+0$

$=\frac{11}{24}e$

Option A is correct