Question

It is given $e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}........$ Find the value of the limit

$\left(\underset{x\to 0}{lim}\frac{e^x-1-x}{x^2}\right)\times 16$

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Answer 1

If we substitute x=0 in the given expression we will get $\frac{0}{0}$ form.

But we are given $e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}........$

We will use this expansion to solve the problem.

$\Rightarrow \left(\underset{x\to 0}{lim}\frac{e^x-1-x}{x^2}\right)=\left(\underset{x\to 0}{lim}\frac{1+x+\frac{x^2}{2!}+\frac{x^3}{3!}..-1-x}{x^2}\right)$

$=\underset{x\to 0}{lim}\frac{\frac{x^2}{2!}+\frac{x^3}{3!}}{x^2}.....$

$=\underset{x\to 0}{lim}(\frac{1}{2}+\frac{x}{3!}+\frac{x^2}{4!}......)$

$=\frac{1}{2}$

$\Rightarrow Answer=\frac{1}{2}\times 16$

$=8$