GATE (TF) Textile 2022 Question Paper Solution | GATE/2022/TF/02

$\lim_{x \rightarrow 0}\frac{e^x - 1}{x^2}$ is equal to

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$\Rightarrow \lim_{x\rightarrow 0} \frac{e^x -1}{x^2}$

As we know the equation :

$\Rightarrow e^x=1+x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+...........+\frac{x^n}{n}$

By putting the equation of $e^x$ we get :

$\Rightarrow \lim_{x\rightarrow 0} \frac{e^x -1}{x^2}$

$\Rightarrow \lim_{x\rightarrow 0} \frac{(1+x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+...........+\frac{x^n}{n})-1}{x^2}$

$\Rightarrow \lim_{x\rightarrow 0} \frac{x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+...........+\frac{x^n}{n}}{x^2}$

$\Rightarrow \lim_{x\rightarrow 0} \frac{x}{x^2} + \frac{1}2{} + \frac{x}{3} + \frac{x^2}{4} + ----+ \frac{x^{n-2}}{n}$

By putting limit $x \to 0$

$\Rightarrow 0 + \frac{1}{2} + 0 + 0 + ----- + 0$

$\Rightarrow \frac{1}{2}$