GATE (TF) Textile 2019 Question Paper Solution | GATE/2019/TF/01

The value of $\lim_{x\rightarrow 0}\frac{e^{x}-1-x}{x^{2}}$ is

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

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 :

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

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

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

By taking limit-

= $\frac{1}{2}$ (Ans)