GATE (TF) Textile 2011 Question Paper Solution | GATE/2011/TF/22

The value of $\lim_{x\to 0} \frac{(1+x)^n -1}{x}$ is

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$\lim_{x\to 0} \frac{(1+x)^n -1}{x}$

By opening the equation-

$(1+x)^n=1+nx+\frac{n(n-1) x^2}{2!}+................$

$\lim_{x\to 0} \frac{1+nx+\frac{n(n-1) x^2}{2!}+.............-1}{x}$

[Show Answer]

$\lim_{x\to 0} \frac{(1+x)^n -1}{x}$

By opening the equation-

$(1+x)^n=1+nx+\frac{n(n-1) x^2}{2!}+................$

$\lim_{x\to 0} \frac{1+nx+\frac{n(n-1) x^2}{2!}+.............-1}{x}$

$\lim_{x\to 0} \frac{nx+\frac{n(n-1) x^2}{2!}+.............}{x}$

By taking limit-
=n (Ans)