Find the value of \(\lim\limits_{x \to 0}\frac{e^x-1}{x}\)

1.	Find the value of \(\lim\limits_{x \to 0}\frac{e^x-1}{x}\)
Answer» \(\lim\limits_{x \to 0}\frac{e^x-1}{x}\) \(=\lim\limits_{x \to 0}\frac{[1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+....]-1}{x}\) \(=\lim\limits_{x \to 0}\frac{1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...}{x}\) \(=\lim\limits_{x \to 0}[\frac{1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...}{x}]\) = 1.

Answer»

\(\lim\limits_{x \to 0}\frac{e^x-1}{x}\)

\(=\lim\limits_{x \to 0}\frac{[1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+....]-1}{x}\)

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

\(=\lim\limits_{x \to 0}[\frac{1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...}{x}]\)

= 1.

Discussion

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