limlimitsx→ 0(ex-1)/(x)= (a) 1 (b) e (c) 0 (d) nxⁿ⁻¹

Question 42 of 49

\lim\limits_{x\to 0}\dfrac{e^x-1}{x}=

(a)

1

(b)

e

(c)

0

(d)

nx^{n-1}

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$\lim\limits_{x\to 0}\dfrac{e^x-1}{x}$ is a standard limit equal to $1$ .

Using the series expansion $e^x=1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\cdots$ , we get

$\frac{e^x-1}{x}=\frac{x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\cdots}{x}=1+\frac{x}{2!}+\frac{x^2}{3!}+\cdots.$

As $x\to 0$ every term after the first vanishes, so …

Q. $\lim\limits_{x\to 0}\dfrac{e^x-1}{x}=$