Show $\lim\limits_{x \to 0}\frac{e^x-1}{x} = 1$ I have a homework problem which asks me to show
$$ \lim\limits_{x \to 0}\frac{e^x-1}{x} = 1 $$

Any help is appreciated.
By the definition of the derivative and if we denote $f(x)=e^x$ then
$$\lim_{x\to0}\frac{e^x-1}{x}=\lim_{x\to0}\frac{f(x)-f(0)}{x-0}=f'(0)=e^0=1$$
You can use l'Hopitals rule. Notice that if we substitute directly we get $\frac{0}{0}$, an indeterminate form. So we can differentiate the numerator and denominator separately then take the limit. Then we have

$$\lim_{x\to 0} \frac{e^x-1}{x}=\lim_{x \to 0}\frac{e^x}{1}=e^0=1$$

Though I've doubt you've seen this yet, we don't even need that because we can use the Taylor series for $e^x$.

$$\lim_{x \to 0}\frac{e^x-1}{x}=\lim_{x\to 0} \frac{\sum_{n=0}^\infty\frac{x^n}{n!}-1 }{x}=\lim_{x \to 0} \frac{1+\sum_{n=1}^\infty \frac{x^n}{n}-1}{x}=\lim_{x \to 0}\frac{\sum_{n=1}^\infty \frac{x^n}{n!}}{x}=\lim_{x\to 0} \sum_{n=1}^\infty \frac{x^{n-1}}{n!}$$

But then we have

$$\lim_{x\to 0} \sum_{n=1}^\infty \frac{x^{n-1}}{n!}=\sum_{n=1}^\infty \frac{0^{n-1}}{n!}=1$$

Though seeing that last step with $0^0=1$ is a bit trickier. Not necessarily the way to do it but another way of thinking of it for those that know the Taylor Series.

EDIT: You wouldn't even need to use the fact that $0^0$ is $1$ as

$$
\lim_{x\to 0} \sum_{n=1}^\infty \frac{x^{n-1}}{n!}=\lim_{x\to 0} 1+\sum_{n=2}^\infty \frac{x^{n-1}}{n!}=1+ \sum_{n=2}^\infty \frac{0^{n-1}}{n!}=1
$$