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

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

This is the standard trigonometric limit form $\dfrac{\sin(\text{angle})}{\text{same angle}}\to 1$, so the value is $1$.

The fundamental trigonometric limit is

$\lim_{t\to 0}\frac{\sin t}{t}=1.$

Here let $t=2\theta$. As $\theta\to 0$, also $t\to 0$, and the numerator's angle ($2\theta$) exactly matches the denominator ($2\theta$): …