Basic L'Hôpital's rule

I suppose Svatejas Shivakumar 's suggestion is as follows:

$\begin{aligned} \lim_{x \to 0} \frac{\color{#3D99F6}{\sin x}}{x} & = \lim_{x \to 0} \left( \frac{1}{x} \color{#3D99F6}{\left(x-\frac{x^3}{3!} + \frac{x^5}{5!} - ... \right)} \right) \quad \quad \small \color{#3D99F6}{\text{Expand into Maclaurin series}} \\ & = \lim_{x \to 0} \left(1-\frac{x^2}{3!} + \frac{x^4}{5!} - ... \right) \\ & = \boxed{1} \end{aligned}$

Let $f(x) = \sin x$ and $g(x) = x$ . at $x = 0$ , $f(x) = \sin 0 = 0$ and $g(x) = x = 0$ . So, $f(x) = g(x) = 0$ . Now, $f'(x) = \cos x$ and $g'(x) = 1$ . Therefore $\lim_{x \to 0} \dfrac{f'(x)}{g'(x)}$ exists which is equal to $1$ . So, applying L'Hôpital's rule. $\lim_{x \to 0} \dfrac{\sin x}{x} = \boxed{1}$ .

$\displaystyle \lim _ { x \to 0 } \frac { \sin x } { x } = \lim _ { x \to 0 } \sin x \degree = \sin 90 \degree = 1$ .