Is it possible to find the limit without l'Hopital's Rule or series Is it possible to find $$\lim_{x\to0}\frac{\sin(1-\cos(x))}{x^2e^x}$$
without using L'Hopital's Rule or Series or anything complex but just basic knowledge (definition of a limit, limit laws, and algebraic expansion / cancelling?)
 A: Multiply and divide the numerator and denominator by $1-\cos(x)$.
Then just basic trigonometric limits to evaluate the answer.
A: \begin{align}
\lim_{x \to 0} \frac{\sin(1 - \cos x)}{\mathrm{e}^x \cdot x^2}
  &= \lim_{x \to 0} \frac{\sin(1 - \cos x)}{1 - \cos x} 
            \cdot \frac{1 - \cos x}{x^2} \cdot \mathrm{e}^{- x} \\
  &= \lim_{u \to 0} \frac{\sin u}{u}
            \cdot \lim_{x \to 0} \frac{1 - \cos x}{x^2}
           \cdot \lim_{x \to 0} \mathrm{e}^{- x} \\
  &= 1\cdot \lim_{x \to 0} \frac{2 \sin^2 \frac{x}{2}}{x^2} \cdot 1 \\
  &= \lim_{u \to 0} \frac{2 \sin^2 u}{4 u^2}
\end{align}
A: We can proceed as follows: $$\begin{aligned}L &= \lim_{x \to 0}\frac{\sin(1 - \cos x)}{x^{2}e^{x}}\\
&= \lim_{x \to 0}\frac{\sin(1 - \cos x)}{1 - \cos x}\cdot\frac{1 - \cos x}{x^{2}e^{x}}\\
&= \lim_{x \to 0}\frac{\sin(1 - \cos x)}{1 - \cos x}\cdot\frac{1 - \cos x}{x^{2}}\cdot \frac{1}{e^{x}}\\
&= \lim_{t \to 0}\frac{\sin t}{t}\cdot\lim_{x \to 0}\frac{1 - \cos x}{x^{2}}\cdot \lim_{x \to 0}\frac{1}{e^{x}}\text{ by putting }t = 1 - \cos x \to 0 \text{ as } x \to 0\\
&= 1\cdot\lim_{x \to 0}\frac{1 - \cos x}{x^{2}}\cdot 1\\
&= \lim_{x \to 0}\frac{2\sin^{2}(x/2)}{(x/2)^{2}}\cdot\frac{(x/2)^{2}}{x^{2}}\\
&= 2\cdot 1\cdot\frac{1}{4} = \frac{1}{2}\end{aligned}$$
