Find $\lim\limits_{x \to 0} \frac{x^2}{\sin^2(x)}$ I think I should use L'Hospital's rule twice, to end up with $$\lim_{x \to 0} \dfrac{x^2}{\sin^2(x)}=\lim_{x \to 0} \dfrac{2}{2(\cos^2(x)-\sin^2(x))} = 1$$
Right?
 A: See this $$ \lim_{x \to 0} \frac{x^2}{\sin^2(x)} = \lim_{x \to 0} \frac{x}{\sin (x)} \cdot \lim_{x \to 0} \frac{x}{\sin(x)} = 1  $$
Another argument can be
$$ f(x) = \frac{x}{\sin(x)}, \quad \lim_{x \to 0} f(x) = 1  \implies \lim_{x \to 0} \big[f(x)\big]^2 = 1   $$
A: Yes you are right. Or you  may  also  use the fact $\lim_{x\to 0}\frac{x}{\sin x} =1 $. Heading to you limit we have $$\lim_{x\to 0}\frac{x^2}{\sin^2 x} =\lim_{x\to 0} \left(\frac{x}{\sin x}\right)^2=1^2=1$$
A: Since we have an indeterminate form of type $(0/0)$, we can apply the l'Hopital's rule:
$$\lim_{x \to 0} \left(\frac{x}{\sin{\left(x \right)}}\right)^{2} = \lim_{x \to 0} \left(\frac{\frac{d}{dx}\left(x\right)}{\frac{d}{dx}\left(\sin{\left(x \right)}\right)}\right)^{2}=\lim_{x \to 0} \left(\frac{1}{\cos(x)}\right)^2=1$$
A: Note that  $$\text{If } \quad f(x):= \frac{x}{\sin(x)} \text{ then }  \lim_{x \to 0} f(x) = 1  \implies \lim_{x \to 0} \big(f(x)\big)^2 = 1   $$
So we only have to show that $$ \lim_{x \to 0}\frac{x}{\sin(x)}=1$$
$\lim_{x \to 0}f(x)=\lim_{x \to 0}\frac{1}{\frac{\sin(x)}{x}}=\frac{1}{\lim_{x \to 0}\frac{\sin(x)}{x}}$
$$\text{Hence its enough to show that}:  \lim_{x \to 0}\frac{\sin(x)}{x}=1$$
In other words
$\forall ε>0$ $\existsδ>0:|x|<δ\Rightarrow\lvert \frac{\sin(x)}{x}-1\rvert<ε$ ,we have :
Since $\cos(x)<\frac{\sin(x)}{x}<1$, namely,
$$\bigg|\frac{\sin(x)}{x}-1\bigg|<1-\cos(x).$$
But $1-\cos(x)=2\sin^2\frac{x}{2}\le\frac{x^2}{2}$ and hence
$$\bigg|\frac{\sin(x)}{x}-1\bigg|\le\frac{x^2}{2}\le ε.$$
Choose $δ=\sqrt{2ε}$
