finding limit $\frac{\cos(2x)-1}{\sin(x^2)}$ for $x \to 0$ I am trying to find the limit of $$\lim_{x \to 0}\frac{\cos(2x)-1}{\sin(x^2)}$$ Can someone give me a hint on how to proceed without applying L'Hôpital's rule. I tried using the trig identity $\cos(2x)-1 = -2\sin^2(x)$ but that doesn't seem to be useful as the denominator is $\sin(x^2)$.
 A: Note that
$$
\lim_{x \to 0}\frac{\cos(2x)-1}{\sin(x^2)} = 
\left(\lim_{x \to 0} \frac{x^2}{\sin(x^2)}\right) \cdot \left(\lim_{x \to 0} \frac{\cos(2x) - 1}{x^2}\right)\\
= \left(\lim_{x \to 0} \frac{x^2}{\sin(x^2)}\right) \cdot \left(\lim_{x \to 0} \frac{-2\sin^2(x)}{x^2}\right)
$$
From there, it suffices to note that $\lim_{u \to 0} \frac{u}{\sin(u)} = \lim_{u \to 0} \frac{\sin(u)}{u} = 1$.
A: As you had, it is $$\lim_{x\to 0}\frac{-2\sin^2(x)}{\sin(x^2)}=\lim_{x\to 0}\left(\frac{x^2}{\sin x^2}\times \frac{-2\sin^2 x}{x^2}\right)=\lim_{x\to 0}\frac{-2\sin^2x}{x^2}=\lim_{x\to 0}\frac{-2\sin x\frac{\sin x}{x}}{x}=-2\lim_{x\to 0}\frac{\sin x}{x}=\boxed{-2}$$
This is simply repeatedly applying $\lim_{u\to 0}\frac{\sin u}{u}=1$ and $\lim_{u\to 0}\frac{u}{\sin u}=1$.
A: $$L=\lim_{x \to 0}\frac{\cos(2x)-1}{\sin(x^2)}$$
With Taylor series at order 2:
$$\cos(2x)-1=-2x^2 +o(x^2)$$
$$ \sin (x^2)=x^2+o(x^2)$$
So that we have :
$$ \frac{\cos(2x)-1}{\sin(x^2)}\sim_0 \dfrac {-2x^2}{x^2}=-2$$
A: $\lim\limits_{x\to0} \dfrac{\cos(2x)-1}{\sin(x^2) } = \dfrac{\frac{\cos(2x)-1}{4x^2 }}{\frac{\sin(x^2) }{4x^2 }}=\dfrac{\frac{\cos(2x)-1}{(2x)^2 }}{\frac{\sin(x^2) }{4x^2 }}=-2$
Because: $\lim_{x\to0} \dfrac{\cos(2x)-1}{2x^2}=-\dfrac{1}{2} $
And
$\lim _{x\to0} \dfrac{\sin(x^2 )}{x^2} =1$.
A: With equivalents and the duplication formula: $\cos 2x=1-2\sin^2x$, we have
$$\sin x\sim_0 x,\qquad \cos 2x-1=-2\sin^2x\sim_0-2x^2,$$
therefore
$$\frac{\cos 2x-1}{\sin^2x}\sim_0\frac{-2\not x^{\not 2}}{\not x^{\not2}}=-2. $$
