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 $\sin(x^2)$.
Finding limit $\frac{\cos(2x)-1}{\sin(x^2)}$ for $x \to 0$
calculuslimitslimits-without-lhopital
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Best Answer
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 $\lim_{u \to 0} \frac{u}{\sin(u)} = \lim_{u \to 0} \frac{\sin(u)}{u} = 1$.