<s>[INST] Using $\lim\limits_{x \to 0} \frac{\sin x}{x} = 1$ evaluate the limit Use $$\lim_{x\to 0} \frac{\sin x}{x} = 1$$
to evaluate the limit
$$\lim_{x\to 0} \frac{x\tan (x^2)}{\cos(3x)\sin^3(2x)}$$
I'm really not sure how to go about this apart from trig identities and using the limit $=1$ to simplify it. Can anyone even get me started on it please?
 [/INST] $$\lim_{x\to0}\frac{x\tan(x^2)}{\cos(3x)\sin^3(2x)}=\lim_{x\to0}\frac{1}{\cos(3x)}\cdot\frac{x^3}{x^2}\cdot\frac{\tan(x^2)}{\sin^3(2x)}=$$
$$\lim_{x\to0}\frac{1}{\cos(3x)}\cdot\frac{x^3}{\sin^3(2x)}\cdot\frac{\tan(x^2)}{x^2}=\lim_{x\to0}\frac{1}{\cos(3x)}\cdot[\frac{x}{\sin(2x)}]^3\cdot\frac{\tan(x^2)}{x^2}$$
Now you have that:
$$\lim_{x\to0}\frac{x}{\sin(2x)}=\lim_{x\to0}\frac{1}{2}\frac{2x}{\sin(2x)}=\frac{1}{2}$$
and, setting $\ x^2=t:$
$$\lim_{x\to0}\frac{\tan(x^2)}{x^2}=\lim_{t\to0}\frac{\tan(t)}{t}=\lim_{t\to0}\frac{1}{\cos(t)}\frac{\sin(t)}{t}=1\cdot1=1$$
so the final result should be: $\ [\frac{1}{2}]^3=\frac{1}{8}$
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