<s>[INST] How to solve $\lim\limits_{x\to 1} \frac{1-\cos(\sin(x^3-1))}{x^3-1}$ without L'Hospital's Rule? I am asked to solve $\lim\limits_{x\to 1} \frac{1-\cos(\sin(x^3-1))}{x^3-1}$ without using L'Hospital's Rule.
I'm not sure how to go about it.
There is an indeterminate form $(\frac{0}{0})$ at $x=1$ and L'Hospital's Rule seems like the best course of action.
I tried to multiply by $\frac{1+\cos(\sin(x^3-1))}{1+\cos(\sin(x^3-1))}$ to get $\lim\limits_{x\to 1} \frac{1-\cos^2(\sin(x^3-1))}{x^3-1}\cdot\frac{1}{1+\cos(\sin(x^3-1))} = \lim\limits_{x\to 1} \frac{\sin(\sin(x^3-1))}{x^3-1}\cdot \sin(\sin(x^3-1))\cdot\frac{1}{1+\cos(\sin(x^3-1))}$
I'd try to get a limit of the form $\lim\limits_{u\to 0} \frac{\sin(u)}{u}$ because I know that that limit is $1$, but the problem is that sine is composed with itself.
Any help would be appreciated!
 [/INST] By standard limits, since as $x \to 1 \iff x^3-1 \to 0$, we have that
$$\frac{1-\cos(\sin(x^3-1))}{x^3-1}=\frac{1-\cos(\sin(x^3-1))}{\sin^2(x^3-1)}\cdot \frac{\sin(x^3-1)}{x^3-1}\cdot \sin(x^3-1)\to \frac12 \cdot 1 \cdot 0 =0$$
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