<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] Let $y=x-1$
so $y \to 0$.
Then
$x^3-1
=(y+1)^3-1
=y^3+3y^2+3y
=y(y^2+3y+3)
=3y+O(y^2)
$
so
$\begin{array}\\
\dfrac{1-\cos(\sin(x^3-1))}{x^3-1}
&=\dfrac{1-\cos(\sin(3y+O(y^2)))}{3y+O(y^2)}\\
&=\dfrac{1-\cos(3y+O(y^2))}{3y+O(y^2)}\\
&=\dfrac{1-(1-\frac{(3y+O(y^2))^2}{2}+O(y^2))}{3y+O(y^2)}\\
&=\dfrac{\frac{(3y+O(y^2))^2}{2}+O(y^2))}{3y+O(y^2)}\\
&=\dfrac{O(y^2)}{3y+O(y^2)}\\
&=O(y)\\
&\to 0\\
\end{array}
$
</s>