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!
 A: Hint: use the following:
$a)$ Put $u = \sin(x^3 - 1)$.
$b)$ $\displaystyle \lim_{u \to 0} \dfrac{1-\cos u}{u}= 0$.
$c)$ Use what you already know $\displaystyle \lim_{ x \to 1} \dfrac{\sin(x^3 - 1)}{x^3 - 1} = 1$.
A: Let $f(t):= \cos ( \sin (t))$ and $u(x):= x^3-1.$ Then $u(x) \to 0$ as $x\to 1$
We get
$$\frac{1-\cos(\sin(x^3-1))}{x^3-1}=-\frac{f(u(x))-f(0)}{u(x)-0} \to -f'(0)=0$$
as $x \to 1.$
A: By standard limits, since as $x \to 1 \iff x^3-1 \to 0$, we have that
$$\frac{1-\cos(\sin(x^3-1))}{x^3-1}=\frac{1-\cos(\sin(x^3-1))}{\sin^2(x^3-1)}\cdot \frac{\sin(x^3-1)}{x^3-1}\cdot \sin(x^3-1)\to \frac12 \cdot 1 \cdot 0 =0$$
A: 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}
$
