How to solve $\lim\limits_{x\to 1} \frac{1-\cos(\sin(x^3-1))}{x^3-1}$ without L'Hospital's Rule?

Question

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!

Answer 1

By standard limits, since as $x \to 1 \iff x^3-1 \to 0$, we have that

$$\frac{1-\cos(\sin(x^3-1))}{x^3-1}=\frac{1-\cos(\sin(x^3-1))}{\sin^2(x^3-1)}\cdot \frac{\sin(x^3-1)}{x^3-1}\cdot \sin(x^3-1)\to \frac12 \cdot 1 \cdot 0 =0$$

Answer 2

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 $\displaystyle \lim_{ x \to 1} \dfrac{\sin(x^3 - 1)}{x^3 - 1} = 1$.

Answer 3

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 $x \to 1.$

Answer 4

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} $