# 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!

• Multiply by $\dfrac{\sin(x^3-1)}{\sin(x^3-1)}$.
– Blue
Oct 6, 2022 at 4:24
• @Blue sweet!!!! Oct 6, 2022 at 4:30

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

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

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 $$x \to 1.$$

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

• The result of the limit of the original poster is $0$, is not it? Oct 6, 2022 at 4:32
• Yes, it should be $0$. Oct 6, 2022 at 4:33

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

$$\begin{array}\\ \dfrac{1-\cos(\sin(x^3-1))}{x^3-1} &=\dfrac{1-\cos(\sin(3y+O(y^2)))}{3y+O(y^2)}\\ &=\dfrac{1-\cos(3y+O(y^2))}{3y+O(y^2)}\\ &=\dfrac{1-(1-\frac{(3y+O(y^2))^2}{2}+O(y^2))}{3y+O(y^2)}\\ &=\dfrac{\frac{(3y+O(y^2))^2}{2}+O(y^2))}{3y+O(y^2)}\\ &=\dfrac{O(y^2)}{3y+O(y^2)}\\ &=O(y)\\ &\to 0\\ \end{array}$$