What techniques I could use to solve this limit
$$\lim_{x \to 0} \frac{x^3-x\cos x }{\sin x}$$
without l'Hopital?
When I use l'Hopital the limit is $-1$.
With l'Hopital $$\lim_{x \to 0} \frac{3x^2-1\cdot \cos x+x\sin x}{\cos x}=$$ $$=\frac{3\cdot 0-1\cdot 1+1\cdot 0}{1}=$$ $$=\frac{-1}{1}=-1$$
As suggested, we can/should proceed as follows
$$\frac{x^3-x\cos x }{\sin x}=\frac{x}{\sin x}(x^2-\cos x)$$
and then use standard limit $\frac{x}{\sin x} \to ?$ and continuity for the other factor.
More in general, I suggest avoiding l'Hopital to solve limits as a first step and always try with, in the order
Refer also to
You can recover a derivative after substituting $x=\arcsin y$ :
$$\lim_{x\to0}\frac{x^3-x\cos x}{\sin x} = \lim_{y\to0}\frac{\arcsin^3y-\sqrt{1-y^2} \arcsin y}{y} \\ = \frac{d}{dy} \left[\arcsin^3y-\sqrt{1-y^2} \arcsin y\right]\bigg|_{y=0}$$
You could also use Taylor series approximation to say as $x \to 0$ we have $\sin x \approx x$ and $\cos x \approx 1$. Then, pluge in these two in the limit
$$\lim_{x \to 0} \frac{x^3-x\cos x }{\sin x}\approx \lim_{x \to 0} \frac{x^3-x }{x} = \lim_{x \to 0} x^2-1 = -1$$
$$\lim_{x \to 0}\frac{x^3-x\cos (x) }{\sin (x)}$$
Factor the numerator: $x^3-x\cos(x)$ into $x(x^2-\cos(x))$
Now we have: $$\lim_{x \to 0}\frac{x(x^2-\cos (x)) }{\sin (x)}$$
We can break this fraction into two: $$\lim_{x \to 0}\frac{x}{\sin(x)} \cdot \frac{x^2-\cos (x) }{1}$$
Now we can use product rule for calculating limits of combinations which states: $$\lim_{x \to c} (f(x)\cdot g(x))= \lim_{x \to c}f(x) \cdot \lim_{x \to c} g(x) $$
So, based on product rule for calculating limits of combinations: $$ \lim_{x \to 0} \frac{x}{\sin(x)} \cdot \lim_{x \to 0} \frac{x^2 - \cos(x)}{1} $$
The limit of $\frac{x}{\sin(x)}$ as x approaches 0 is a standard limit which equals 1
Therefore, we get: $$1 \cdot \lim_{x \to 0} \frac{x^2 - \cos(x)}{1}$$
Now we can plug in 0 for x: $$1 \cdot \frac{0^2 - \cos(0)}{1}$$
This simplifies into: $$-\cos(0)$$
The cosine of 0 is 1 (refer to the unit circle) therefore: $$-\cos(0)=-1$$
So,$$\lim_{x \to 0}\frac{x^3-x\cos (x) }{\sin (x)}=-1$$
Another simple way is using the trial and error method.
Denote $f(x)=\frac{x^3-x\cos x}{\sin x}$
First use $x=1$ and you find that $f(1)=0.546...$
Then use $x=0.1$ and you see that $f(0.1)=-0.986...$
Then use $x=0.01$ and you see that $f(0.01)=-0.999$
Therefore, $\lim_{x \to 0} \frac{x^3-x\cos x }{\sin x}=-1$.
You can also use a graphical method to show that it is equal to $-1$.
You can clearly see a discontinuity when $x=0$ since $\sin0$ is undefined. And also, the graph shows that as $x\to 0, y\to -1$.
These are two ways of finding out the limit without using L'Hopital's Rule.
