1

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 $$\lim_{x \to 0} \frac{3x^2-1\cdot \cos x+x\sin x}{\cos x}=$$ $$=\frac{3\cdot 0-1\cdot 1+1\cdot 0}{1}=$$ $$=\frac{-1}{1}=-1$$

Blue
  • 75,673
Rosario A
  • 9

3 Answers3

3

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 to avoid l'Hopital to solve limits as a first step and always try with, in the order

  • continuity/sustituition (not intederminate expression)
  • manipulation to get standard limits
  • Taylor expansion
  • l'Hopital only for some particualr limits (e.g. integral functions)

Refer also to

user
  • 154,566
0

You can recover a derivative after substituting $x=\arcsin y$ :

$$\lim_{x\to0}\frac{x^3-x\cos x}{\sin x} = \lim_{y\to0}\frac{\arcsin^3y-\sqrt{1-y^2} \arcsin y}{y} \\ = \frac{d}{dy} \left[\arcsin^3y-\sqrt{1-y^2} \arcsin y\right]\bigg|_{y=0}$$

user170231
  • 19,334
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

$$\lim_{x \to 0} \frac{x^3-x\cos x }{\sin x}\approx \lim_{x \to 0} \frac{x^3-x }{x} = \lim_{x \to 0} x^2-1 = -1$$

Bob
  • 438