Find the two limits without the use of l'Hospital's rule or series expansion.

I was asked to evaluate these two limits:

$$\lim_{x\rightarrow0}\frac{x^3}{x-\sin x}$$

$$\lim_{x\rightarrow0}\frac{e^{-x^2}+x^2-1}{\sin(3x^4)}$$

For the first one I tried to divide the numerator and denominator by $x^3$, but I can't get the answer unless I apply l'Hospital's rule or using a series expansion.

I also tried to use a substitution $u=x^2$ for the second limit, but I can't seem to relate anything between the exponential function and sine function.

Answer

As shown here we have that

• $\lim_{x\to0}\frac{\sin x-x}{x^3}=-\frac16$



• $\lim_{x\to0}\frac{e^x-x-1}{x^2}=\frac12$

For the second one we can use that

$$\frac{e^{-x^2}+x^2-1}{\sin(3x^4)}=\frac{e^{-x^2}+x^2-1}{-x^4}\cdot \frac{-3x^4}{\sin(3x^4)}\cdot\frac13$$