Calculate limit without L'Hôpital's theorem

Question

I'm trying to solve the following limit:

$$\lim_{x\rightarrow 0^+} \frac{\displaystyle\sqrt{x^2+x^3} - \sin(x)}{\displaystyle 2x^2 - {e}^{-1/x}}$$

For WolframAlpha the result is: $ \frac14 $, while, according to my calculations, it is: $0$.

The text forbids me to use L'Hôpital's rule. Is the answer given by WolframAlpha wrong? or am I?

Answer 1

I'll systematise using equivalents:

Since $\mathrm e^{-1/x}=o\bigl(x^2\bigr)$, we know $2x^2+\mathrm e^{-1/x}\sim_0 2x^2$.

Let's expand the numerator with Taylor's formula at order $2$: $$\sqrt{x^2+x^3}-\sin x=x\sqrt{1+x}-\sin x=x\Bigl(1+\frac x2\Bigr)-x+o\bigl(x^2\bigr)=\frac{x^2}2+o\bigl(x^2\bigr)\sim_0\frac{x^2}2,$$ so the function is equivalent, near $0$, to $\;\dfrac{\dfrac{x^2}2}{2x^2}=\dfrac14$.