Calculation of limits - indetermination

Question

I am having trouble trying to calculate

$$\lim_{x\to 0} \frac{\sqrt{x}}{x^2+x}$$

I tried go above the $\frac{0}{0}$ indetermination by doing $\lim_{x\to 0} \frac{x^{\frac{1}{2}}}{x^2(1+\frac{1}{x})}$ but I am confuse about what to do next.

Can you guys give me a hint? Thank you.

Answer 1

Divide both term by $\sqrt{x}$ and get:

$$\lim_{x\to 0^+} \frac{\sqrt{x}}{x^2+x}=\lim_{x\to 0^+} \frac{1}{x^{3/2}+x^{1/2}}=\infty$$

Look that you just can calculate the limit when $x\to 0^+$

Answer 2

Use equivalents: $x^2+x\sim_0x $, hence $$\frac{\sqrt x}{x^2+x}\sim_0\frac{\sqrt x}{x}=\frac1{\sqrt x}\to+\infty.$$