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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.

  • 1
    *Hint* $$\frac{\sqrt x}{x^2+x}=\frac{1}{\sqrt x( x +1)}\to +\infty $$ when $x\to 0^+$ – 2017-02-12
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    The two-sided limit does not exist. Should it be $\lim\limits_{\color{red}{x\to 0^+}} \frac{\sqrt{x}}{x^2+x}$ instead? – 2017-02-12
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    It's only possible with 0+ because in (√x) x have to be always positive, right? Now I see that you multiplied by √x and you cut the x's above and below. Thank you very much. :) – 2017-02-12

2 Answers 2

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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 $x\to 0^+$

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

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    The brave anonymous downvoter struck again! – 2017-02-13