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Can you please help me out with this limit problem. Actually, I tried to solve it by the conjugate method but it didn't work with me.

Thank you.

$$\lim_{x \to 0}\; \bigg( \frac{1}{\sqrt{x}} - \frac{1}{\sqrt{x²+x}} \bigg)$$

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    $\begingroup$ I edited your question. Please check and make sure that I did not alter what you are asking. $\endgroup$ – Thomas Jan 24 '13 at 15:13
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    $\begingroup$ what was the result from your conjugate method? $\endgroup$ – still_learning Jan 24 '13 at 15:15
  • $\begingroup$ The limit is not defined as the function is not defined in a deleted neighbourhood of $0$. The function is defined on $(0, \infty)$ so you could instead consider the corresponding one-sided limit. $\endgroup$ – Michael Albanese Jan 24 '13 at 15:41
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This is probably a bit too straightforward, there should be smarter way to do it, but nevertheless...

$$ \lim_{x \to 0} \frac{\sqrt{x^2+x} - \sqrt{x}}{\sqrt{x(x^2+x)}}= \lim_{x \to 0} \frac{\sqrt{x}(\sqrt{x+1} - 1)}{x\sqrt{x+1}}= \lim_{x \to 0} \frac{x^{\frac{3}{2}}}{x\sqrt{x+1}(\sqrt{x+1}+1)}=0 $$ The third step is due to mupltiplying both numerator and denominator by $\sqrt{x+1}+1$

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  • $\begingroup$ You are welcome. You can upvote answers and accept the one you find most useful. PEople usually appreciate it) $\endgroup$ – Alex Jan 25 '13 at 14:48
  • $\begingroup$ Thank you very match for the answer. Actually this is my first time using this site and I do not no how to write the math symbols. I'll try to learn it. $\endgroup$ – Hassan Alkhamis Jan 25 '13 at 15:54
  • $\begingroup$ Indeed it’s a satisfied answer and thanks to you and to all that’s gave me a comments. $\endgroup$ – Hassan Alkhamis Jan 25 '13 at 15:57
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Take $\frac{1}{\sqrt{x}}$ common and then try the conjugate method.

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Also, $$\dfrac{1}{\sqrt{x}}-\dfrac{1}{\sqrt{x^2+x}}=\dfrac{1}{\sqrt{x}}\left(1-(1+x)^{-1/2}\right)=\dfrac{1}{\sqrt{x}}\left(\dfrac{1}{2}x+o(x)\right)\to 0 \mbox{ as } x\to 0^+$$

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