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The question is $$\lim_{x \to\infty} \sqrt{x^2+3x}-x$$ I divided and multiplied by $x^2$ but it gave me answer $0\cdot\infty$ which is undefined I suppose. any hint !! (PS: avoid using L Hospital's Rule)

Martin Sleziak
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Archis Welankar
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    Just multiply on $\sqrt{x^{2}+3x}+x$ – openspace Jan 11 '16 at 06:59
  • That would lead to $3.\infty$ so thats infinity please explain further – Archis Welankar Jan 11 '16 at 07:02
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    Related : http://math.stackexchange.com/questions/1470634/finding-the-limit-without-lhopitals-rule%C2%B4s http://math.stackexchange.com/questions/1454627/calculating-lim-x-to-infty-left-sqrt4x2-6-sqrt4x2x-right http://math.stackexchange.com/questions/1457529/find-the-limit-of-x-sqrtx2-8x-as-x-to-infty – lab bhattacharjee Jan 11 '16 at 07:32
  • This question is almost the same (only the sing in $\sqrt{x^2-3x}$ is different.) And there is also this post which deals with a more general version of this problem. Many similar limits can be found in the posts which are linked there. – Martin Sleziak Jan 11 '16 at 13:30
  • Please, try to make the titles of your questions more informative. E.g., Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here. – Martin Sleziak Jan 11 '16 at 13:43
  • I try making it descriptive and tge bot says include it in your bidy or says its a low quality title – Archis Welankar Jan 11 '16 at 13:51
  • @ArchisWelankar I am not sure who you mean by tge bot but they are right. If you take the time to read the post How to ask a good question?, you will find out there this advice: "Your question should be clear without the title." The fact that the question should be clear from the title does not mean that it cannot be included in the body, too. – Martin Sleziak Jan 13 '16 at 14:33

4 Answers4

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HINT: Multiply by

$$\frac{\sqrt{x^2+3x}+x}{\sqrt{x^2+3x}+x}\;,$$

One more step will be required, but you can probably work it out. Just in case, I’ve included it in the spoiler-protected block below.

Then divide numerator and denominator by $x$.

Brian M. Scott
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Hint

Just another way (among so many). Rewrite $$\sqrt{x^2+3x}-x=x\sqrt{1+\frac 3x}-x$$ Remember that, for small $y$, $\sqrt{1+y}\sim1+\frac y2$. Replace $y$ by $\frac 3 x$ and conclude.

Claude Leibovici
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    I am really curious to know why downvotes. I gave that as another way and it is probably the fastest way to solve the problem. – Claude Leibovici Jan 11 '16 at 08:14
  • The approximation $\sqrt{1+y}\sim1 + \frac{y}{2}$ is the reason I downvoted this. I think that one needs to be very cautious with such statements: How small does $y$ need to be? What if I multiply it with,e.g. $y^2$? – Roland Jan 12 '16 at 10:57
  • @Roland. Thanks for explaining your downvote ! It is so exceptional. As I wrote : for small $y$. In the case ot the post $x\to\infty$ so $y\to 0$. Moreover, if you take an extra term, you will get the asymptotics of the expression (what other methods cannot do). Cheers and again congratulations for you nice attitude ! – Claude Leibovici Jan 12 '16 at 11:02
  • A rigorous statement (which would avoid qualms by scrupulous readers and suffices to solve the exercise) is that $\sqrt{1+y}=1+\frac12y+o(y)$ when $y\to0$. – Did Jan 25 '16 at 14:30
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As I said this limit is equals to: $$\lim \frac{3x}{\sqrt{x^{2}+3x}+x}$$ Now you can finish by yourself.

Martin Sleziak
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openspace
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Another way:

As $x^2+3x=\dfrac{(2x+3)^2-3^2}4,$

Let $2x+3=\sqrt3\csc2y\implies y\to0^+$

$$\lim_{x\to\infty}(\sqrt{x^2+3x}-x)=\lim_{y\to0^+}\dfrac{\sqrt3\cot2y-(\sqrt3\csc2y-3)}2$$

$$=\dfrac32-\dfrac{\sqrt3}2\lim_{y\to0^+}\dfrac{1-\cos2y}{\sin2y}$$

$$\lim_{y\to0^+}\dfrac{1-\cos2y}{\sin2y}=\lim_{y\to0^+}\dfrac{2\sin^2y}{2\sin y\cos y}=\lim_{y\to0^+}\tan y=0$$

lab bhattacharjee
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