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I have solved two limits, but only found one answer while the solution contains two. What am I missing?

First limit

$$\lim_{x\to\infty} \sqrt{x(x+a)}-x = \lim_{x\to\infty} \sqrt{x^2(1+\frac{a}{x}})-x = \lim_{x\to\infty} x(\sqrt{(1+\frac{a}{x}})-1) = \lim_{x\to\infty} \frac{x(\sqrt{1+\frac{a}{x}}-1) (\sqrt{1+\frac{a}{x}}+1)}{(\sqrt{1+\frac{a}{x}}+1)} = \lim_{x\to\infty} \frac{a}{(\sqrt{1+\frac{a}{x}}+1)} = \frac{a}{2}$$

However, $+\infty$ is also an answer?

Second limit

$$\lim_{x\to\infty} x(\sqrt{x^2+1}-x) = \lim_{x\to\infty} \frac{x(\sqrt{x^2+1}-x)(\sqrt{x^2+1}+x)}{(\sqrt{x^2+1}+x)} = \lim_{x\to\infty} \frac{x}{(\sqrt{x^2+1}+x)}$$

But $x\to\infty$ so we can say $\sqrt{x^2+1} \approx \sqrt{x^2}$, which gives:

$$ = \lim_{x\to\infty} \frac{x}{|x|+x} = \lim_{x\to+\infty} \frac{x}{2x} = \frac{1}{2}$$

But it should also be $-\infty$.

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    $\begingroup$ Your two answers look fine. Where do $\infty$ and $-\infty$ come from? $\endgroup$ – Olivier Oloa Jul 26 '16 at 13:49
  • $\begingroup$ From the answers in my course book. We don't get the full solution though, so I don't know how they got to them. $\endgroup$ – Aaron Jul 26 '16 at 13:53
  • $\begingroup$ The $-\infty$ for the second limit is definitely wrong because the expression is positive for $x>0$ $\endgroup$ – Peter Jul 26 '16 at 13:58
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    $\begingroup$ I don't see how you did the first one. Specifically, the first equality $$\lim_{x \to \infty} x\left( \sqrt{x(x+a)} -x \right) = \lim_{x\to \infty} x\left( \sqrt{1 + \tfrac a x} - 1 \right).$$ It seems you lost a factor of $x$ somewhere. I think that limit should be infinite. $\endgroup$ – User8128 Jul 26 '16 at 13:59
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    $\begingroup$ @Peter Yes, but I introduced the $+\infty$ myself, maybe there is another possibility when $x\to-\infty$? $\endgroup$ – Aaron Jul 26 '16 at 14:14
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Hint:

The first factorisation is false if $x<0$: $$\sqrt{x(x+a)}-x = \sqrt{x^2\Bigl(1+\frac{a}{x}\Bigr)}-x=\lvert x\rvert\sqrt{1+\frac{a}{x}}-x=\begin{cases}x\biggl(\sqrt{1+\dfrac{a}{x}}-1\biggr)&\text{if }x>0,\\-x\biggl(\sqrt{1+\dfrac{a}{x}}+1\biggr)&\text{if }x<0.\end{cases}$$

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  • $\begingroup$ Sorry, there was a typo at the start, the x in the beginning. Just edited that. $\endgroup$ – Aaron Jul 26 '16 at 14:08
  • $\begingroup$ @Bernard .since $x\to \infty$ we shall concentrate only in the neighborhood where $x>0$ .so you need not define your second case that is x<0 $\endgroup$ – Sathasivam K Jul 26 '16 at 14:20
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    $\begingroup$ For me, $x\to\infty$ comprises both $+\infty$ and $-\infty$. $\endgroup$ – Bernard Jul 26 '16 at 14:30
  • $\begingroup$ OK sir.I think you assumed $\infty$ to be either + or - infinity,Right? $\endgroup$ – Sathasivam K Jul 26 '16 at 14:33
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    $\begingroup$ Yes. I don't write the sign only in the case it's clear from the contest (e.g. sequences, for wich the variable is a natural integer). $\endgroup$ – Bernard Jul 26 '16 at 14:34
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Setting $1/x=h,$ $$\lim_{x\to\infty} \sqrt{x(x+a)}-x=\lim_{h\to0^+}\dfrac{\sqrt{1+ah}-1}h=\lim_{h\to0^+}\dfrac{1+ah-1}{h(\sqrt{1+ah}+1)}=\dfrac a{\sqrt1+1}\text{ for finite }a$$

Setting $1/x=h,$ $$\lim_{x\to\infty} x(\sqrt{x^2+1}-x) =\lim_{h\to0^+}\dfrac{\sqrt{1+h^2}-1}{h^2}$$

$$=\lim_{h\to0^+}\dfrac{1+h^2-1}{h^2(\sqrt{1+h^2}+1)}=\dfrac1{\sqrt1+1}=?$$

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