# How to find all answers for these limits

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 $-\infty$.

• Your two answers look fine. Where do $\infty$ and $-\infty$ come from? – Olivier Oloa Jul 26 '16 at 13:49
• 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. – Aaron Jul 26 '16 at 13:53
• The $-\infty$ for the second limit is definitely wrong because the expression is positive for $x>0$ – Peter Jul 26 '16 at 13:58
• 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. – User8128 Jul 26 '16 at 13:59
• @Peter Yes, but I introduced the $+\infty$ myself, maybe there is another possibility when $x\to-\infty$? – Aaron Jul 26 '16 at 14:14

Hint:

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

• Sorry, there was a typo at the start, the x in the beginning. Just edited that. – Aaron Jul 26 '16 at 14:08
• @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 – Sathasivam K Jul 26 '16 at 14:20
• For me, $x\to\infty$ comprises both $+\infty$ and $-\infty$. – Bernard Jul 26 '16 at 14:30
• OK sir.I think you assumed $\infty$ to be either + or - infinity,Right? – Sathasivam K Jul 26 '16 at 14:33
• 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). – Bernard Jul 26 '16 at 14:34

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

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