How to find all answers for these limits

Question

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 $-\infty$.

Answer 1

Hint:

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

Answer 2

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

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