Find the limit of $\sqrt{x^2+x-1} +x$ as $x\to-\infty$

Question

Find the limit of $\sqrt{x^2+x-1} +x$ as $x\to-\infty$.

My solution:

multiplying by: $\displaystyle\frac{\sqrt{x^2+x-1}-x}{\sqrt{x^2+x-1}-x}$

Which gives us: $\displaystyle\frac{x-1}{\sqrt{x^2+x-1}-x}$

dividing by $\sqrt{x^2}$ gives:

$\displaystyle \frac{1}{\sqrt{1}-1}$

which equals $1/0$

However, I double checked my answers, and this does not seem to be correct, am I making a mistake (perhaps when I take the $\sqrt{1}$ in the denominator of the last step?

Answer 1

$\sqrt{x^2}$ is a positive number and since we are taking the limit as $x\to -\infty$, we have $\sqrt{x^2} = -x$. Your limit will be negative since $x^2>x^2+x-1$ as $x\to-\infty$.

Answer 2

I don't really like negative numbers, they are so negative. Let $x=-p$ where $p$ is positive. We want $$\lim_{p\to\infty}\left(\sqrt{p^2-p-1}-p\right).$$ A mistake is now much less likely.

Answer 3

Your mistake at this step:

dividing by $\sqrt{x^2}$ gives: $\displaystyle \frac{1}{\sqrt{1}-1}$

Since $x\to -\infty$, then $\sqrt{x^2}=-x$. So $\frac{x-1}{\sqrt{x^2+x-1}-x}=\frac{-1+\frac 1x}{\sqrt{1+\frac 1x-\frac{1}{x^2}}+1}.$ Take careful!

Answer 4

Writing $-\dfrac1h=x,$

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

$$=\lim_{h\to0^+}\frac{(1+h-h^2)-1}{h(\sqrt{1+h-h^2}+1)}$$

$$=\lim_{h\to0^+}\frac{1-h}{\sqrt{1+h-h^2}+1}=?$$