Finding $\lim_{x\to -\infty}\sqrt{x^2+9x+1}-x$ In this question,
$$\lim_{x\to -\infty}\sqrt{x^2+9x+1}-x=\lim_{x\to -\infty}\left(\sqrt{x^2+9x+1}-x\right)\cdot \frac{\sqrt{x^2+9x+1}+x}{\sqrt{x^2+9x+1}+x}=$$
$$=\lim_{x\to -\infty}\frac{9x+1}{\sqrt{x^2+9x+1}+x}=\lim_{x\to -\infty}\frac{\frac{9x+1}{\sqrt{x^2}}}{\frac{\sqrt{x^2+9x+1}}{\sqrt{x^2}}+\frac x{\sqrt{x^2}}}=\frac{-9}{\sqrt{1}-1}$$
I got an answer that is $-9/0$ where that $0$ is just indefinitely approaching $0$. Therefore, I got the answer $-\infty$. However, the correct answer is $+\infty$.
May I know what mistakes have I made?

Answer :
$\lim_{x \to - \infty } \sqrt{x^2+9x+1 }-x$ = $\lim_{x \to - \infty } - x(\sqrt{1+\frac{9}{x}+\frac{1}{x^2 }}+1)$=$-\infty$
Because
$\lim_{x \to - \infty } \frac{9}{x}$=$0$
And
$\lim_{x \to - \infty } \frac{1}{x^2 }=0$
