How to solve this limit

Question

How can I solve this: $$\lim_{x\to\infty} x\left(\frac{\sqrt{x^2+6x+10}}{x^2}-1\right)$$

What is the easiest method to solve it? (Without the use of Hopital)

Answer 1

Hint: $\large{x\frac{\sqrt{x^2+6x+10}}{x^2}=\sqrt{1+\frac{6}{x}+\frac{10}{x^2}}}$

Answer 2

Note that $$ \lim_{x\to\infty} x\left(\frac{\sqrt{x^2+6x+10}}{x^2}-1\right)=\lim_{x\to\infty}\sqrt{1+\frac{6}{x}+\frac{10}{x^2}}-x $$

Answer 3

Write your term in the form $$\frac{1}{x}\frac{(\sqrt{x^2+6x+10}-x^2)(\sqrt{x^2+6x+10}+x^2)}{\sqrt{x^2+6x+10}+x^2}$$

Answer 4

As $x^2$ goes to $\infty$ faster than $6x$, we have that $$\frac{\sqrt{x^2+6x+10}}{x^2}\underbrace{\to}_{x\to \infty} \frac {\sqrt{x^2}}{x^2} =\frac x{x^2}=\frac1x$$ Thus $$\lim_{x\to\infty} x\left(\frac1x-1\right)=\lim_{x\to\infty}1-x=-\infty$$