Limit $\lim\limits_{x\to\infty} \frac{5x^2}{\sqrt{7x^2-3}}$ Evaluate the following limit: $$\lim_{x\to\infty} \frac{5x^2}{\sqrt{7x^2-3}}$$
I'm not really sure what to do when there is a square root for an infinity limit.
Please Help!

You can consider $$\lim_{x\to\infty}\frac{5x^2}{\sqrt{7x^2-3}}=\lim_{x\to\infty}\frac{5x^2}{\sqrt{7x^2-3}}\frac{\sqrt{7x^2-3}}{\sqrt{7x^2-3}}=\lim_{x\to\infty}\frac{5x^2\sqrt{7x^2-3}}{7x^2-3}=
\lim_{x\to\infty}\frac{5\sqrt{7x^2-3}}{7-\frac{3}{x^2}}$$
Now for $x\to\infty$ we have $\sqrt{7x^2-3}\to\infty,\frac{3}{x^2}\to{}0$ so by using the arithmetics of limits rule, we get
$$\lim_{x\to\infty}\frac{5\sqrt{7x^2-3}}{7-\frac{3}{x^2}}=\frac{5\cdot\infty}{7-0}=\infty$$
