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!
 A: You can divide numerator and denominator by $x$ to find that the limit $\to + \infty$.
$$ \large \frac{5x^2}{\sqrt{7x^2-3}}\cdot \frac{\frac {1}{x}}{\frac{1}{\sqrt{x^2}}} =  \frac {\frac {5x^2}{x}}{\sqrt {\frac {7x^2}{x^2} - \frac 3{x^2}}} = \frac {5x}{\sqrt{7 - \frac 3{x^2}}} = \frac {5x}{7}$$
This gives us that $$\lim_{x\to \infty} \frac{5x^2}{\sqrt{7x^2-3}} =\lim_{x \to \infty}\frac {5x}{7}$$
and clearly, $\dfrac {5x}{7} \to +\infty$ as $x \to \infty$.
A: Informally, the denominator looks like $\sqrt{x^2} = |x|$, while the numerator is $x^2$; hence, the limit is infinity.
More formally, we know that $7x^2 - 3 < 9x^2$ for all $x$. Hence,
$$\frac{1}{7x^2 - 3} > \frac{1}{9x^2}$$
for sufficiently large $x$ and so 
$$\frac{5x^2}{\sqrt{7x^2 - 3}} > \frac{5x^2}{\sqrt{9x^2}} = \frac{5x}{3}$$
for all $x$ large. Hence, the relevant quantity can be bounded below by something tending to infinity.
A: Hint: try to show that $3x\ge \sqrt{7x^2-3}\ge x$ for $x$ sufficiently large.
A: Hint: Observe that for $|x|\ge\sqrt{3/7},$ we have $$\sqrt{7x^2-3}=\sqrt{x^2}\sqrt{7-\frac3{x^2}}=|x|\sqrt{7-\frac3{x^2}},$$ so for $x\ge\sqrt{3/7},$ we have $$\sqrt{7x^2-3}=x\sqrt{7-\frac3{x^2}}.$$
A: 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
$$\lim_{x\to\infty}\frac{5\sqrt{7x^2-3}}{7-\frac{3}{x^2}}=\frac{5\cdot\infty}{7-0}=\infty$$
