Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
$\sqrt{7x^2-3}\sim\sqrt{7x^2}=\sqrt7x$ –  oldrinb Sep 8 '13 at 22:22

6 Answers 6

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 $x \to \infty$.

share|improve this answer
Clearly shown steps +1 –  Amzoti Sep 9 '13 at 0:29

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.

share|improve this answer

Hint: try to show that $3x\ge \sqrt{7x^2-3}\ge x$ for $x$ sufficiently large.

share|improve this answer

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 $$\sqrt{7x^2-3}=x\sqrt{7-\frac3{x^2}}.$$

share|improve this answer

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 $$\lim_{x\to\infty}\frac{5\sqrt{7x^2-3}}{7-\frac{3}{x^2}}=\frac{5\cdot\infty}{7-0}=\infty$$

share|improve this answer

Answer is infinity

Here is another step by step solution: http://www.symbolab.com/solutions/limits?query=%5Clim_%7Bx%5Cto%5Cinfty%20%7D%5Cleft(%5Cfrac%7B5x%5E%7B2%7D%7D%7B%5Csqrt%7B7x%5E%7B2%7D-3%7D%7D%5Cright)

share|improve this answer
Answers on math.SE should be as self-contained as possible. This answer consists of little more than a link to an external site, meaning that if that site were to ever disappear (or even just change its link URLs) your answer would be entirely meaningless. Please STOP posting link-only answers, and begin including more of the pertinent information in your posts (with appropriately cited sources, of course). –  Arthur Fischer Sep 9 '13 at 18:54

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.