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!
[Math] Limit $\lim\limits_{x\to\infty} \frac{5x^2}{\sqrt{7x^2-3}}$
calculuslimits
Best Answer
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 $x \to \infty$.