"Find the limits: limx→∞8x2−32x2+x" was answered by our community

alesterp

Answered question

2021-11-10

Find the limits:

lim_{x \to \infty} \sqrt{\frac{8 x^{2} - 3}{2 x^{2} + x}}

FieniChoonin

Skilled2021-11-11Added 102 answers

Given:
Consider rational function inside the radical. The highest power of x in the denominator is 2.
So divide numerator and denominator by x^2,

lim_{x \to \infty} \sqrt{\frac{8 x^{2} - 3}{2 x^{2} + x}} = lim_{x \to \infty} \sqrt{\frac{\frac{8 x^{2}}{x^{2}} - \frac{3}{x^{2}}}{\frac{2 x^{2}}{x^{2}} + \frac{x}{x^{2}}}}

= lim_{x \to \infty} \sqrt{\frac{8 - \frac{3}{x^{2}}}{2 + \frac{1}{x}}}

Use the limit laws,

lim_{x \to \infty} \sqrt{\frac{8 x^{2} - 3}{2 x^{2} + x}} = \sqrt{lim_{x \to \infty} \frac{(8 - \frac{3}{x^{2}})}{(2 + \frac{1}{x})}}

= \sqrt{\frac{lim_{x \to \infty} (8 - \frac{3}{x^{2}})}{lim_{x \to \infty} (2 + \frac{1}{x})}}

= \sqrt{\frac{8 - 0}{2 + 0}}

= \sqrt{4}

= 2

Therefore,

lim_{x \to \infty} \sqrt{\frac{8 x^{2} - 3}{2 x^{2} + x}} = 2

Jeffrey Jordon

Expert2022-06-26Added 2605 answers

Answer is given below (on video)

Jeffrey Jordon

Expert2022-08-30Added 2605 answers

Answer is given below (on video)

Do you have a similar question?

Recalculate according to your conditions!

Didn't find what you were looking for?