Limits at Infinity Question 3

$ \lim\limits_{x\to\infty} \dfrac{\sqrt{9x^4-2}}{4x^2-2x+5} $

$\dfrac{3}{4} $

$\dfrac{9}{4} $

$9$

0

Approach 1

When determining limits at infinity, only the highest power terms in the numerator and denominator are relevant.

$$ \lim\limits_{x\to\infty} \dfrac{\sqrt{9x^4-2}}{4x^2-2x+5} \approx \lim\limits_{x\to\infty} \frac{\sqrt{9x^4}}{4x^2} $$ $$ = \lim\limits_{x\to\infty}\frac{3x^2}{4x^2} $$ $$ = \lim\limits_{x\to\infty}\frac{3}{4} $$ $$ = \boxed{\frac{3}{4}}$$

Approach 2

We can divide the numerator and denominator by $x^2$.

Numerator

$$ \sqrt{9x^4-2} \div x^2 $$ $$ = \sqrt{\frac{9x^4-2}{x^4}} $$ $$ = \sqrt{9-\frac{2}{x^4}} $$

Denominator

$$ 4x^2-2x+5 \div x^2 $$ $$ = \frac{4x^2-2x+5}{x^2} $$ $$ = 4-\frac{2}{x}+\frac{5}{x^2} $$

$$ \lim\limits_{x\to\infty} \dfrac{\sqrt{9-\frac{2}{x^4}}}{4-\frac{2}{x}+\frac{5}{x^2}} $$

Recall that as $x$ approaches large values, fractions in the form of $\dfrac{c}{x^r}$ approach 0.

$$ = \frac{\sqrt{9-0}}{4-0+0} $$ $$ = \boxed{\frac{3}{4}} $$