How do you evaluate the limit (3x^4-x^2+5)/(10-2x^4) as x approaches oo?

Dec 15, 2016

${\lim}_{x \rightarrow \infty} \frac{3 {x}^{4} - {x}^{2} + 5}{10 - 2 {x}^{4}} = - \frac{3}{2}$

Explanation:

Note that as $x \rightarrow \infty$ then $\frac{1}{x} , \frac{1}{x} ^ 2 , \frac{1}{x} ^ 3 \rightarrow 0$

We can therefore multiply numerator and denominator by $\frac{1}{x} ^ 4$ (the reciprocal of the largest power in the denominator) as follows:

${\lim}_{x \rightarrow \infty} \frac{3 {x}^{4} - {x}^{2} + 5}{10 - 2 {x}^{4}} = {\lim}_{x \rightarrow \infty} \frac{3 {x}^{4} - {x}^{2} + 5}{10 - 2 {x}^{4}} \cdot \frac{\frac{1}{x} ^ 4}{\frac{1}{x} ^ 4}$
$\text{ } = {\lim}_{x \rightarrow \infty} \frac{\left(\frac{1}{x} ^ 4\right) \left(3 {x}^{4} - {x}^{2} + 5\right)}{\left(\frac{1}{x} ^ 4\right) \left(10 - 2 {x}^{4}\right)}$
$\text{ } = {\lim}_{x \rightarrow \infty} \frac{3 - \frac{1}{x} ^ 2 + \frac{5}{x} ^ 4}{\frac{10}{x} ^ 4 - 2}$
$\text{ } = \frac{3 - 0 + 0}{0 - 2}$
$\text{ } = - \frac{3}{2}$

We can verify this result by looking at the graph of $y = \frac{3 {x}^{4} - {x}^{2} + 5}{10 - 2 {x}^{4}}$
graph{(3x^4-x^2+5)/(10-2x^4) [-10, 10, -5, 5]}
and indeed it does appear that for large $x$ the function is approaching a horizontal asymptote $y = - \frac{3}{2}$