# Find the limit?

Jan 22, 2018

${\lim}_{x \to \infty} \frac{\frac{1}{x} - 10 x}{2 x + \frac{1}{3 x}} = - 5$

#### Explanation:

A helpful procedure would be to divide by the highest denominator power.

But First, let's simplify the function given

$- - - - - - - - - - - - - - - - - - - -$
$\frac{\frac{1}{x} - 10 x}{2 x + \frac{1}{3 x}} \implies \frac{\frac{1 - 10 {x}^{2}}{x}}{\frac{6 {x}^{2} + 1}{3 x}} \implies \frac{1 - 10 {x}^{2}}{x} \cdot \frac{3 x}{6 {x}^{2} + 1} = \frac{3 \cancel{x} \left(1 - 10 {x}^{2}\right)}{\cancel{x} \left(6 {x}^{2} + 1\right)} = \frac{3 - 30 {x}^{2}}{6 {x}^{2} + 1}$
$- - - - - - - - - - - - - - - - - - - -$

So now we have ${\lim}_{x \to \infty} \frac{3 - 30 {x}^{2}}{6 {x}^{2} + 1}$

${x}^{2}$ is the highest denominator power so we can divide ${\lim}_{x \to \infty} \frac{3 - 30 {x}^{2}}{6 {x}^{2} + 1}$ by ${x}^{2}$ or multiply it by $\frac{1}{x} ^ 2$. It's the same thing.

${\lim}_{x \to \infty} \frac{3 - 30 {x}^{2}}{6 {x}^{2} + 1} \cdot \frac{\frac{1}{x} ^ 2}{\frac{1}{x} ^ 2} \implies {\lim}_{x \to} \frac{\frac{3}{x} ^ 2 - \frac{30 {x}^{2}}{x} ^ 2}{\frac{6 {x}^{2}}{x} ^ 2 + \frac{1}{{x}^{2}}}$

=>lim_(x->oo)(3/x^2-30)/(6+1/(x^2)

Plugging in $\infty$

$= \frac{\frac{3}{\textcolor{red}{\infty}} ^ 2 - 30}{6 + \frac{1}{{\textcolor{red}{\infty}}^{2}}}$

$= \frac{0 - 30}{6 + 0}$

$= - \frac{30}{6}$

$= - 5$

$\therefore {\lim}_{x \to \infty} \frac{\frac{1}{x} - 10 x}{2 x + \frac{1}{3 x}} = - 5$