# How do you find the limit (t+5-2/t-1/t^3)/(3t+12-1/t^2) as x->oo?

Nov 4, 2016

${\lim}_{x \rightarrow \infty} \frac{x + 5 - \frac{2}{x} - \frac{1}{x} ^ 3}{3 x + 12 - \frac{1}{x} ^ 2} = \frac{1}{3}$

#### Explanation:

I assume that you mean ${\lim}_{x \rightarrow \infty} \frac{x + 5 - \frac{2}{x} - \frac{1}{x} ^ 3}{3 x + 12 - \frac{1}{x} ^ 2}$ and that the expression should not contain the variable $t$!!

Now, As $x \to \infty$ then $\frac{1}{x} \to 0$

So, it would be better if we could replace $x$ with $\frac{1}{x}$

${\lim}_{x \rightarrow \infty} \frac{x + 5 - \frac{2}{x} - \frac{1}{x} ^ 3}{3 x + 12 - \frac{1}{x} ^ 2} = {\lim}_{x \rightarrow \infty} \frac{\frac{1}{x} \left(x + 5 - \frac{2}{x} - \frac{1}{x} ^ 3\right)}{\frac{1}{x} \left(3 x + 12 - \frac{1}{x} ^ 2\right)}$

$\therefore {\lim}_{x \rightarrow \infty} \frac{x + 5 - \frac{2}{x} - \frac{1}{x} ^ 3}{3 x + 12 - \frac{1}{x} ^ 2} = {\lim}_{x \rightarrow \infty} \frac{1 + \frac{5}{x} - \frac{2}{x} ^ 2 - \frac{1}{x} ^ 4}{3 + \frac{12}{x} - \frac{1}{x} ^ 3}$

$\therefore {\lim}_{x \rightarrow \infty} \frac{x + 5 - \frac{2}{x} - \frac{1}{x} ^ 3}{3 x + 12 - \frac{1}{x} ^ 2} = \frac{1 + 0 - 0 - 0}{3 + 0 - 0} = \frac{1}{3}$