# Math Help - Proving a limit of a function

1. ## Proving a limit of a function

Prove that $(x^3-8)/(x^2+x-6)$ $\rightarrow$ $\infty$ as $x\rightarrow\infty$

2. Originally Posted by Slazenger3
Prove that $(x^3-8)/(x^2+x-6)$ $\rightarrow$ $\infty$ as $x\rightarrow\infty$
For large enough $x,\; x^2+x-6 > x^2 \implies \frac{x^3-8}{x^2+x-6} < \frac{x^3-8}{x^2}=x-\frac8x$.

Now apply the limit comparison test.

3. Originally Posted by chiph588@
For large enough $x,\; x^2+x-6 > x^2 \implies \frac{x^3-8}{x^2+x-6} < \frac{x^3-8}{x^2}=x-\frac8x$.

Now apply the limit comparison test.
Except what you've shown is that our limit is less than infinity

4. Originally Posted by Drexel28
Except what you've shown is that our limit is less than infinity
Haha!

Let's do this again:

$\frac{x^3-8}{x^2+x-6} = \frac{x^2+2x+4}{x+3}$

Let $x=t-3$:

$\frac{(t-3)^2+2(t-3)+4}{t} = \frac{t^2-4t+7}{t} = t-4+\frac7t$

So $\lim_{x\to\infty} \frac{x^3-8}{x^2+x-6} = \lim_{t\to\infty} \left(t-4+\frac7t\right) = \infty$

5. Originally Posted by Slazenger3
Prove that $(x^3-8)/(x^2+x-6)$ $\rightarrow$ $\infty$ as $x\rightarrow\infty$
A common proof (valid for any coefficients) would go along these lines: dividing numerator and denominator by $x^2$, we have $\frac{x^3-8}{x^2+x-6}= \frac{x-\frac{8}{x^2}}{1+\frac{1}{x}-\frac{6}{x^2}}$. Under this form, the limits of the terms are immediate to see; in particular, the numerator goes to $+\infty$ and the denominator converges to 1, hence the ratio goes to $+\infty$.