# Proving a limit of a function

• Apr 15th 2010, 11:40 AM
Slazenger3
Proving a limit of a function
Prove that $(x^3-8)/(x^2+x-6)$ $\rightarrow$ $\infty$ as $x\rightarrow\infty$
• Apr 15th 2010, 11:48 AM
chiph588@
Quote:

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.
• Apr 15th 2010, 11:58 AM
Drexel28
Quote:

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 ;)
• Apr 15th 2010, 12:42 PM
chiph588@
Quote:

Originally Posted by Drexel28
Except what you've shown is that our limit is less than infinity ;)

Haha! (Rofl)

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 $\lim_{x\to\infty} \frac{x^3-8}{x^2+x-6} = \lim_{t\to\infty} \left(t-4+\frac7t\right) = \infty$
• Apr 15th 2010, 04:56 PM
Laurent
Quote:

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 $+\infty$.