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

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- Apr 15th 2010, 11:40 AMSlazenger3Proving a limit of a function
Prove that $\displaystyle (x^3-8)/(x^2+x-6)$$\displaystyle \rightarrow$$\displaystyle \infty$ as $\displaystyle x\rightarrow\infty$

- Apr 15th 2010, 11:48 AMchiph588@
- Apr 15th 2010, 11:58 AMDrexel28
- Apr 15th 2010, 12:42 PMchiph588@
Haha! (Rofl)

Let's do this again:

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

Let $\displaystyle x=t-3 $:

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

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