2

Using the $\epsilon-M $ definition of the limit, calculate $$\lim_{x\to\infty}\frac{3x^2+7}{x^2+x+8}.$$

Working so far:

$$\lim_{x\to\infty}\frac{3x^2+7}{x^2+x+8}=3$$

Given $\epsilon>0$, I want M s.t. $x>M \implies \left|\frac{3x^2+7}{x^2+x+8}-3 \right|<\epsilon$

$$\left|\frac{3x^2+7}{x^2+x+8}-3 \right|<\epsilon$$

$$\left|\frac{3x^2+7-3(x^2+x+8)}{x^2+x+8} \right|<\epsilon$$

$$\left|\frac{-3x-17}{x^2+x+8} \right|<\epsilon$$

And now I'm stuck.. Any help would be great, thanks.

Siminore
  • 35,136
user74856
  • 21

2 Answers2

3

use the fact that $ \left | \frac 1 x - 0\right| < \delta $ $$\left| \frac{-3x - 17}{x^2 + x + 8}\right| < \left| \frac{-3x - 17}{x^2 }\right| \le 3\left |\frac 1 x \right | + 17\left |\frac 1 {x^2} \right | < 20 \delta = \epsilon $$

S L
  • 11,731
  • 1
    Since the OP only writes $x \to \infty$, I don't really know if we can assume $\infty$ to be $+\infty$? What if $x \to -\infty$, in that case, I'm afraid we cannot drop the $x+8$ part like that. – user49685 Apr 28 '13 at 18:41
  • 1
    @user49685 yes ... thank you for pointing out, in that case we would use $x \to +\infty$ and change $x = -x$ and we would have $\left| \frac{3x - 17}{x^2 - x + 8}\right| < \left| \frac{3(x-1) - 14}{(x-1)^2 }\right| \le 3\left |\frac 1 {x-1} \right | + 14\left |\frac 1 {(x-1)^2} \right | < 17 \delta = \epsilon$ – S L Apr 28 '13 at 18:56
2

Let $x>\max\{\sqrt{17},\frac{3}{\epsilon-1}\}$. We have $x>0$. This gives us

$$\left|\frac{-3x-17}{x^2+x+8} \right|= \left|\frac{3x+17}{x^2+x+8} \right|=\frac{3x+17}{x^2+x+8}<\frac{3x+17}{x^2}<\frac{3}{x}+1<\epsilon$$

This is incomplete as it assumes $\epsilon\neq 1$. When $\epsilon=1$ you can solve a quadratic and find which $x$'s work.

mathemagician
  • 3,225