# Homework Help: Epsilon-delta limit definition trouble

1. Aug 14, 2011

### rudders93

1. The problem statement, all variables and given/known data
Guess the limit and use the $\epsilon$-$\delta$ definition to prove that your guess is correct.

$\lim_{x \to 9}\frac{x+1}{x^2+1}$

2. The attempt at a solution

Guess limit to be $\frac{10}{82}=\frac{5}{41}$

Therefore:

$|\frac{x+1}{x^2+1}-\frac{5}{41}| = |\frac{(x-9)(5x+4)}{41(x^2+1)}|$

Restrict attention to $|x-9|<1$

Therefore: $|\frac{(x-9)(5x+4)}{41(x^2+1)}|<|\frac{54|x-9|}{4141}|<\epsilon$

By taking $\delta=min(\frac{4141\epsilon}{54},1)$ we get that:

$|\frac{x+1}{x^2+1}-\frac{5}{41}| < \epsilon$ whenever $0<|x-9|<\delta$

That's my working, but I think I've made a mistake, as when I check my work by using $\epsilon=0.01$ my $\delta$ does not satisfy. This is because I get $0<|x-9|<0.76686 \Rightarrow 8.23314<x<9.76686$. But then I take say 9.76 and it doesn't hold as I get $|\frac{9.76+1}{9.76^2+1}-\frac{5}{41}| = 0.010168$ which is greater than my $\epsilon$ of 0.01.

Can anyone help me out with where I went wrong? Thanks!!

Last edited: Aug 14, 2011
2. Aug 14, 2011

### HallsofIvy

41(11)= 451, not 4141.

3. Aug 14, 2011

### rudders93

So it becomes: $|\frac{(x-9)(5x+4)}{41(x^2+1)}|<|\frac{54|x-9|}{2665}|<\epsilon$