Epsilon-delta limit definition trouble

[rudders93]

Homework Statement

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|<\deltaThat'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!

 

[HallsofIvy]

41(11)= 451, not 4141.

 

[rudders93]

Hi, thanks for reply!

Sorry, but where is the calculation 41 * 11? I can only think of 41*(10^2+1) :(

 

[rudders93]

Found my error. Was I subbed in the largest value of the restricted x range in the denominator, which makes the expression smaller, instead of subbing in the smallest value of the restricted x range (x = 8) into the denominator

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