(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Guess the limit and use the [itex]\epsilon[/itex]-[itex]\delta[/itex] definition to prove that your guess is correct.

[itex]\lim_{x \to 9}\frac{x+1}{x^2+1}[/itex]

2. The attempt at a solution

Guess limit to be [itex]\frac{10}{82}=\frac{5}{41}[/itex]

Therefore:

[itex]|\frac{x+1}{x^2+1}-\frac{5}{41}| = |\frac{(x-9)(5x+4)}{41(x^2+1)}|[/itex]

Restrict attention to [itex]|x-9|<1[/itex]

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

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

[itex]|\frac{x+1}{x^2+1}-\frac{5}{41}| < \epsilon[/itex] whenever [itex]0<|x-9|<\delta[/itex]

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

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

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# Homework Help: Epsilon-delta limit definition trouble

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