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I would like to know how to prove this limit via the epsilon delta definition:

$$\lim\limits_{x\to3}\frac{x+1}{(x-3)^2}=\infty$$

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    $\begingroup$ What have you tried? $\endgroup$ – Don Thousand Mar 30 at 21:30
  • $\begingroup$ I wrote out the definition: $\forall M>0 \exists \delta>0: 0<|x-3|<\delta \rightarrow \frac{x+1}{(x-3)^{2}}>M$ And tried solving the inequality for x. I arrived at this conclusion: $x=-\frac{-6*M+\sqrt{16*M+1}-1}{2*M}$ but I'm not sure where to go from there. $\endgroup$ – user760529 Mar 30 at 21:32
  • $\begingroup$ Well if $|x-3| < \delta$ then $\frac 1{|x-3|} > \frac 1{\delta}$ and $(\frac 1{|x-3|})^k > (\frac 1{\delta})^k$..... Just sayin'. $\endgroup$ – fleablood Mar 30 at 21:36
  • $\begingroup$ I realize that my goal is to find a delta such that the inequality is true. At the same time I need to find it such that the delta is as small as possible. I am however stuck. $\endgroup$ – user760529 Mar 30 at 21:37
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    $\begingroup$ it does, thank you @DonThousand $\endgroup$ – user760529 Mar 30 at 21:48
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$\forall M >0, \exists \delta > 0 : |x-3|<\delta \implies |\frac {x+1}{(x-3)^2}|> M$

Lets force $\delta < 1$ which will force $x+1>2$

$|\frac {x+1}{(x-3)^2}| > \frac {2}{\delta^2} > M$

When $\delta< \min (\sqrt {\frac {2}{M}}, 1),\frac {x+1}{(x-3)^2}> M$

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