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 July 17th, 2013, 12:52 PM #1 Senior Member   Joined: Aug 2011 Posts: 149 Thanks: 0 Why my solution is different from sample solution? Hello. When I solved problem I found that my solution is different from one in textbook. I am wondering if I am doing anything wrong or if sample solution is wrong. Task: Prove equality by using definition. $\lim\limits_{x\rightarrow -2} \dfrac{1}{(x+2)^2}=\infty$ My solution. 1)We have to show that $\forall M > 0 \;\;\;\; \exists \delta > 0 \;\;\;\; :\;\;\;\; \forall x \;\;\;\; 0 < |x+2| < \delta \Rightarrow \dfrac{1}{(x+2)^2} > M$ 2)Let $M > 0$. Implications are valid\ $\dfrac{1}{(x+2)^2} > M \Leftarrow(x+2)^2 > \dfrac{1}{M} \Leftarrow |x+2| < \dfrac{1}{\sqrt{M}}$ 3)(provided that $x \neq -2$ so $0 < |x+2|$). By taking $\delta= \dfrac{1}{\sqrt{M}}$ ,we get $0<|x+2|<\delta \Rightarrow 0<|x+2|< \dfrac{1}{\sqrt{M}} \Rightarrow \dfrac{1}{(x+2)^2} > M.$ Sample solution 1)We have to show that $\forall M > 0 \;\;\;\; \exists \delta > 0 \;\;\;\; :\;\;\;\; \forall x \;\;\;\; 0 < |x-2| < \delta \Rightarrow \dfrac{1}{(x-2)^2} > M$ 2)Let $M > 0$. Implications are valid\ $\dfrac{1}{(x-2)^2} > M \Leftarrow(x-2)^2 > \dfrac{1}{M} \Leftarrow |x-2| < \dfrac{1}{\sqrt{M}}$ 3)(provided that $x \neq 2$ so $0 < |x-2|$). By taking $\delta= \dfrac{1}{\sqrt{M}}$ ,we get $0<|x-2|<\delta \Rightarrow 0<|x-2|< \dfrac{1}{\sqrt{M}} \Rightarrow \dfrac{1}{(x-2)^2} > M.$ Only difference is that sample solition has - where I have + and + where I have -. Is there any reason why sample solution should have - instead of + or is it just they're typo?
 July 17th, 2013, 01:28 PM #2 Global Moderator   Joined: May 2007 Posts: 6,643 Thanks: 628 Re: Why my solution is different from sample solution? It looks like a typo.

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