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$$ \frac{-|x-a|}{a|x|} < \varepsilon $$

Then i did was assume $\delta = a/2$ and got the values where $x \in (a/2,3a/2)$. Idk what to do next and im sorry idk how to format.

Eric Towers
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Bob
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  • I think the downvotes are because the body of your question is incomprehensible. Start from the beginning and use real sentences please. – user320832 Oct 06 '16 at 02:39
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    @Bob Don't worry about the negs. It takes a while to get the hang of this place. It's not like the usual online forums. Start by writing down exactly what you need to prove according to the definition of a limit. – user4894 Oct 06 '16 at 02:39

1 Answers1

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You are on the right track. Let's assume that $x>0$ and $a>0$.

We first take $a/2<x<3a/2$, which means that $|x-a|<a/2$.

Next, let $\epsilon>0$ be given. Then, we can write

$$\begin{align} \left|\frac1x-\frac1a\right|&=\frac{|x-a|}{ax}\\\\ &<\frac{|x-a|}{a(a/2)}\\\\ &<\epsilon \end{align}$$

whenever $|x-a|<\min\left(a/2,\frac12 a^2\epsilon\right)$.

The case for $x<0$, $a<0$ follows an analogous line and is left as an exercise.

Mark Viola
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  • im confused how did u get a/2 for x? – Bob Oct 06 '16 at 02:56
  • Bob, if we fix the point $a$, then certainly $\delta$ has to less than $a$ else there would be values of $x$ such that $x$ is within $\delta$ from $a$, but $x<0$. So, we choose to confine the problem (arbitrarily) by choosing the larges $\delta $ can be to ensure $x>0$ for all $|x-a|<\delta$. So, and number less than $a$ suffices. Here, $a/2$ was selected. – Mark Viola Oct 06 '16 at 03:03