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I need to prove that $\frac{1}{x-3} \to 1$ as $x \to 4$

So I did $\left | \frac{1}{x-3} - 1 \right | = \left | \frac{4 - x}{x-3} \right | = \left | \frac{x-4}{x-3} \right | = \frac{|x-4|}{|x-3|} < K|x-4|$. So I need to pick $| x - 4| < \delta = \frac{\epsilon}{K}$

Now the problem is that I can't bound my $\frac{1}{|x-3|}$

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Let $\epsilon>0$. Suppose $|x-4|<1/2$. Then $-1/2<x-4<1/2$ which implies that $1/2<x-3<3/2$. Hence, $|x-3|\ge x-3>1/2$, that is, if $x\ne 3$ then $\frac{1}{|x-3|}<2$. Define $\delta=\min\left\{\frac{1}{2},\frac{\epsilon}{2}\right\}$. Let $0<|x-4|<\delta$. Then certainly $x\ne 3$. Hence,

$\left| \frac{1}{x-3}-1\right|=\left| \frac{x-4}{x-3}\right|<|x-4|\cdot 2 < \frac{\epsilon}{2}\cdot 2=\epsilon$.

The result follows.

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Can I do this instead? $|x - 4| < 1/2 \implies 1/2 < x - 3 < 3/2 \implies 2 > \frac{1}{x-3} > 3/2$ and take K = 2? – Hawk Dec 7 '12 at 7:21
    
Yes. Sorry for my previous answer. That was wrong. An edited solution is already reflected.:) – juniven Dec 7 '12 at 8:32

You can choose $x$ near 4: $|x-4|<1/2$ then $|x-3|>1/2$.

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This is precisely correct.:) – juniven Dec 7 '12 at 8:08
    
Sorry for my previous answer. That was wrong.:) – juniven Dec 7 '12 at 8:26

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