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prove $\lim\limits_{x\to\frac{5}{2}}\frac{1}{4x-8}=0.5$ via delta epsilon I have the following

$|\frac{1}{4x-8}-\frac{1}{2}|<\epsilon$

$|\frac{5-2x}{4x-8}|<\epsilon$ which I got after simplification

But I want to have $|x-\frac{5}{2}|$ somewhere but I can't seem to be able to get that

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1 Answer 1

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Guide:

Dividing the numerator and denominator by $2$ and since $|a|=|-a|$, we have \begin{align} \left|\frac{5-2x}{4x-8}\right| = \left|\frac{x-\frac52}{2x-4}\right| \end{align}

Choose a small $\delta$ to control the magnitude of $\frac1{|2x-4|}$ and you should be able to prove it.

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