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A while ago I posted another one like this with a incorrect approach, please see this one!

Is this an accurate proof for limits for the function $\frac{1}{x}$

$\displaystyle \lim_{x\to1} \frac{1}{x} = \frac{1}{1} = 1$

Using $\epsilon-\delta$ So,

$\displaystyle \frac{|x-1|}{|x|} < \epsilon$ for some $\displaystyle |x-1| < \delta$

Lets assume $|x - 1| < \frac{1}{3}$

$\displaystyle \frac{2}{3} < |x| < \frac{4}{3}$

$= \frac{3}{2} > \frac{1}{|x|} > \frac{3}{4}$

So we have,

$|x - 1| < \delta$

$\frac{1}{x} < \frac{4}{3}$

Therefore we get: $\frac{|x-1|}{|x|} < \frac{3\delta}{4}$

It is the equation: $\frac{3\delta}{4} = \epsilon$, which is possible only if, $\delta_1 = \frac{4\epsilon}{3}$

Therefore, $\delta = \min(\frac{1}{3},\frac{4\epsilon}{3})$

gebruiker
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Amad27
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1 Answers1

1

I think you have the right idea, but it is a bit unorganized, separate the scratch work from the proof. This is how I would approach it.

Let $\epsilon>0$ be arbitrary, let $\delta=\min(1/2,\epsilon/2)$ such that $\vert x−1\vert<\delta$, then $$\left\vert\frac{1}{x}−1\right\vert= \frac{\vert x−1\vert}{\vert x\vert}< 2\vert x−1\vert <2\epsilon/2=\epsilon.$$

gebruiker
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mrdifficult
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