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I hit a block when discovering a negative $\delta$. This is how:

I need to show that$$\forall \epsilon>0 \; \exists \delta>0 \text{ s.t. } \mid x-1 \mid < \delta \Rightarrow \Bigl| \frac{2-x}{4-x}-\frac{1}{3} \Bigr| < \epsilon$$

To find such a $\delta$: $$\left| \frac{3(2-x)-(4-x)}{3(4-x)} \right|=\left|\frac{2-2x}{3(4-x)} \right|=\left| \frac{2(x-1)}{3(4-x)} \right|.$$ Since $|x-1|<\delta$, consider $\frac{2}{3}\Bigl|\frac{1}{4-x} \Bigr|$ further for $\delta=1$, $$-1<x-1<1,\ -4<x-4<-2 \Rightarrow \frac{1}{x-4}<-\frac{1}{4}.$$

How can I get an appropriate positive $\delta$?

Ѕᴀᴀᴅ
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alortimor
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1 Answers1

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If $\delta < 1$ then $|x-1| < \delta$ implies $$x-1 \in [-1,1] \implies x-4 \in [-4,-2] \implies |x-4| \in [2,4] \implies \frac1{|x-4|} \in \left[\frac14,\frac12\right]$$ so $\frac1{|x-4|} \le \frac12$.

Therefore we have

$$\left|\frac{2(x-1)}{3(4-x)}\right| = \frac23 \cdot \frac1{|x-4|} \cdot |x-1| < \frac23 \cdot \frac12 \cdot \delta = \frac{\delta}3 < \delta.$$

We want this to be $\le \varepsilon$ so we can take $\delta = \min\{\varepsilon, 1\}$.

mechanodroid
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