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i am asked to prove with $\epsilon$-$\delta$-argument that $x\rightarrow |-2x+3|$ is continuous

my steps: Definition of $\epsilon-\delta$-argument:
$\forall \epsilon >0 \exists \delta>0$ with $|x-x_0|<\delta \Longrightarrow |f(x)-f(x_0)|<\epsilon$

so: $|f(x)-f(x_0)|=||-2x+3|-|-2x_0+3|| = |(2x+3)-(2x_0+3)| = |2x-2x_0|=\ldots\text{help}\ldots<\epsilon$

i am stuck there again

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The step $$\left|\left|-2x+3\right|-\left|-2x_0+3\right|\right| = \left|(2x+3)-(2x_0+3)\right| $$ is wrong (how did you get that?). We have $$\left|f(x)-f(x_0)\right|=\left|\left|-2x+3\right|-\left|-2x_0+3\right|\right| \le \left|(-2x+3)-(-2x_0+3)\right| =2\left|x-x_0\right| $$ by a variant of the triangle inequality $$\left|\left|a\right|-\left|b\right|\right| \le \left|a-b\right| $$ We must have $$\left|f(x)-f(x_0)\right|<\epsilon$$ Because $$\left|f(x)-f(x_0)\right|<2\left|x-x_0\right|$$ it suffices $$\left|x-x_0\right|<\frac{\epsilon}{2}$$ whenever $$\left|x-x_0\right|<\delta$$ How will you choose your $\delta_{\epsilon}$?

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  • $\begingroup$ @doniyor You mean $\delta=\frac{\epsilon}2$ right? $\endgroup$ – Nameless Dec 23 '12 at 11:14
  • $\begingroup$ Yes so that the last two equations are the same $\endgroup$ – Nameless Dec 23 '12 at 11:16
  • $\begingroup$ so, generally, $\delta$ and $\epsilon$ should stay equal? visually, they are the small areas in $x$ and $y$ planes, right? $\endgroup$ – doniyor Dec 23 '12 at 11:18
  • $\begingroup$ @doniyor No. Here $\delta=\frac{\epsilon}2<\epsilon$ $\endgroup$ – Nameless Dec 23 '12 at 11:22

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