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$$\lim_{x\to 1} \frac{100}{x} = 100$$

I'm trying to proving the limit using epsilon-delta definition and here's what I've come up with so

My attempt:

$\forall ε>0, \exists δ>0$, such that if for x, $0<|x-1|<δ$ then $|\frac{100}{x} - 100| < ε$

$$100*|\frac{1}{x} - 1| < ε$$ $$100*\frac{1}{|x|} * |x-1| < ε$$

δ : $$100*\frac{1}{|x|} * |x-1| < ε$$ $$|x-1| < δ$$, choose δ = 1/2

then,

$$|x-1| < \frac{1}{2} \Rightarrow 0 < x < \frac{3}{2}$$

From here I'm not too sure how to proceed on. I was wondering if anyone can help me on what to do next.

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  • $\begingroup$ Your $\delta$ should be a function of $\epsilon$ $\endgroup$ – user76568 Mar 2 '14 at 21:52
  • $\begingroup$ You made a mistake on the last line, $|x-1| < { 1\over 2}$ gives $x > { 1 \over 2}$, which you can use to get a $\delta$ since ${1\over x} < 2$. $\endgroup$ – copper.hat Mar 2 '14 at 21:56
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From $|x-1|<\dfrac12$, \begin{align} |1|-|x|\leq |1-x|<\dfrac12\tag{triangle inequality}\\ \implies \dfrac12<|x|\end{align}

So, $100\cdot\dfrac{|x-1|}{|x|}<100\cdot2\cdot|x-1|$.

Thus you may use $\delta=\min(\dfrac12,\dfrac{\epsilon}{200})$.

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  • $\begingroup$ Thank you Alraxite this was very helpful! $\endgroup$ – Sc4r Mar 2 '14 at 22:10
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You want $$\left|\frac{100}{x} - 100\right| < \varepsilon$$ whenever $\left|x-1\right| < \delta$.

Your task is to find a $\delta$ that makes this true for a given $\varepsilon.$

Let $\varepsilon > 0$, we have \begin{align} \left|\frac{100}{x} - 100\right| &= 100\cdot\left|\frac{1}{x} - 1\right| = 100\cdot\left|\frac{1-x}{x}\right| = 100\cdot\frac{\left|1-x\right|}{\left|x\right|}. \end{align} If we choose $\delta\le\frac 1 2$ we have $\left|x-1\right|<\frac 1 2$ which says $$-\frac 1 2 < x - 1 < \frac 1 2$$ or equivalently $$ \frac 1 2 < x < \frac 3 2.$$ In particular $\left|x\right|=x > \frac 1 2$, so we continue with \begin{align} \left|\frac{100}{x} - 100\right| &= 100\cdot\frac{\left|1-x\right|}{\left|x\right|} = 100\cdot\frac{\left|1-x\right|}{\left|x\right|} \\&< 100\cdot\frac{\left|1-x\right|}{1/2} = 200 \cdot \left|1-x\right| < 200\cdot \delta. \end{align} For this to be less than or equal to $\varepsilon$ we need $200\cdot\delta\le\varepsilon$, which is equivalent to $\delta \le \frac{\varepsilon}{200}$.

So if we have both $\delta\le\frac 1 2$ and $\delta<\frac{\varepsilon}{200}$, which can be achieved by picking $$ \delta = \min \left\{\frac 1 2, \frac{\varepsilon}{200}\right\},$$ we get \begin{align} \left|\frac{100}{x} - 100\right| &= 100\cdot\frac{\left|1-x\right|}{\left|x\right|} < 200\cdot \delta \le 200\cdot \frac{\varepsilon}{200} = \varepsilon. \end{align}


You get this when you look at $$\frac{\left|1-x\right|}{\left|x\right|},$$ and realize that to get this fraction small you need $\left|1-x\right|$ to be small, which can be achievied directly by choosing $\delta$ small, but you also want $\left|x\right|$ to don't be small, since a small denominator gives a large fraction. So you need a way to make $\left|x\right|>c>0$ by making $\left|x-1\right|$ small. I chose to make $\left|x-1\right|<\frac 1 2$ which gave us $\left|x\right|>\frac 1 2$ so we can finish the proof.

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