How to find a $\delta>0$ from information in a graph so that $0<|x-2|<\delta$ implies $|\frac1x-0.5|<0.2$?

Question

Use the graph below to find a $\delta \gt 0$ such that $$ 0 \lt \lvert x -2 \rvert \lt \delta \Rightarrow \lvert \frac {1}{x}-0.5\rvert \lt 0.2 $$

Provided answer is $\delta = \frac {4}{7}$

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What I have done:

Given $\varepsilon \gt 0, \varepsilon = 0.2$, for $0 \lt \lvert x -2 \rvert \lt \delta \Rightarrow \lvert \frac {1}{x}-0.5\rvert \lt \varepsilon $

$\lvert x-2 \rvert \\= \lvert (x)(1- \frac{2}{x}) \rvert \\= \lvert (-2x)(\frac{1}{x}-0.5) \rvert$

Since $\lvert x-2 \rvert \lt \delta$ and $\lvert \frac {1}{x}-0.5 \rvert \lt 0.2$

$0.2\lvert -2x \rvert \lt \delta$

$\lvert 2x \rvert \lt 5\delta$

$-5\delta \lt 2x \lt 5\delta$

Answer 1

The questions says "use the graph ...". We are looking for a value $\delta$ such that if $2-\delta<x<2+\delta$ then the graph of the function is between $0.3$ and $0.7.$ It is clear that if $$\dfrac{1}{0.7}<x<\dfrac{1}{0.3}$$ then the graph satisfies the conditions. Now $\delta$ is given by the minimum distance from $2$ to $\frac{10}{7}$ and $\frac{10}{3}.$ Since

$$\left|2-\dfrac{10}{7}\right|=\dfrac 47<\dfrac 43=\left|2-\dfrac{10}{3}\right|$$

we get that $\delta =\dfrac 47.$

Answer 2

$$|\frac {1}{x}-0,5|=|\frac {x-2}{\color {green}{2}x}|$$

as we are near $2$, we can assume that

$|x-2| <\color {red}{1}$ or $1<x <3$.

thus $$|\frac {1}{x}-0,5|<\frac {|x-2|}{\color {green}{2}} $$

hence, to get $$|\frac{1}{x}-0,5|<\epsilon$$

we just need to have

$|x-2|<\color {red}{1}$ and $|x-2|<\color {green}{2}\epsilon $.

So we will take $$\delta=\min (\color {red}{1},\color {green}{2}\epsilon) $$

Answer 3

We will prove the stronger statement

For every $\epsilon > 0$, there exists $\delta > 0$ such that, if $|x-2| < \delta$, then $\left| \dfrac 1x - \dfrac 12 \right| < \epsilon$

Let $\delta = \min\{ 1, \epsilon\}$.

Then $0 < \delta < 1$. It follows that $2-1 < 2 - \delta$ and $2 + \delta < 2+1$.

\begin{align} |x-2| < \delta &\implies 2-\delta < x < 2 + \delta \\ &\implies 1 < x < 3 \\ &\implies \frac 13 < \frac 1x < 1 \\ &\implies \frac 16 < \frac{1}{2x} < \frac 12 \end{align} It follow that

\begin{align} \left| \dfrac 1x - \dfrac 12 \right| &= \left| \dfrac 12 - \dfrac 1x \right| \\ &= \dfrac{1}{2x}|x - 2| \\ &< \dfrac 12|x - 2| \\ &< \dfrac 12 \delta \\ &< \dfrac 12 \epsilon \\ &< \epsilon \end{align}

Answer 4

Such exercise suggests students to think about the geometric meaning of the $\epsilon$-$\delta$ proof instead of blind algebraic manipulations.

Find a $\delta \gt 0$ such that $0 \lt \lvert x -2 \rvert \lt \delta \Rightarrow \lvert \frac {1}{x}-0.5\rvert \lt 0.2 $

Note that $0<|x-2|<\delta$ means $$ x\in(2-\delta,2+\delta)\quad\hbox{and }\ \ \ x\neq 2, $$ which means that $x$ falls in the interval $(2-\delta,2+\delta)$ (except the center of the interval: $2$).

On the other hand, $\lvert \frac {1}{x}-0.5\rvert \lt 0.2$ means $$ \frac1x\in(0.5-0.2,0.5+0.2)=(0.3,0.7). $$ (Can you see that how this coincides with the interval on the $y$-axis of the graph?)

Now look at the graph, which suggests (make sure that you understand this!) that if $x\in(\frac{1}{0.7},\frac{1}{0.3})$, then $$ \frac{1}{x}\in(0.3,0.7)\tag{*} $$ But wait, we are looking for $(2-\delta,2+\delta)$ so that if $x\in (2-\delta,2+\delta)$ then (*) is true. ($x\neq 2$ is really irrelevant.) Now observe that if we can make $$ (2-\delta,2+\delta)\subset (\frac{1}{0.7},\frac{1}{0.3}) $$ then we are done. Can you see how to go on?