- #1

danago

Gold Member

- 1,123

- 4

**Let [tex]**

f(x) = \left\{ {\begin{array}{*{20}c}

x & {x < 1} \\

{x + 1} & {x > 1} \\

\end{array}} \right.

[/tex], and let [tex]\varepsilon = 0.5[/tex]. Show that no possible [tex]\delta > 0[/tex] satisfies the following condition:

For all x:

[tex]0 < \left| {x - 1} \right| < \delta \quad \Rightarrow \quad \left| {f(x) - 2} \right| < 0.5[/tex].

That is, for each [tex]\delta > 0[/tex] show that there is a value of x such that:

[tex]

0 < \left| {x - 1} \right| < \delta \quad \Rightarrow \quad \left| {f(x) - 2} \right| \ge 0.5

[/tex]

This will show that [tex]\mathop {\lim }\limits_{x \to 2} f(x) \ne 2[/tex]

f(x) = \left\{ {\begin{array}{*{20}c}

x & {x < 1} \\

{x + 1} & {x > 1} \\

\end{array}} \right.

[/tex], and let [tex]\varepsilon = 0.5[/tex]. Show that no possible [tex]\delta > 0[/tex] satisfies the following condition:

For all x:

[tex]0 < \left| {x - 1} \right| < \delta \quad \Rightarrow \quad \left| {f(x) - 2} \right| < 0.5[/tex].

That is, for each [tex]\delta > 0[/tex] show that there is a value of x such that:

[tex]

0 < \left| {x - 1} \right| < \delta \quad \Rightarrow \quad \left| {f(x) - 2} \right| \ge 0.5

[/tex]

This will show that [tex]\mathop {\lim }\limits_{x \to 2} f(x) \ne 2[/tex]

I started by finding the set of x values for which the function produces values between 1.5 and 2.5.

For x<1:

[tex]

\begin{array}{l}

\left| {f(x) - 2} \right| < 0.5 \Rightarrow \left| {x - 2} \right| < 0.5 \Rightarrow x \in (1.5,2.5) \\

(1.5,2.5) \cap ( - \infty ,1) = \emptyset \\

\end{array}

[/tex]

Therefore there are no values of x<1 for which f(x) is within 0.5 of 2.

For x>1:

[tex]

\begin{array}{l}

\left| {f(x) - 2} \right| < 0.5 \Rightarrow \left| {x - 1} \right| < 0.5 \Rightarrow x \in (0.5,1.5) \\

(0.5,1.5) \cap (1,\infty ) = (1,1.5) \\

\end{array}

[/tex]

Therefore, f(x) is within 0.5 units of 2 if and only if [tex]x \in (1,1.5)[/tex]. This means that the only possible value for delta is 0, and since [tex]\delta > 0[/tex], this value of 0 is not possible, thus making no possible value for delta.

Is that how i should go about proving such a limit does not exist? Are there flaws in what i have done?

Thanks in advance,

Dan.