Epsilon-delta question with limits

1. Feb 5, 2008

danago

Let $$f(x) = \left\{ {\begin{array}{*{20}c} x & {x < 1} \\ {x + 1} & {x > 1} \\ \end{array}} \right.$$, and let $$\varepsilon = 0.5$$. Show that no possible $$\delta > 0$$ satisfies the following condition:

For all x:
$$0 < \left| {x - 1} \right| < \delta \quad \Rightarrow \quad \left| {f(x) - 2} \right| < 0.5$$.

That is, for each $$\delta > 0$$ show that there is a value of x such that:

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

This will show that $$\mathop {\lim }\limits_{x \to 2} f(x) \ne 2$$

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

For x<1:
$$\begin{array}{l} \left| {f(x) - 2} \right| < 0.5 \Rightarrow \left| {x - 2} \right| < 0.5 \Rightarrow x \in (1.5,2.5) \\ (1.5,2.5) \cap ( - \infty ,1) = \emptyset \\ \end{array}$$

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

For x>1:
$$\begin{array}{l} \left| {f(x) - 2} \right| < 0.5 \Rightarrow \left| {x - 1} \right| < 0.5 \Rightarrow x \in (0.5,1.5) \\ (0.5,1.5) \cap (1,\infty ) = (1,1.5) \\ \end{array}$$

Therefore, f(x) is within 0.5 units of 2 if and only if $$x \in (1,1.5)$$. This means that the only possible value for delta is 0, and since $$\delta > 0$$, 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?

Dan.

2. Feb 6, 2008

sutupidmath

are u required to use only delta epsylon to show that the lim of this function as x-->1 does not exist, because if not there is a nicer way using sequences to show this?

3. Feb 6, 2008

sutupidmath

I can show you a general way using sequences, and also incorporating delta and epsylon in it, to show that this limit does not exist. YOu might want to try to use the Bolcano-Cauchy criterion for the existence of the limit. Are u familiar with it?

4. Feb 6, 2008

danago

No i cant say i am familiar with it :( Im only just about to begin my first year in an undergraduate course in engineering, and thought id try and get a head start in the calculus.

5. Feb 6, 2008

sutupidmath

well it basically says that a function f is said to have a limit A, as x-->a if and only if for every epsylon there exists some delta such that for any two points from its domain, that is x',x"
that satisfy the condition abs(x'-a)<delta and abs(x"-a)<delta , than the following is fullfilled:

abs( f(x')-f(x"))<epsylon.

so the idea to show that your function does not have a limit as x-->1 is to find any two points x', and x'' ( also sequences if you like to) and to show that

abs(f(x')-f(x")) will always be greater than epsylon =.5

such numbers would be x'=(1-1/n) where n is a natural number, and x"=(1+1/n)
look these two numbers are one at the right of 1 and the other at the left. Do you see how to go now?

6. Feb 6, 2008

HallsofIvy

Staff Emeritus
Obviously, if x< 1, there exist x arbitrarily close to 1, that is |x-1|< $\delta$, such that f(x)= x is close to 1 and |f(x)- 2| is close to 1 not < 0.5.

7. Feb 6, 2008

sutupidmath

yeah, x'=(1-1/n) is close to 1 and x'<1, so obviously

abs(x'-1)=abs(1/n)<delta, whatever you chose delta to be but on the contrary

abs(f(x')-2)=abs(-1) which is obviously greater than 0.5.

But you can use the bolchano-cauchy's criterion as well, as i explained above, and you will get to the same result, depends which one you like more!

8. Feb 6, 2008

sutupidmath

Because in these cases where u are given to prove that the limit of some function is not, say A, the method above will work, because it simply contradicts the deffinition of a limit given my Cauchy. But if you are asked to prove whether the limit of some function exists or not, in general withoug given to prove whether the limit of f is A, than the method above wont be of any help. SO, in those cases in order to determine whether the limit of that particular function exists or not as x-->a, you need to use the Bolcano-Cauchy's criterion for the existence of the limit, as i explained in my previous posts.