Epsilon-delta question with limits

[danago]

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

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

That is, for each \delta &gt; 0 show that there is a value of x such that:

<br /> 0 &lt; \left| {x - 1} \right| &lt; \delta \quad \Rightarrow \quad \left| {f(x) - 2} \right| \ge 0.5<br />

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:
<br /> \begin{array}{l}<br /> \left| {f(x) - 2} \right| &lt; 0.5 \Rightarrow \left| {x - 2} \right| &lt; 0.5 \Rightarrow x \in (1.5,2.5) \\ <br /> (1.5,2.5) \cap ( - \infty ,1) = \emptyset \\ <br /> \end{array}<br />

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

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

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 &gt; 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?

Thanks in advance,
Dan.

 

[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?

 

[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?

 

[danago]

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?

No i can't say i am familiar with it :( I am 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.

 

[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?

 

[HallsofIvy]

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.

 

[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!

 

[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 won't 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.