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I have an exercise, it says proof that the limit at $\dfrac{1}{2}$ of the function $f(x) = \dfrac{1}{(2x-1)}$ doesn't exist.

1.- For any $L$ number, proof that a $K$ number exists so. $|L|+1=\dfrac{1}{2K-1}$

Here, what I did was find $K$, from that expression.

What I got was $K= \dfrac{2}{2|L|+1}$

But I'm not sure if that means the $K$ number I found satisfies what the are asking for. if not, then how can i found the value of $K$?

Then, it continues.

2.- Proof that $x\leq K$ implicates $|f(x)| \geq |L|+1$

Here, from the definition I suposed,that it must be $|f(x)| \ge \epsilon$ And if $\epsilon$ is equal to $1$ then $|f(x)| \ge |L| +1$ by canceling the absolute value from $|f(x)-L|\lt \epsilon$

They are similar but, how can i define the value of $x$?

3.- Show an $=x$ number so that $|x-\frac{1}{2}| \lt \delta$ and $|f(x)-L|\ge 1$

Actually I didn't knew what to do here.

4.- Finish the proof by choosing $\epsilon = 1$ and conclude the limit of the function doesn't exists.

When I was trying to solve it, I enclose $\delta$, so I could find the min of $\delta$. When I gave $\delta$ a value $1$, then by enclosing $|x-\frac{1}{2}| \lt \delta$, the result was $\delta = \min (1, -\frac{2\epsilon}{1})$ and that doesn't have sense. I thought that, the result of that was another way to proof the limit doesn't exist.

If someone can help me to solve my dudes, i would really appreciate it! Thanks

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2 Answers 2

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To show that $\lim_{x \to a}f(x)$ does not exist it is necessary to use the negation of definition of a limit and this means that we need to ensure that the following statement is true:

For any given number $L$ there is an $\epsilon > 0$ such that the implication $0 < |x - a| < \delta \Rightarrow |f(x) - L| < \epsilon$ is false for any value of $\delta > 0$.

Here the challenge to find an $\epsilon$ based on $L$ such that the desired implication is false no matter what value of $\delta > 0$ is chosen.

Here the function $f(x) = 1/(2x - 1)$ and $a = 1/2$ and your textbook desires to falsify the implication by trying to ensure that $|f(x) - L| \geq \epsilon$ for some value of $\epsilon > 0$. In order to do so it uses the inequality $|f(x) - L| \geq |f(x)| - |L|$ and thus we need to ensure that $|f(x)| \geq |L| + \epsilon$. Note that the textbook has chosen $\epsilon = 1$ so that $|f(x)| \geq |L| + 1$. This means that $|1/(2x - 1)| \geq |L| + 1$ or $|x - 1/2| \leq 1/2(|L| + 1)$. Now whatever $\delta > 0$ is chosen the inequality $0 < |x - 1/2| < \delta$ will also include values of $x$ such that $|x - 1/2| \leq 1/2(|L| + 1)$. And this will ensure that $|f(x) - L| \geq 1 = \epsilon$. Note that for this simple problem any value of $\epsilon > 0$ will work.

You should try to use similar technique to show that $\lim_{x \to 0}\sin(1/x)$ does not exist and here you will need to carefully choose value of $\epsilon$ based on $L$.

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If $x = \frac12(1+y)$, then $\dfrac1{2x-1} =\dfrac1{2(\frac12(1+y))-1} =\dfrac1{y} $, so $\lim_{x \to \frac12} \dfrac1{2x-1} =\lim_{y \to 0} \dfrac1{y} $.

See if you can prove that this simpler limit does not exist.

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