Question

How do you prove that the limit of $\dfrac{1}{x-3}$ does not exist as $x$ approaches $3$ using the epsilon delta proof?

Answer 1

Problems on proving limits can be easily done by using the epsilon delta proof. To use the epsilon delta proof to show that limit of $\dfrac{1}{x-3}$ does not exist, it will be sufficient to prove that limit of $\dfrac{1}{x-3}\ne L$ . Here, $L$ belongs to any number from the set of real numbers.

Complete step by step solution:
According to the epsilon delta proof, $\displaystyle \lim_{x \to 3}\dfrac{1}{x-3}\ne L$ means that there is an $\varepsilon > 0$ , such that for any $\delta > 0$ , there is an $0 < |x-3| < \delta ~$ so that $\left| \dfrac{1}{x-3}-L \right|\ge \varepsilon$ .
To use the epsilon delta proof to show that limit of $\dfrac{1}{x-3}$ does not exist, it will be sufficient to prove that limit of $\dfrac{1}{x-3}\ne L$ , for any real number $L$ .
Let $\varepsilon =1$ (arbitrarily)
We first, consider $L\ge 0$
We can see that when $x=3+\dfrac{1}{1+L} > 3$
$\left| \dfrac{1}{x-3}-L \right|=1=\varepsilon$
Now, if we see that $x=3+\dfrac{1}{1+L}$ is not in the interval $0 < |x-3| < \delta ~$ , then $\delta \le \dfrac{1}{1+L}$ .
Hence, we can write that
$\dfrac{1}{\left( 3+\dfrac{\delta }{2} \right)-3}-L > \dfrac{1}{\left( 3+\delta \right)-3}-L > \dfrac{1}{\left( 3+\dfrac{1}{1+L} \right)-3}-L=1=\varepsilon$
Also, $x=3+\dfrac{\delta }{2}$ is guaranteed to be in the interval $0 < |x-3| < \delta ~$ .
Thus, let $x=\min \left( 3+\dfrac{\delta }{2},3+\dfrac{1}{1+L} \right)$
This ensures that $0 < |x-3| < \delta ~$ and $\left| \dfrac{1}{x-3}-L \right|\ge \varepsilon$ .
Therefore, $\displaystyle \lim_{x \to 3}\dfrac{1}{x-3}\ne L$
Now, we again consider $L < 0$
Similarly, we see that $x=3+\dfrac{1}{1+L} < 3$
$\left| \dfrac{1}{x-3}-L \right|=\left| -1 \right|=1=\varepsilon$
Now, if $x=3-\dfrac{1}{1+L}$ is not in the interval $0 < |x-3| < \delta ~$ , then $\delta \le \dfrac{1}{1-L}$
This means that
$\dfrac{1}{\left( 3-\dfrac{\delta }{2} \right)-3}-L < \dfrac{1}{\left( 3-\delta \right)-3}-L\le \dfrac{1}{\left( 3+\dfrac{1}{1-L} \right)-3}-L=-1$
$\Rightarrow \left| \dfrac{1}{\left( 3-\dfrac{\delta }{2} \right)-3}-L \right| > 1=\varepsilon$
And $x=3-\dfrac{\delta }{2}$ is guaranteed to be in the interval $0 < |x-3| < \delta ~$ .
Thus, let $x=\min \left( 3-\dfrac{\delta }{2},3-\dfrac{1}{1-L} \right)$
Hence, we can write that $0 < |x-3| < \delta ~$ and $\left| \dfrac{1}{x-3}-L \right|\ge \varepsilon$ .
Therefore, $\displaystyle \lim_{x \to 3}\dfrac{1}{x-3}\ne L$
Hence, $\displaystyle \lim_{x \to 3}\dfrac{1}{x-3}\ne L$ for all values of $L$ and therefore, $\displaystyle \lim_{x \to 3}\dfrac{1}{x-3}$ doesn’t exist.

Note: While showing that the limit of $\dfrac{1}{x-3}$ does not exist we must keep in mind that we must properly use the epsilon delta proof so that mistakes are avoided. Also, we must consider both the cases such as $L\ge 0$ and $L < 0$ , otherwise the solution will not be a complete one.