# Math Help - [SOLVED] A D.N.E limit proofs

1. ## [SOLVED] A D.N.E limit proofs

Hello,

I need help to write a formal proof to show that the following limits do not exist.

1. $\lim \limits_{x \rightarrow 0} \frac{1}{x}$. i.e. show that $\lim \limits_{x \rightarrow 0} \frac{1}{x}=L$ is false for $\forall L$.

2. $\lim \limits_{x \rightarrow 1} \frac{1}{x-1}$ does not exist. i.e. show that $\lim \limits_{x \rightarrow 1} \frac{1}{x-1}$ is false for $\forall L$.

2. Consider the right-hand limit $\lim _{x \to 1^ + } \frac{1}{{x - 1}}$.
The points $\left( {\forall n} \right)\left[ {x = 1 + \frac{1}{n}} \right]$ each can be used in that limit. But $\frac{1}{{1 + \frac{1}{n} - 1}} = n$.
Thus that limit will get greater and greater. So the limit cannot exist.

3. $\forall \epsilon > 0$, $\exists \delta > 0$ such that whenever $0 < |x - 0| < \delta$ it necessarily follows that $\left|\frac{1}{x} - L\right| < e$

$\left|\frac{1}{x} - L\right| < \epsilon$
$L - \epsilon < \frac{1}{x} < L + \epsilon$
$\frac{1}{L+\epsilon} < x < \frac{1}{L-\epsilon}$

Let $\delta = \frac{1}{L - \epsilon}$

So, we must find an $\epsilon$ such that the above will not be satisfied for any $\delta > 0$. Let $\epsilon = |L| + 1> 0$.

Looking at L > 0
$\delta = \frac{1}{L - (|L| + 1)} = \frac{1}{L - (L + 1)} = \frac{1}{-1} = -1 < 0$. Not possible.

Or looking at L < 0
$\delta = \frac{1}{L - (|L| + 1)} = \frac{1}{L - (-L + 1)} = \frac{1}{2L - 1} < 0$

4. Let's do #2:

Let's assume there is a number L such that

$\lim_{x\rightarrow{1}}\frac{1}{x-1}=L$.

Then there exists a number $\delta>0$ such that

$|\frac{1}{x-1}-L|<1$ whenever $0<|x-1|<{\delta}$

In particular, $x=1+\frac{\delta}{\delta+1}$ and

$x=1-\frac{\delta}{\delta+1}$ are two values of x.

Therefore, $|\frac{\delta+1}{\delta}-L|<1$ and

$|-\frac{\delta+1}{\delta}-L|<1$ or

$\frac{1}{\delta} and

$-2-\frac{1}{\delta},

which is impossible. Therefore, the limit does not exist.

5. Thank you very much for helping me out