All the cases in which a limit cannot exist

**AMaccy** · June 7th 2009, 08:17 PM

My prof said there was three ways, so far I've found two:

When the limit goes up to infinity, and when the left and right limits don't match. What are the other ways a limit cannot exist?

**mr fantastic** · June 7th 2009, 08:21 PM

Originally Posted by AMaccy

My prof said there was three ways, so far I've found two:

When the limit goes up to infinity, and when the left and right limits don't match. What are the other ways a limit cannot exist?

An example for you to consider: $\lim_{x \rightarrow 0} \sin \left(\frac{1}{x}\right)$ .

**pankaj** · June 7th 2009, 09:22 PM

There are indeed three ways in which the $\lim_{x\to a}f(x)$ cannot exist.

1. $f(x)$ is not well defined inthe neighbourhood of $x=a$ .

For example $\lim_{x\to 1}\sec^{-1}(\sin x)$ does not exist

since $\sec^{-1}(\sin x)$ is not defined in the neighbourhood

of $x=1$ though it is defined at $x=1$ .The same

is true for $\lim_{x\to 1}\sin^{-1}(\frac{1+x^2}{2x})$ .

2. $f(x)$ does not display tendency to approach a fixed numerical value.

For e.g. it is not known as to what value

$\lim_{x\to 0}\sin (\frac{1}{x})$ will assume though we are

sure that it is a finite quantity on the interval $[-1,1].$

Similarly, $f(x)=\begin{cases}0,\mbox { if } x\in irrationals\\1,\mbox{ if } x\in rationals\end{cases}$

Here, $f(3) = 1$ .But if you try to evaluate $\lim_{x\to 3}f(x)$ we will not be able to decide to what value should f(x)

assume on the interval $[3-h,3+h]\notin {3}$ .

3.Of course ,the last condition being Left hand limit $\neq$ Right hand limit

Thread: All the cases in which a limit cannot exist

LinkBack

Thread Tools

Display

All the cases in which a limit cannot exist

Similar Math Help Forum Discussions

When does a limit not exist?

Why does this limit not exist

Does this limit exist?

Limit does not exist

Does a limit exist?

Search Tags