# Limits

• May 13th 2010, 06:37 PM
riemann
Limits

We know that since the right hand limit and left hand limits are not same for the following limit, therefore

$
\lim_{x \to 0} 1/x\ does\ not\ exist
$

Now rewriting the above limit in another way,

$
\lim_{x \to \infty} x
$

does it exist? It should be infinity, thus possessing a limit. But how can rewriting a non-existing limit give an existing limit?
I am somewhere wrong here, can anyone point where? (thanks)

We are totally clear about the non-existance of

$
\lim_{x \to 0} 1/x
$

yet I came to know that

$
\lim_{x \to 0} sin(1/x)= -1\ to\ 1\
$

Therefore the above limit exists. But how can sine of an undefined function give a definite value? (The word 'definite value' does not mean in its exact sense, i.e., I tried to mean (i).how sine of an undefined function gave some value between -1 and 1 and (ii).though we are not sure of any definite value, yet we say it's limit exists?) My natural instincts tell that the above limit should not exist, yet it exists. Can someone please explain?

Your replies would be highly appreciated. Thank you.(Happy)
• May 13th 2010, 06:58 PM
pickslides
Why $\lim_{x \to 0 }\frac{1}{x} = \lim_{x \to \infty }x$ ?
• May 13th 2010, 07:19 PM
Bruno J.
Quote:

Originally Posted by riemann

We know that since the right hand limit and left hand limits are not same for the following limit, therefore

$
\lim_{x \to 0} 1/x\ does\ not\ exist
$

Now rewriting the above limit in another way,

$
\lim_{x \to \infty} x
$

does it exist? It should be infinity, thus possessing a limit. But how can rewriting a non-existing limit give an existing limit?
I am somewhere wrong here, can anyone point where? (thanks)

We are totally clear about the non-existance of

$
\lim_{x \to 0} 1/x
$

yet I came to know that

$
\lim_{x \to 0} sin(1/x)= -1\ to\ 1\
$

Therefore the above limit exists. But how can sine of an undefined function give a definite value? (The word 'definite value' does not mean in its exact sense, i.e., I tried to mean (i).how sine of an undefined function gave some value between -1 and 1 and (ii).though we are not sure of any definite value, yet we say it's limit exists?) My natural instincts tell that the above limit should not exist, yet it exists. Can someone please explain?

Your replies would be highly appreciated. Thank you.(Happy)

Good question! In a sense, the limit of $x^{-1}$ as $x \rightarrow 0$ does exist; it's equal to $\infty$, but you have to be clear about what $\infty$ is. There is made no distinction between $+\infty$ and $-\infty$ here. There is only one point at infinity; read this!

The limit $\lim_{x \rightarrow 0}\sin(1/x)$, on the other hand, does not exist; the function oscillates between $-1$ and $1$ infinitely many times in any neighbourhood of $0$.