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

\(\displaystyle

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

\)

Now rewriting the above limit in another way,

\(\displaystyle

\lim_{x \to \infty} x

\)

\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

yet I came to know that

I am somewhere wrong here, can anyone point where? (thanks)

We are totally clear about the non-existance of

\(\displaystyle

\lim_{x \to 0} 1/x

\)

\lim_{x \to 0} 1/x

\)

yet I came to know that

\(\displaystyle

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

\)

\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)

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