Need help with the following proof:

prove that if lim x->c 1/f(x)= 0 then
lim x->c f(x) does not exist.I think i need to use the delta epsilon definition i am not sure how to set it up.

1 answer

In general,

A implies B

is equivalent to

Not(B) implies not (A)

This means that the statement:

lim x->c 1/f(x)= 0 ------->

lim x->c f(x) does not exist,

is equivalent to the statement:

lim x->c f(x) exists ------->

lim x->c 1/f(x) is not zero

Now, we can make the statement more precise.

If lim x->c f(x) exists, then there is some rela number y such that:

lim x->c f(x) = y

Then, it is easy to prove that if y is not zero, we have:

lim x->c 1/f(x) = 1/y

If y = 0, then:

lim x->c 1/f(x) does not exist.

So, we always have that the limit of
1/f(x) is not equal to zero.