Question

Is the lim exists when `lim_(x->0) f(x)= +- oo?`

Answer 1

No, the limit does not exist.

Explanation:

Consider the example `f(x) =1/x`

`lim_(x->0^+) f(x) = +oo`

and

`lim_(x->0^-) f(x)= -oo`

Hence, one could conclude that `lim_(x->0) f(x) =+-oo`

However, since the limit tends to different results from above and from below the limit does not exist.

Answer 2

Many of us teach that so-called "infinite limits" are not limits that exist.

Explanation:

`lim_(xrarra) f(x) = L` is and only if for every positive `epsilon`, there is a positive `delta` such that, for all `x`, if `0 < abs(x-a) < delta`, then `f(x) - L) < epsilon`.

`L` cannot be a symbol that does not represent a number. The distance between `f(x)` and `L`, (`abs(f(x) -L)`) must be defined.

There are various ways that a limit can fail to exist.

One way is exemplified by `lim_(xrarr0) 1/x^2`.

In this case, the limit fails to exist because as `xrarr0`, `1/x^2` increases without bound.

We indicate this be writing `lim_(xrarr0)1/x^2 = oo`.

The definition is:

We write `lim_(xrarra) f(a) = oo` if and only if For any `M`, there is a delta such that is `0 < abs(x-a) < delta`, then `f(x) > M`.

Refusing to say that this limit exists allows us to simplify statements like:

If both `lim_(xrarra)f(x)` and `lim_(xrarra)g(x)` exist, then

`lim_(xrarra)f(x)g(x) = lim_(xrarra)f(x) lim_(xrarra)g(x)`.

This theorem fails if we include `lim_(xrarroo) f(x) = oo` as a limit that exists.