Question

What is the limit as x approaches infinity of `e^x`?

Answer 1

Another perspective...

Explanation:

`color(white)()`
As a Real function

Treating `e^x` as a function of Real values of `x`, it has the following properties:

• The domain of `e^x` is the whole of `RR`.

• The range of `e^x` is `(0, oo)`.

• `e^x` is continuous on the whole of `RR` and infinitely differentiable, with `d/(dx) e^x = e^x`.

• `e^x` is one to one, so has a well defined inverse function (`ln x`) from `(0, oo)` onto `RR`.

• `lim_(x->+oo) e^x = +oo`

• `lim_(x->-oo) e^x = 0`

At first sight this answers the question, but what about Complex values of `x`?

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As a Complex function

Treated as a function of Complex values of `x`, `e^x` has the properties:

• The domain of `e^x` is the whole of `CC`.

• The range of `e^x` is `CC "\" { 0 }`.

• `e^x` is continuous on the whole of `CC` and infinitely differentiable, with `d/(dx) e^x = e^x`.

• `e^x` is many to one, so has no inverse function. The definition of `ln x` can be extended to a function from `CC "\" { 0 }` into `CC`, typically onto `{ x + iy : x in RR, y in (- pi, pi] }`.

What do we mean by the limit of `e^x` as `x -> "infinity"` in this context?

From the origin, we can head off towards "infinity" in all sorts of ways.

For example, if we just set off along the imaginary axis, the value of `e^x` just goes round and around the unit circle.

If we choose any complex number `c = r(cos theta + i sin theta)`, then following the line `ln r + it` for `t in RR` as `t->+oo`, the value of `e^(ln r + it)` will take the value `c` infinitely many times.

We can project the Complex plane onto a sphere called the Riemann sphere `CC_oo`, with an additional point called `oo`. This allows us to picture the "neighbourhood of `oo`" and think about the behaviour of the function `e^x` there.

From our preceding observations, `e^x` takes every non-zero complex value infinitely many times in any arbitrarily small neighbourhood of `oo`. That is called an essential singularity at infinity.

Answer 2

Exaplanation using logarithms.

Explanation:

The limit does not exist because as `x` increases without bond, `e^x` also increases without bound. `lim_(xrarroo)e^x = oo`.

Te xplanation of why will depand a great deal on the definitions of `e^x` and `lnx` with which you are working.

I like to define `lnx = int_1^x 1/t dt` for `x > 0`, then prove that `lnx` is invertible (has an inverse) and define `e^x` as the inverse of `lnx`.

Since `d/dx(lnx) = 1/x`, it is clear that `lnx` is increasing#

In the process, I also prove to my students that that `lim_(xrarr00)lnx = oo`.
(The proof uses ideas similar to those used in showing that the harmonic series diverges.)

Knowing these things allows us to reason as follows.

For any positive Real number `T`, `lnT` is defined and for all `x >= lnT`, we have `e^x >= e^(lnT) = T`.

That is, for every `T`, there is an `M` (namely `lnT` such that if `x >= M, then`f(x) >= T#

Therefore, by definition, `lim_(xrarroo)e^x = oo`