# Limit of exp -> exp (limit) ?

1. Oct 25, 2014

### CeilingFan

1. The problem statement, all variables and given/known data

Okay, so the problem is to find lim (x→∞) (e^x + x) ^ (1/x)

I was given the solution in the assignment in which the first step was to take the natural log of the function, then exponentiate it.

i.e.

lim (x→∞) [exp ln( (e^x + x) ^ (1/x))]

which I understand.

However the next step, the solution equated this with:

exp [ lim (x→∞) ln( (e^x + x) ^ (1/x))]

which I became confused with. Are the two equal? Am I missing out on something?

This assignment we're doing l'Hospital's Rule, if that's of any help...

I probably won't be able to understand too abstract math, currently doing my first year in university...

Thanks.

2. Oct 25, 2014

### pasmith

There is a theorem which states that if $f$ is continuous then $$\lim_{x \to a} f(g(x)) = f( \lim_{x \to a} g(x)).$$ (This implicitly asserts that either both limits exist and are equal, or neither of them exist.)

3. Oct 25, 2014

### robotpie3000

Remember that before you can apply L'Hospital's rule you must convert the function into a 0/0 or ∞/∞ form.

4. Oct 25, 2014

### Ray Vickson

I would do it differently: (x+e^x)^(1/x) = (a b)^c = (a^c) (b^c), where a = e^x, b = 1+ x e^(-x), and c = 1/x.

5. Oct 25, 2014

### Staff: Mentor

Then you still have to show x e^(-x) --> 0 for x->inf, but it is still an elegant solution as this limit was probably studied before.

6. Oct 26, 2014

### CeilingFan

Ah. All make sense now. Is there anywhere I can read up on this rule? I don't think I was taught this, unless it's derivable from what I already know. Either that or I've been a horrible student. Haha..

So it doesn't have to be a limit at infinity for this rule to hold, can be any limit right?

Also didn't know this was considered calculus... Shows that I don't even know what I'm studying.

7. Oct 26, 2014

### Ray Vickson

The concept you want is "continuity", or "continuous functions". If your textbook does not have it, use Google. Basically, a function $f(x)$ is said to be continuous at $x_0$ if $\lim_{x \to x_0} f(x) = f(x_0)$. Note that if $f(x)$ is continuous at $x_0$ and $g(y)$ is continuous at $y_0 = f(x_0)$, then $g(y) \to g(y_0)$ as $y \to y_0$, and $y = f(x) \to y_0 = f(x_0)$ as $x \to x_0$, so we have $g(f(x)) \to g(f(x_0))$ as $x \to x_0$. This just needs the continuity properties of $f$ and $g$ mentioned before.

However, the less stringent result
$$\lim_{x \to x_0} g(f(x)) = g(\lim_{x \to x_0} f(x))$$
requires weaker conditions: you need only have $g(y)$ continuous at $y_0 = \lim_{x \to x_0} f(x)$; the function $f(x)$ need not be continuous (or even defined) at $x = x_0$. Basically, that is the case you have, with $x_0 = \infty$.

After that (or, maybe before that) you need to worry about the question "are my functions continuous?""

Last edited: Oct 26, 2014