Homework Help: 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