# Having trouble solving this limit

1. Sep 4, 2010

### pyrosilver

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

lim (1+(a/x))^(bx) as x-->infinity

2. Relevant equations

3. The attempt at a solution

so, i raised the limit to e, and said e^lim(as x->inf) bxlog((a/x)+1). Then I pulled the constant b out and put it outside of the lim... I don't know how to do the rest though :( Was doing all my lim problems just fine until i came to this one

2. Sep 4, 2010

### Staff: Mentor

Raising things as a power of e is the wrong direction to go. Instead, let y = (1 + a/x)^(bx).

Now take the natural log of both sides, and then take the limit. You should get something you can use L'Hopital's rule on.

3. Sep 4, 2010

### Mentallic

Or, from where you left off, let y=a/x. Remember that this changes your limit from x approaches infinite to y approaches 0.

4. Sep 4, 2010

### Mentallic

Sorry, a rescaling would be much better. Let x=an

Last edited: Sep 5, 2010
5. Sep 4, 2010

### Hurkyl

Staff Emeritus
That's more or less exactly what he did, he just organized the work differently.

6. Sep 5, 2010

### hunt_mat

Do you know the power series expansion for log (1+x)? This will help you solve your problem.

7. Sep 5, 2010

### Mentallic

I don't think that's really necessary. It would be assumed obvious to let $$\lim_{x\to \infty}\left(1+\frac{1}{x}\right)^{x}=e$$ so all that is required is to transform the question into a form that leaves this as the limit.

8. Sep 5, 2010

### hunt_mat

I'd be uneasy saying that as e is defined using the natural numbers not the real numbers

9. Sep 5, 2010

### Mentallic

What does x approaching infinite have to do with either the natural or the real numbers?

10. Sep 5, 2010

### hunt_mat

It still looks dodgy to me, I don't think you can say that. The route with expanding the log function is by far the safer route to go down.

11. Sep 5, 2010

### gomunkul51

Here you go, the equivalent real value definition:

$$\lim_{x\to0}\left(1+x\right)^{\frac{1}{x}}=e$$

just don't use it all in one place! :)

12. Sep 5, 2010

### hunt_mat

You then have to show that this in indeed the same value as the normal definition. No, I am convinced that there is far more work this way than my way.

13. Sep 5, 2010