# Homework Help: Exponential limits

1. Feb 27, 2008

### javi438

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

The limit of (1+a/x)^x as x goes to infinity, where a>0

2. Relevant equations

3. The attempt at a solution

I started with saying e^[xln(1+a/x)], but i can't get to the answer e^a

2. Feb 27, 2008

### rocomath

$$\lim_{x\rightarrow\infty}\left(1+\frac a x\right)^x$$

Indeterminate power: $$\lim_{x\rightarrow\infty}1^{\infty}$$

$$y=\left(1+\frac a x\right)^x$$

$$\ln y=x\ln{\left(1+\frac a x\right)}$$

$$\ln y=\frac{\ln{\left(1+\frac a x\right)}}{\frac 1 x}$$

$$\lim_{x\rightarrow\infty}\ln y=\frac{\ln{\left(\lim_{x\rightarrow\infty}1+\lim_{x\rightarrow\infty}\frac a x\right)}}{\lim_{x\rightarrow\infty}\frac 1 x}\left[\frac 0 0\right]$$

Last edited: Feb 27, 2008
3. Feb 27, 2008

### sutupidmath

if you could only go from that step to this, are u sure that your right hand side is okay?

4. Feb 27, 2008

### sutupidmath

after you came to this part
$$\ln y=\frac{\ln{\left(1+\frac a x\right)}}{\frac 1 x}$$
probbably this would do better

$$\ lny=\frac{x}{\frac{1}{\ln{\left(1+\frac a x\right)}}}$$ so when you take the lim as x-->infinity, then you will end up with an intermediate form of $$\frac{\infty}{\infty}$$ so you can apply l'hopitals rule.

5. Feb 27, 2008

### sutupidmath

well you still have an intermediate form

6. Feb 27, 2008

### rocomath

Yep, I know that. What I want to know is if I can do what I did. I don't like the way you set yours up, but I'm not sure if I can do what I did.

7. Feb 27, 2008

### sutupidmath

to the OP: you can either chose to apply the l'hopitals rule here $$\ lny=\lim_{x\rightarrow\infty}\frac{x}{\frac{1}{\ln{\left(1+\frac a x\right)}}}$$

or here:$$\ln y=\lim_{x\rightarrow\infty} \frac{\ln{\left(1+\frac a x\right)}}{\frac 1 x}$$ it is basically the same, besides at the former u have an intermediate form of the type $$\frac{\infty}{\infty} \ while \ on \ the \ latter \ you \ have \ \frac{0}{0}$$

8. Feb 27, 2008

### rocomath

Ok, taking too long to respond. Sleep time, and I bet Hall's will rip me on this post, lol.

9. Feb 27, 2008

### sutupidmath

yes indeed, but you went some steps further,while you should have stopped before, and apply the l'hopitals rule!

10. Feb 27, 2008

### sutupidmath

and nothing is wrong with how i set it up, both ways work!

11. Feb 27, 2008

### sutupidmath

i think i am going to bed too, it is getting to late now!

12. Feb 27, 2008

### rocomath

I can't exactly remember what to do, I will need to consult my book tomorrow. Don't feel like turning on the lights to look, my eyes will explode, lol. But it definitely doesn't make sense to apply the limit inside while holding it on the outside. But I did it anyways :p

13. Feb 27, 2008

### sutupidmath

the steps above are completely fine, if you are reffering when you entered with the limit in, but doing that does not do you any good, since you will get zero over zero which is undefined, so what we end up with is nothing! and we need a solution! lol

14. Feb 27, 2008

### sutupidmath

well, you keep editing your original post all the time,
how the heck do you think that the denominator of what u did is okay? it is wrong now, disregard my post#13, since when i wrote it what u did was okay, now what u did is completely wrong!

15. Feb 27, 2008

### sutupidmath

well there is another way of doing it, at least another that i can think of, but more or less at some point it comes to the same problem!

16. Feb 27, 2008

### rocomath

My computer freezes. Not that you've noticed, but I always send in my posts early which may be incomplete or incorrect since I'm worried my comp will freeze and all that typing was for nothing.

Hmm ... I just re-editted it, I think it's fine now.

Last edited: Feb 27, 2008
17. Feb 27, 2008

### sutupidmath

$$\lim_{x\rightarrow\infty}\frac{1}{x}=0$$ and not 1/x like you are claiming

18. Feb 27, 2008

### sutupidmath

you have to use l'hopital rule, i cannot think of any other way. What's the big deal here, the OP merely needs a solution, and this way you will defenetely get one.

19. Feb 27, 2008

### sutupidmath

what was not needed to do here is this very las step. because the previous step you should have applyed the l'hopital rule,and everything would be okay, and i would have been doing other stuff right now!

edit: i just noticed, the other way i was talking about is the way originally the OP started, but at some point we would end up with the same problem as here,so both ways would work!

Last edited: Feb 27, 2008