# Limit as x approaches Infinity

1. May 20, 2008

### caesius

1. The problem statement, all variables and given/known data
Evaluate $$lim (1+a/x)^x$$
(that's limit as x tends to +infinity, sorry don't know how to latex that)

2. Relevant equations

3. The attempt at a solution
Stumped. Having that $$x$$ exponent has my confused. As x tends to infinity I know what's inside the brackets will tend to one, but the exponent will make it tend to infinity. But the question has three marks to it, it can't be that simple.

2. May 20, 2008

### lurflurf

so
f(a)
-show the the limit exists (the a=1 case should be familar)
-show f(x)f(y)=f(x+y)
-show f is continous and differentiable at some point
now you should be able to identify f

3. May 20, 2008

### DavidWhitbeck

Caesius, have you seen all definitions of e?

Oh and your latex that you wanted to do is \lim_{x \rightarrow \infty} which becomes $$\lim_{x \rightarrow \infty}$$

4. May 20, 2008

### rocomath

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

Set it equal to y, take the natural log of both sides.

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

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

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

Now you have an Indeterminate form of $$\infty\cdot0$$

Keep solving till you can apply L'Hopital's Rule on the right side, then you will have "solved for $$\ln y$$, so use algebra to solve for "y" and you're pretty much done.

5. May 20, 2008

### lurflurf

6. May 22, 2008

### mubashirmansoor

Thats a great solution but I think no need to go that long;

as x approaches infinity "(a/x)" equates 0 .

Hence : 1+0 = 1

7. May 22, 2008

### Gib Z

And that is just wrong. Try reading the other posts, they did mention "e" several times for a good reason.

8. May 22, 2008

### rocomath

Indeterminate Power of form: $$1^{\infty}$$ so yes it is necessary.