Exponential Limit: Find the Limit of (1+a/x)^x as x→∞

[javi438]

Homework Statement

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

Homework Equations

The Attempt at a Solution

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

 

[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]

 

[sutupidmath]

\lim_{x\rightarrow\infty}\ln y=\infty\cdot\ln 1

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

\lim_{x\rightarrow\infty}\ln y=0

Now what would you do?

 

[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'hospital's rule.

 

[sutupidmath]

I think this may be ok, hopefully someone can clarify.

\lim_{x\rightarrow\infty}\ln y=\frac{\ln 1}{\lim_{x\rightarrow\infty}\frac 1 x}

Now what would you do?

well you still have an intermediate form

 

[rocomath]

well you still have an intermediate form

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.

 

[sutupidmath]

to the OP: you can either chose to apply the l'hospital's 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}

 

[rocomath]

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

 

[sutupidmath]

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.

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

 

[sutupidmath]

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.

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

 

[sutupidmath]

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

 

[rocomath]

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

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

 

[sutupidmath]

\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]

I think this may be ok, hopefully someone can clarify. I know you can bring the limits in, but I forget how to properly apply them.

\lim_{x\rightarrow\infty}\ln y=\frac{\ln 1}{\frac 1 x} You won't even need to use L'Hopital's rule here if the step above is okay!

the steps above are completely fine, if you are referring 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

 

[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!

 

[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!

 

[rocomath]

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!

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.

 

[sutupidmath]

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

 

[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.

 

[sutupidmath]

\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]

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!