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

• javi438
In summary, the homework statement is trying to solve an equation e^[xln(1+a/x)], but can't seem to get to the answer. The attempt at a solution is using e^a and finding y as lny=x/(ln(1+a/x)). The limit of y as x goes to infinity is y=ln(1+a/x) and y=0 when x=infinity. The conclusion is either that the original equation is wrong or that the steps taken so far are not sufficient.
javi438

## Homework Statement

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

## The Attempt at a Solution

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

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

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

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

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

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

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

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rocophysics said:
$$\lim_{x\rightarrow\infty}\ln y=\infty\cdot\ln 1$$
if you could only go from that step to this, are u sure that your right hand side is okay?
rocophysics said:
$$\lim_{x\rightarrow\infty}\ln y=0$$

Now what would you do?

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.

rocophysics said:
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

sutupidmath said:
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.

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 $$\frac{\infty}{\infty} \ while \ on \ the \ latter \ you \ have \ \frac{0}{0}$$

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

rocophysics said:
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'hopitals rule!

rocophysics said:
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!

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

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

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

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 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

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!

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!

sutupidmath said:
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.

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

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.

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

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!

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## 1. What is an exponential limit?

An exponential limit is a mathematical concept that refers to the value that a function approaches as its input variable becomes infinitely large.

## 2. How is the limit of (1+a/x)^x as x→∞ calculated?

The limit of (1+a/x)^x as x→∞ can be calculated using the formula e^a, where e is the base of the natural logarithm and a is the constant term in the function.

## 3. What does the value of the limit represent?

The value of the limit represents the maximum value that the function can approach as its input variable becomes infinitely large.

## 4. Can the limit of (1+a/x)^x as x→∞ be negative?

No, the limit of (1+a/x)^x as x→∞ can never be negative because the function is always increasing as x approaches infinity.

## 5. Are there any real-life applications of exponential limits?

Yes, exponential limits are used in various fields of science and engineering, such as in the study of population growth, radioactive decay, and compound interest in finance.

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