# I Is it possible to find the limit of (1+1/x)^x as x approaches -infinity?

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1. May 28, 2017

### manjuvenamma

Is it possible to find the limit of (1+1/x)^x as x approaches minus infinity using only the fact that it is e if x approaches infinity?

2. May 28, 2017

1/e??

3. May 28, 2017

### manjuvenamma

No, the answer is e. We have to find a reason without using LHospital's rule.

4. May 28, 2017

### Staff: Mentor

What happens, if you write $y=-x$ and consider $y \to +\infty$?

5. May 28, 2017

### manjuvenamma

I thought about it but got stuck up. Then it becomes (1-1/y)-y as y approaches infinity which is not a standard formula.

6. May 28, 2017

### Staff: Mentor

Maybe, but $a^{-y}=(\frac{1}{a})^y$ is.

7. May 28, 2017

### manjuvenamma

but 1+1/y is different from 1-1/y. That is where I am stuck.

8. May 28, 2017

### Staff: Mentor

For $y \to \infty$? Isn't it only the difference between left and right?

9. May 28, 2017

### dgambh

I am not sure if this helps but what if you multiply by the conjugate and then try to think why $\lim_{y \to \infty} \left(1 - \frac{1}{y^2}\right)^y$ should be 1.

10. May 28, 2017

### manjuvenamma

A great idea but I am not able to think of a proof that it is 1. If this is proved, then I agree that the original question is proved. But how to prove that? It raises other similar interesting questions like what about (1-1/y^3)^y, (1-1/y^4)^(y^2) etc.

11. May 28, 2017

### Aryamaan Thakur

I think this problem can be solved in this way. Here is what I think:-

let x = -1/y
then, Lt (1+1/x)x where x→ -∞
=> Lt (1 - y)-1/y where y→ 0
Now let -y = 1/p
then, Lt (1 - y)-1/y where y→ 0
=> Lt (1+1/p)p where p→∞
And this is equal to e.

12. May 29, 2017

### ehild

You have
$$A=\lim _{x\rightarrow\infty} (1+1/x)^x=e$$
$$B=\lim _{x\rightarrow\infty} \left(\frac{1}{1-1/x}\right)^x$$
$$AB=\lim _{x\rightarrow\infty}\left(\frac{1+1/x}{1-1/x}\right)^x=\lim _{x\rightarrow\infty}(1+2/x)^x=\lim _{x\rightarrow\infty}\left((1+2/x)^{x/2}\right)^2=e^2$$

13. May 29, 2017

### manjuvenamma

I am not able to figure out how 1+2/x came. Is it an approx?

14. May 29, 2017

### ehild

Yes, it is an approximation. 1/(1-a) ≈ 1+a and (1+a)2≈1+2a for |a|<<1 . As x-->∞, 1/x -->0 and you can substitute 1/(1-1/x) with (1+1/x). The second approximation is not needed as you have (1+1/x)x already.
1/(1-a) is the sum of the geometric series 1+a+a2+.... . If a <<1, the higher order terms can be neglected.

15. May 29, 2017

### manjuvenamma

@ ehild, Great explanation. Thanks to all the respondents who took time to clarify things to me. I would like to know how to write math equations here, is there a link? Is there a way I can accept the answer and close as the raiser of this question?

16. May 29, 2017

### ehild

To write nice equations use LaTeX : https://www.physicsforums.com/help/latexhelp/
There are useful symbols also under the ∑ key, and you can make subscript and superscript with the x2 and x2 keys.
I think, closing a thread is possible in the Homework Forum only.

17. May 29, 2017

### jack action

This question has been intriguing me, so I've gave it a shot:
$$\left(1+\frac{1}{x}\right)^x = \left(\frac{1+x}{x}\right)^x$$
With $\infty$:
$$\left(\frac{1+\infty}{\infty}\right)^{\infty}$$
With $-\infty$:
$$\left(\frac{1-\infty}{-\infty}\right)^{-\infty} = \left(\frac{\infty - 1}{\infty}\right)^{-\infty} = \left(\frac{\infty}{\infty - 1}\right)^{\infty}$$
And one can see that:
$$\frac{\infty}{\infty - 1} = \frac{1+\infty}{\infty}$$
Where the numerator is just 1 + the denominator (like $\frac{101}{100}$ or $\frac{1000}{999}$). Thus it should give the same answer when at infinity.

18. May 29, 2017

### Staff: Mentor

This shouldn't get on my desk for correction.

19. May 29, 2017

@fresh_42 I agree. They could do a better job of solving this. Suggestion in computing $y=$ limit $(1+\frac{1}{x})^x$ in a couple of the above posts is to take the natural log of this expression, (so that you have $ln(y)=x \, ln(1+\frac{1}{x})$), use a Taylor expansion of $ln(1+\frac{1}{x})$, (in the form of $ln(1+u )$ where $u$ is small), and then take $e^{ln(y )}=y$.

Last edited: May 29, 2017
20. May 30, 2017

### pwsnafu

Here's a direct method
$\lim_{x\to-\infty}\left(1+\frac{1}{x}\right)^x = \lim_{y\to\infty}\left(1-\frac{1}{y}\right)^{-y} = \lim_{y\to\infty}\frac{1}{\left(1-\frac{1}{y}\right)^y}= \lim_{y\to\infty}\left(\frac{y}{y-1}\right)^{y}.$
Now let $w = y-1$. Clearly $y\to\infty$ is equivalent to $w\to\infty$. So
$\lim_{w\to\infty}\left(\frac{w+1}{w}\right)^{w+1}= \lim_{w\to\infty}\left[ \left(1+\frac{1}{w}\right)^w\left(1+\frac{1}{w}\right) \right] = e\times 1 = e$ as required.