Trying the solve lim x->0 (1+1/x)^x with L'Hôpital's rule

[pineapplewithmouse]

I want to solve the limit lim x->0 (1+x)^1/x from scratch (so assuming I don't know it's equal to e).
This limit has the indeterminate form of 1^infinity, and according to Wikipedia, you can transform 1^infinity into e to the power of 0/0 (or infinity/infinity for that matter), and then you can use L'Hôpital's rule on the limit that has the 0/0 form.
As I understand it, the transformation was done in this way:

And then:

So the equality in the second picture is a particular case of the equality:

And as I understand it, this equality hold true if and only if p(x) is continuous, and if the limit lim x -> c q(x) exists (i.e. is a real number).

But when I solve the limit from scratch, I don't actually know the answer, so I don't know if the limit lim x -> c q(x) exists or not (in this case, the limit lim x -> 0 (ln(1+1/x))/(1/x)), so how can I make the transformation above? Is there some indication that the limit does exist? Or am I misunderstanding something about the transformation?

 

[fresh_42]

Let us assume that $\displaystyle{ \lim_{x \to c} q(x) }$ does not exist. Then how can $p(x)$ be continuous at $q(c)$?

 

[Dr.Peterson]

But when I solve the limit from scratch, I don't actually know the answer, so I don't know if the limit lim x -> c q(x) exists or not (in this case, the limit lim x -> 0 (ln(1+1/x))/(1/x)), so how can I make the transformation above?

You try to find the limit of q(x). If it exists, you can use it. If it doesn't exist, then the limit you want doesn't exist.

What have you done to find it?

 

[pineapplewithmouse]

Let us assume that $\displaystyle{ \lim_{x \to c} q(x) }$ does not exist. Then how can $p(x)$ be continuous at $q(c)$?

It made sense to me, but then I thought about it again.
q(c) doesn't exist, so it doesn't make sense that because p(q(c)) doesn't exist, then p(x) isn't continuous, since q(c) doesn't exist.
For example, p(x)=x^2 is for sure continuous for every real x.
q(x)=1/x, the limit x -> 0 q(x) doesn't exist.
p(q(x))=(1/x)^2, so the limit x -> 0 p(q(x)) doesn't exist, but obviously p(x) is still continuous everywhere.

 

[pineapplewithmouse]

You try to find the limit of q(x). If it exists, you can use it. If it doesn't exist, then the limit you want doesn't exist.

What have you done to find it?

I used L'Hopitals rule, but I don't see how does that matter.

If I assume the limit exists and use a transformation that requires the limit to exist, and then find that it exists, isn't that circular reasoning?

 

[logistic_guy]

The title says:

$\displaystyle \bold{1.} \lim_{x \rightarrow 0}\left(1 + \frac{1}{x}\right)^x$

The first post says:

$\displaystyle \bold{2.} \large\lim_{x \rightarrow 0}\left(1 + x\right)^\frac{1}{x}$


Which limit are you trying to solve?

 

[fresh_42]

It made sense to me, but then I thought about it again.
q(c) doesn't exist, so it doesn't make sense that because p(q(c)) doesn't exist, then p(x) isn't continuous, since q(c) doesn't exist.
For example, p(x)=x^2 is for sure continuous for every real x.
q(x)=1/x, the limit x -> 0 q(x) doesn't exist.
p(q(x))=(1/x)^2, so the limit x -> 0 p(q(x)) doesn't exist, but obviously p(x) is still continuous everywhere.

It was too late to edit as I noted that we actually need the continuity of $p\circ q$ because what we want is
$$\lim_{x\to c}(p \circ q)(x)=\lim_{x\to c}p(q(x))=p(q(\lim_{x\to c}x))=p(q(c)).$$
Your example fails since we need continuity of $p$ at $q(c)$ and $x^2$ isn't defined at "$q(0)$" because $q$ isn't defined at $x=0.$

 

[lookagain]

I want to solve the limit lim x->0 (1+x)^1/x from scratch (so assuming I don't know it's equal to e).

pineapplewithmouse, before you go on, before the helpers go on, tell us which problem for sure
you are working on, as post #6 pointed out. Those are not equivalent problems.

 

[Dr.Peterson]

I used L'Hopitals rule, but I don't see how does that matter.

If I assume the limit exists and use a transformation that requires the limit to exist, and then find that it exists, isn't that circular reasoning?

Show your work for the actual problem (whichever it is), and we can all talk about that specifically.

 

[pineapplewithmouse]

The title says:

$\displaystyle \bold{1.} \lim_{x \rightarrow 0}\left(1 + \frac{1}{x}\right)^x$

The first post says:

$\displaystyle \bold{2.} \large\lim_{x \rightarrow 0}\left(1 + x\right)^\frac{1}{x}$


Which limit are you trying to solve?

Honestly it doesn't matter, in both limits I stumble into the same problem I described.
But let's say it is $\displaystyle \lim_{x \rightarrow 0}\left(1 + \frac{1}{x}\right)^x$ this limit

 

[pineapplewithmouse]

It was too late to edit as I noted that we actually need the continuity of $p\circ q$ because what we want is
$$\lim_{x\to c}(p \circ q)(x)=\lim_{x\to c}p(q(x))=p(q(\lim_{x\to c}x))=p(q(c)).$$
Your example fails since we need continuity of $p$ at $q(c)$ and $x^2$ isn't defined at "$q(0)$" because $q$ isn't defined at $x=0.$

So you are saying the the requirement for the transformation:

to work is that p(q(x)) would be continuous and not just p(x)?

 

[pineapplewithmouse]

pineapplewithmouse, before you go on, before the helpers go on, tell us which problem for sure
you are working on, as post #6 pointed out. Those are not equivalent problems.

See post #10

 

[pineapplewithmouse]

Show your work for the actual problem (whichever it is), and we can all talk about that specifically.

Fine.
I have this limit:
$\displaystyle \lim_{x \rightarrow 0}\left(1 + \frac{1}{x}\right)^x$
which has the indeterminate form 1^infinity.
So I use the transformation I described to rewrite the limit as:

Now the limit has the indeterminate form e^(0/0), so I use l'hopital's rule on the 0/0 part:

Now I have infinity/infinity, so I use l'hopital's rule again:

I substitute the value I got into the original limit:

So I get:
$\displaystyle \lim_{x \rightarrow 0}\left(1 + \frac{1}{x}\right)^x=e$

Now as I said, this solution assumes that I can make the transformation I mentioned in post #1 at all, which for me sounds like circular reasoning.

 

[fresh_42]

So you are saying the the requirement for the transformation:
View attachment 39296
to work is that p(q(x)) would be continuous and not just p(x)?

Yes. The proposition (in my book the definition of continuity) says: $f(x)$ is continuous at $x=c$ if $\lim_{x\to c}f(x)=f(c).$ In your case, you just have $f(x)=(p\circ q)(x)=p(q(x)).$ This implicitly requires that $c$ is in the domain of $f(x).$ It automatically guarantees that $f(c)$ exists. If $f=p\circ q$ with $p(x)=x^2$ and $q(x)=x^{-1}$ then $0$ is not in the domain of $f.$

That L'Hôpital allows infinities is a different subject and should be discussed separately from continuity.

 

[lookagain]

I have this limit:
$\displaystyle \lim_{x \rightarrow 0}\left(1 + \frac{1}{x}\right)^x$
which has the indeterminate form 1^infinity.

No, if x were approaching 0 from the right, then it would have the
indeterminate form $\displaystyle \ \infty^0.$

Also, regarding your work in post #13, the limit is supposed be as x approaches
0 (preferably from the right I would recommend), and not as x approaches
$\displaystyle \infty, \$ as you did. It will also give a different result. The solution of e at the end is
incorrect.

 

[logistic_guy]

$\displaystyle \lim_{x \rightarrow 0}\left(1 + \frac{1}{x}\right)^x$

This limit has an interesting result.

From the right side, we have:
$\displaystyle \lim_{x \rightarrow 0^+}\left(1 + \frac{1}{x}\right)^x = 1$

While from the left side, we have:
$\displaystyle \lim_{x \rightarrow 0^-}\left(1 + \frac{1}{x}\right)^x \rightarrow$ needs complex analysis.

If I was studying calculus, I would say:

$\displaystyle \lim_{x \rightarrow 0^+}\left(1 + \frac{1}{x}\right)^x \neq \lim_{x \rightarrow 0^-}\left(1 + \frac{1}{x}\right)^x$

Therefore,

$\displaystyle \lim_{x \rightarrow 0}\left(1 + \frac{1}{x}\right)^x = \text{DNE}$

If I was studying complex analysis, I would say:

$\displaystyle \lim_{x \rightarrow 0}\left(1 + \frac{1}{x}\right)^x = 1$

Why?

Let us analyze the left side.

$\displaystyle \lim_{x \rightarrow 0^-}\left(1 + \frac{1}{-0.001}\right)^{-0.001} \approx 0.9931 - 0.0031 \ \bold{i}$

$\displaystyle \lim_{x \rightarrow 0^-}\left(1 + \frac{1}{-0.000001}\right)^{-0.000001} \approx 0.999986 - 3.141549 \times 10^{-6} \ \bold{i}$

This suggests that:

$\displaystyle \lim_{x \rightarrow 0^-}\left(1 + \frac{1}{x}\right)^x = 1 - 0 \ \bold{i} = 1$

Then,

$\displaystyle \lim_{x \rightarrow 0}\left(1 + \frac{1}{x}\right)^x = 1$

$\displaystyle \text{WolframAlpha}$ is a respectful computational website. It gives the answer as:

$\displaystyle \lim_{x \rightarrow 0}\left(1 + \frac{1}{x}\right)^x = 1$

This means that this machine is assuming that we are studying complex analysis!

Why didn't this $\displaystyle \text{Alpha}$ say
$\displaystyle \lim_{x \rightarrow 0}\left(1 + \frac{1}{x}\right)^x = \text{DNE}$?

@fresh_42
@Dr.Peterson
@lookagain

What is your opinion?

 

[Dr.Peterson]

$\displaystyle \text{WolframAlpha}$ is a respectful computational website. It gives the answer as:

$\displaystyle \lim_{x \rightarrow 0}\left(1 + \frac{1}{x}\right)^x = 1$

This means that this machine is assuming that we are studying complex analysis!

Why didn't this $\displaystyle \text{Alpha}$ say
$\displaystyle \lim_{x \rightarrow 0}\left(1 + \frac{1}{x}\right)^x = \text{DNE}$?

I assume you mean the site is respected, not respectful?

Limit[Power[(1+Divide[1,x]),x],x->0] - Wolfram|Alpha

Wolfram|Alpha brings expert-level knowledge and capabilities to the broadest possible range of people—spanning all professions and education levels.

www.wolframalpha.com

Interestingly, when you ask it for complex-valued plots, it shows graphs that indicate that the limit is indeed 1:

In fact, it quite often does assume complex analysis (and mentions that specifically). Sometimes its answers are inappropriate for questions intended to be about functions of a real variable.

Here, it's conceivable that it doesn't mention that because you get the same result if you consider the limit to be at an endpoint of the (real-valued) domain of the function (one-sided). But most likely it's because it doesn't care what the user might be thinking, and it does whatever it does, including complex numbers.

 

[Dr.Peterson]

Now as I said, this solution assumes that I can make the transformation I mentioned in post #1 at all, which for me sounds like circular reasoning.

Pretending that all your work were correct, hasn't it shown that the inner function has a limit, and that the outer function has a limit at that value? What is circular?

And as I understand it, this equality hold true if and only if p(x) is continuous, and if the limit lim x -> c q(x) exists (i.e. is a real number).

But when I solve the limit from scratch, I don't actually know the answer, so I don't know if the limit lim x -> c q(x) exists or not (in this case, the limit lim x -> 0 (ln(1+1/x))/(1/x)), so how can I make the transformation above? Is there some indication that the limit does exist? Or am I misunderstanding something about the transformation?

You don't know the limit of q before you do any work , but you do know it after doing the middle part of the work, so the theorem applies then! Your work is the indication that the limit exists; you don't actually use the theorem until then (though you are hoping you can).

 

[logistic_guy]

Thanks a lot professor Dave for sharing your thoughts. I really appreciate your reply.

I assume you mean the site is respected, not respectful?

Indeed, respected.

I have always taken the results of this $\displaystyle \text{Alpha}$ for granted. It seems that from now on, I will have to be careful.

 

[logistic_guy]

No, if x were approaching 0 from the right, then it would have the
indeterminate form $\displaystyle \ \infty^0.$

Also, regarding your work in post #13, the limit is supposed be as x approaches
0 (preferably from the right I would recommend), and not as x approaches
$\displaystyle \infty, \$ as you did. It will also give a different result. The solution of e at the end is
incorrect.

My intuition tells me that the original limit of the OP was $\displaystyle \large\lim_{x \rightarrow 0}\left(1 + x\right)^\frac{1}{x}$.

Then, he changed it to $\displaystyle \large\lim_{x \rightarrow 0}\left(1 + x\right)^\frac{1}{x} = \lim_{x \rightarrow \infty}\left(1 + \frac{1}{x}\right)^x$ because working with $\displaystyle \infty$ was easier.

But when he wrote the title, he forgot to replace the zero by $\displaystyle \infty$. Later, he was more focused on continuity and circular reasoning, so his eyes could not see that the zero was still there. Also copying and pasting Latex makes it very difficult to spot errors, but at least it is faster than reinventing the wheel.