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

[pineapplewithmouse]

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.

I really didn't catch that wow, the limit I was solving was with x approaching infinity and not 0 as I wrote.

 

[pineapplewithmouse]

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.

Thanks!

 

[pineapplewithmouse]

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.

You are absolutely correct in your assumption. Btw, that picture of Levy you have as profile picture is just god tier, I love it.

 

[pineapplewithmouse]

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?

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).

I see, thank you so much!

 

[logistic_guy]

You are absolutely correct in your assumption. Btw, that picture of Levy you have as profile picture is just god tier, I love it.

If you know Levy Rozman, then you are a legend yourself.

saaac the Roooooooooooooook!