Is the L'Hopital's rule true when using limits on $\mathbb{C}$ (the complex field)?  I don't know if it is only valid using the real numbers $\mathbb{R}$... is that the case?
59303
I have a question about something I'm wondering about. I've read somewhere that L'Hopitals rule can also be applied to complex functions, when they are analytic. So if have for instance:  $$ \lim_{z \rightarrow 0} \frac{\log(1+z)}{z} \stackrel{?}{=} \lim_{z \rightarrow 0} \frac{1}{(1+z)} = 1 $$  Now i'm wondering if this is correct? Also if we take $|z|&lt;1$, is it then correct?  Thanks,
107333
This is an exercise from Remmert.  Let D be  domain in $\mathbb{C} \ $, and $f : D \rightarrow \mathbb{C} \ $ a real-differentiable function. Suppose that for some $ c \in D $ the limit  $\lim_{h \to 0} |\frac{f(c + h) - f(c)}{h}|$  exists. Prove that either $f \ $ or $\bar{f} \ $ is complex-differentiable at c.   Any hint ?
66681
The limit is $$\lim_{x\to0}\left(\frac{1}{\sin^2x}-\frac{1}{x^2}\right)$$ which I'm aware can be rearranged to obtain the indeterminate $\dfrac{0}{0}$, but in an attempt to avoid L'Hopital's rule (just for fun) I tried using the fact that $\sin x\approx x$ near $x=0$. However, the actual limit is $\dfrac{1}{3}$, not $0$.  In this , the approximation reasoning works out.
24800
I am stuck on how to calculate the following limit: $$\lim_{x\to\infty}x\left(\left(1 + \frac{1}{x}\right)^x - e\right).$$  Definitely, it has to be through l'Hôpital's rule. We know that $\lim_{x\to\infty} (1+(1/x))^x = e$. So, I wrote the above expression as $$\lim_{x\to\infty}\frac{\left(1 + \frac{1}{x}\right)^x - e}{1/x}.$$Both numerator and denominator tend to zero as x tends to infinity. I applied l'Hôpital's rule twice, but I got the limit equal to infinity which is clearly wrong. In the book, it says that the limit should be $-e/2$.  Any help please? Also, in case someone managed to solve it, please tell me which numerator and denominator did you apply your l'Hôpital's rule to in both cases? Thanks a lot  Also, guys can you tell me how to write equations in this forum? Thanks.
40617
$$ \lim_{(x,y)\to (0,0)} \frac {x^3y^2}{x^4+y^6}  $$ Does this limit exist?   I've tried substituting y=x^0.5 and y=x^(2/3) which both goes to 0.
6393
I need to find simple limit: $$\lim_{n\to\infty}{\sqrt{n+1}-\sqrt{n}}$$  I can't use L'Hôpital's rule, which I think is natural for such problems. How to calculate such limit using only basic methods (definition and simple transformations etc.)?  It's one problem (from 50 problems) I don't know how to solve from my homework.  Thanks for help.
95279
I'm self-studying a bit of complex analysis, and I'm attempting to figure out the following.   Suppose $\lim_{n\to\infty}z_n=A$. How can I show that $$ \lim_{n\to\infty}\frac{1}{n}(z_1+\cdots+z_n)=A. $$  Is there a clever way to write the limit to make it more approachable? Thank you.
68437
Small question, I'm trying to solve this limit but I just can't wrap my head around this problem.  $$\lim_{x\to 0} \frac {e^{-1/x^2}}{x} $$  L'Hopital just seems to make it messier.   It's probably pretty simple - I'd like to hear what I'm missing.
108666
Prove that   $$\lim_{x\to \frac{\pi}{2}} \frac{1}{\big(x-\frac{\pi}{2}\big)}+{\tan(x)}=0.$$   I'm not really sure how to proceed. I know that I should not try L'Hôpital's rule (tried that) but not sure how I would incorporate into the Squeeze Theorem or how I would use continuity.  Thanks!  Edit: Turns out I was really dumb and you do use L'Hôpital's rule twice. I made the mistake of differentiating the whole quotient rather than the function on top and the bottom of the vinculum separately.
96491
I have a strong feeling that the following limit is zero, can anybody help me prove it.    $ \lim\limits_{(x,y,z) \to (0,0,0)} \frac{{xyz}}{{x+y+z}}$   Thanks!
19755
How to prove that $\lim_{(x,y) \to (0,0)} \dfrac{x^3y}{x^4+y^2} = 0?$   First I tried to contradict by using $y = mx$ , but I found that the limit exists.  Secondly I tried to use polar coordinates, $x =  \cos\theta $ and $y = \sin\theta$,  And failed .. How would you prove this limit equals $0$?
46576
What is the limit of $0/x$ as x goes to $0$, without using L'Hopital's rule? Clearly it should be $0$, but I'm not sure how it is any different from something like the divergence of $1/r^2$ which yields a dirac delta.
11394
Prove that the $\lim_{n\to \infty} r^n = 0$ for $|r|\lt 1$.   I can't think of a sequence to compare this to that'll work.  L'Hopital's rule doesn't apply.  I know there's some simple way of doing this, but it just isn't coming to me. :(
30273
I have a homework in real analysis and I'm very confused about it. I would be very thankful, if you could give me any ideas or solutions how to get this task done. The task is as follows:     Find the limit:   $$\lim_{x\to 0} \frac{1}{x} \int_{0}^{x} (1+ \sin(t))^{1/t} dt.$$   A hint that has been given is to use L'Hospital's Rule. Moreover, it must be substantiated why we can use L'Hospital' Rule in this situation. It is also important to mention that $(1 + \sin(t))$ has a removable point of discontinuity at $0$ and therefore the integral can be interpreted as a Riemann integral.
107847
How to check if this limit exists: $$\lim_{(x,y)\rightarrow (0,0)}{ \frac{x^4y^4}{(x^2 + y^4)^3}}$$  Can I convert it to polar form ? $$\frac{r^8 (\cos^4 (\theta)\sin^4(\theta)}{r^6 (\cos^6 (\theta) + r^6 \sin^{12} (\theta))} = \frac{r^2}{\cos^6(\theta) + 0} = 0$$
78897
$\lim_{n \to \infty} n[(1+\frac{1}{n})^n - e]$  I let, $x = \frac{1}{n}$, then as   $\lim_{x \to 0} \frac{1}{x}[(1+x)^\frac{1}{x} - e] = \infty$  L'hopital's: $\lim_{x \to 0} \frac{\frac{1}{x}(1+x)^{\frac{1}{x}-1}}{1} = \frac{1}{0}(1+0)^{\frac{1}{0}-1} = \infty$  Again, if we apply L'hopital's: $\lim_{x \to 0} \frac{\frac{1}{x}(1+x)^{\frac{1}{x}-1}}{x+(1+x)}$. This is also going to $\infty$ as $x \to 0$. But, I know the answer is $\frac{-e}{2}$, and I am not even close. Can anyone please find me the mistakes here. Oh, I am supposed to use L'Hopital's rule.
122021
Does the limit $\lim _{x\to 0} \frac 1x \int_0^x \left|\cos \frac 1t \right| dt$ exists ? If it does then what is the value ?   I don't think even L'Hospital's rule can be applied . Please help . Thanks in advance