All Questions
11
questions
3
votes
3
answers
189
views
To compute $\lim\limits_{n\to +\infty} \log_2\left(\frac{n^3}{n^4+2}\right)=-\infty, \lim\limits_{x\to+\infty}\log_2\left(\frac{x^3}{x^4+2}\right)$
On my textbook of Maths for students of my high school (there are really only two such limits) I have found, for example, a limit of a succession with this solution:
$$\color{magenta}{\lim_{n\to +\...
- 7,792
1
vote
0
answers
148
views
On using logarithms to evaluate limits
The general method of using logs to evaluate limits is $$\lim_{x\to a}f(x)=e^{lim_{x\to a}\ln{f(x)}}$$
And this makes sense usually; I assume you can only use this method when you can see the limit ...
- 1,076
1
vote
2
answers
275
views
Radioactive Decay Equations and Some Related Confusion on Discrete vs. Continuous Growth/Decay, Continuously Compounding Interest, etc.
UPDATE: I've gotten some great answers regarding the first part of my question, related to radioactive decay specifically - thanks! However, I'd still very much appreciate a response to the second ...
- 830
2
votes
2
answers
157
views
Computing $\lim\limits_{n\to \infty} \frac{\ln^2n}{n}\sum\limits_{k=2}^{n-2}\frac{1}{\ln k\ln(n-k)}$
Computing the following limit $$\lim_{n\to \infty} \frac{\ln^2n}{n}\sum_{k=2}^{n-2}\frac{1}{\ln k\ln(n-k)}$$
I first tried to write it as a Riemann sum as follows
$$\frac{\ln^2n}{n}\sum_{k=2}^{n-2}\...
- 24.2k
3
votes
3
answers
208
views
Proving that $\lim\limits_{x \to 0} x \cdot (\log(x))^\alpha = 0$ for every $\alpha > 0$?
$\lim_{x \to 0} x \cdot (\log(x))^\alpha = 0$ for every $\alpha > 0$?
It should tend to zero for every $\alpha \in \Bbb R_+$, though I can't find a rigorous way to prove it.
- 501
3
votes
1
answer
2k
views
Evaluate $\lim_{x \to 0^+} x^{x^x}$
Evaluate $$\lim_{x \to 0^+} x^{x^x}$$
I have assumed $$L=\lim_{h \to 0}h^{(h^h)}$$ taking $x=0+h$ where $h$ is a very small positive real number.
Now taking natural Log on both sides we get
$$\ln L=...
- 10.2k
2
votes
4
answers
77
views
Find $\lim_{n\to\infty}\ln(n^{n}\cdot(n+1)^{-n-1})$
I have to solve this limit:
$$\lim_{n\to\infty}\ln(n^{n}\cdot(n+1)^{-n-1})$$
I know the answer is $-\infty$. My question is, can I do this:
$$\ln[\lim_{n\to\infty}n^{n}\cdot(n+1)^{-n-1}]$$
If not, how ...
- 2,084
3
votes
1
answer
76
views
Stuck with understanding transformation step in calculating limit of $n(\sqrt[n]{a}-1)$
Although this question has already been asked in general
( How to find $\lim\limits_{n\to\infty} n·(\sqrt[n]{a}-1)$?)
, my question is different, because I am stuck with a specific ...
- 307
1
vote
1
answer
105
views
Is it true that $x^n <\epsilon \Rightarrow n < \frac{\ln \epsilon}{\ln x}$?
Let:
$0 \lt x \lt 1$
$\epsilon > 0$
I need to show that there exists an $N(\epsilon,x)$ such that:
$n\ge N(\epsilon,x) \Rightarrow x^n < \epsilon$
This is what I've tried:
$x^n <\epsilon ...
- 3,600
2
votes
1
answer
624
views
Reasoning behind multiplying by conjugates
What is the reason behind multiplying by conjugates?
I am currently studying single variable calculus and throughout the lessons from the text I'm using, the reasoning as to why one would multiply by ...
user144809
3
votes
4
answers
304
views
Limit $\lim\limits_{x\to0}\frac1{\ln(x+1)}-\frac1x$
The limit is
$$\lim\limits_{x\to0}\frac1{\ln(x+1)}-\frac1x$$
The problem is I don't know if I can calculate it normally like with a change of variables or not. Keep in mind that I'm not allowed to ...
- 137