Question

Just trying out... (the math there is just for me see the entirety of my latex)

Answer 1

\(\large\color{black}{ \displaystyle \left.\int_{a}^{b} f(x)={\rm F}(x)\right|^{b}_{a}={\rm F}(b)-{\rm F}(a) }\)

Answer 2

Successful!
```
\(\large\color{black}{ \displaystyle \left.\int_{a}^{b} f(x)
={\rm F}(x)\right|^{b}_{a}={\rm F}(b)-{\rm F}(a) }\)
```
\(\color{red}{\text{\left.}}\) integral, blah blah \(\color{red}{\text{\right|}}\)
\(\color{black}{\Uparrow}\)
(Placed \(\color{red}{\text{\left.}}\) before the integral, and that
works to make the line higher to make it look nicer)

Answer 3

@SolomonZelman Can you teach me how to do this if you have some time?

Answer 4

I am not that good :)

Answer 5

It seems hard to do haha, so you seemed to do really good

Answer 6

\(\large\color{#A44999}{ \displaystyle \oint F(x){~\sf d}{\bf \ell}}\)

Nice colors,
http://www.w3schools.com/tags/ref_colorpicker.asp
choose or make up.

Again I am just trying out things on here

Answer 7

\(\large{\bbox[4px,#fff1ff ,border:2px solid red ]{\color{#660033}{ \left.\begin{matrix} ~~~x &~~ & i^x~ \\
\\[0.9em]
~~4k & | & i^{4k}=(i^4)^x=1^x=\color{blue}{1} \\[0.9em]
~~4k+1 & | & i^{4k+1}=i^{4k}\cdot i^{1} =1\cdot i =\color{blue}{i}\\[0.9em]
~~4k+2& | & i^{4k+2}=i^{4k} \cdot i^{2}=1\cdot(-1)=\color{blue}{-1}\\[0.9em]
~~ 4k+3 & | & i^{4k+3}=i^{4k} \cdot i^{3}=i^{4k} \cdot i^{2}\times i^{9} \\[0.9em]&&=1\cdot(-1)\cdot i =\color{blue}{-i} \\[0.9em]
\end{matrix}\right. }}}\)

Answer 8

\(\large\color{#003366}{\displaystyle\lim_{x \rightarrow ~\infty }\left[\frac{x}{\ln(x!)}\right] }\)

\(\large\color{#003366}{\displaystyle\lim_{x \rightarrow ~\infty }\left[\frac{x\ln(e)}{\ln(x!)}\right] }\)

\(\large\color{#003366}{\displaystyle\lim_{x \rightarrow ~\infty }\left[\frac{\ln(e^x)}{\ln(x!)}\right] }\)

\(\large\color{#003366}{\displaystyle \ln \left[\lim_{x \rightarrow ~\infty }\frac{e^x}{x!}\right]=0 }\)

Answer 9

\(\large\color{#660033}{\displaystyle\lim_{x \rightarrow ~\infty }\left[\frac{\left( \ln x\right)^{k+2}}{x}\right]\quad ;\quad k\in \mathbb{N} }\)

\(\large\color{#660033}{\displaystyle\lim_{x \rightarrow ~\infty }\left[\frac{~\frac{d}{dx}~\left[\left( \ln x\right)^{k+2} \right]~}{\frac{dx}{dx}}\right] }\)

\(\large\color{#660033}{\displaystyle\lim_{x \rightarrow ~\infty }\left[\frac{~(k+2)\left( \ln x\right)^{k+1}~\times \dfrac{1}{x} }{1}\right] }\)

\(\large\color{#660033}{\displaystyle\lim_{x \rightarrow ~\infty }\left[\frac{~(k+2)\left( \ln x\right)^{k+1} }{x}\right]}\)

and differentiate a (k+2) number of times...
and this is how it goes.

\(\large\color{#339966}{\displaystyle\lim_{x \rightarrow ~\infty }\left[\frac{~\left( \ln x\right)^{k+2} }{x}\right]}\)

\(\large\color{#339966}{\displaystyle\lim_{x \rightarrow ~\infty }\left[\frac{~\dfrac{d^{k+2}}{dx^{k+2}}~\left( \ln x\right)^{k+2} }{1}\right]}\)

\(\large\color{#339966}{\displaystyle\lim_{x \rightarrow ~\infty }\left[\frac{(k+2)! }{x}\right]=0}\)

Answer 10

the colors are awesome...

(squeeze trm to prove this for non integer k,

suppose that there is a positive noninteger S,
and a positive integer A that is greater than S,
such that: k<k+S<k+A
and since when k+A is in the power of ln x,
and when k is in the power of ln x, the limit
is 0, therefore when any nuninteger positive number is in the power of ln, the limit is 0)

Answer 11

just in case, for math completion.