The limit will be pi/2

see the graph

I like you manners  

Collin gets 3/10 so there is 7/10 left

Dave gets 2 shares out of 5 shares of what is left.

this means that he gets 2 fifths of 7 tenths

$$\left({\frac{{\mathtt{2}}}{{\mathtt{5}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{7}}}{{\mathtt{10}}}}\right) = {\frac{{\mathtt{7}}}{{\mathtt{25}}}} = {\mathtt{0.28}}$$

Anyway Dave ends up it 7 twenty fifths  or  28% of the entire amount.  

You question looks a bit fishy.    

How can you have

 $$\begin{pmatrix}
1 \\
k
\end{pmatrix}$$

The big one has to be on the top.  

You can't 'solve' it.

There is no equal sign!

Mon 27/10/14

1) This one looks really interesting.  Thanks Chris and anonymous.

@@ End of Day Wrap          Sun 26/10/14           Sydney, Australia            Time 9:50pm      ♬

Hello everybody,

Fantastic answers today from Geno3141, Kitty<3 and Alan.

I just pulled a few interest questions:

1) Vectors.   Thanks Alan

$$\\LHS\\
=(cis60)^{1991}+(cis(-60))^{1991}+1\\
=(e^{60i})^{1991}+(e^{-60i})^{1991}+1\\
=(e^{60*1991*i})+(e^{-60*{1991}*i})+1\\
=(cos(60*1991)+isin(60*1991))+(cos(-60*1991)+isin(-60*1991))+1\\
Now\;\; cos(-\theta)=cos(\theta)\;\;and\;\;sin(-\theta)=-sin(\theta)\\
so\\
=cos(60*1991)+cos(-60*1991)+isin(60*1991)+isin(-60*1991)+1\\
=cos(60*1991)+cos(-60*1991)+i[sin(60*1991)+sin(-60*1991)]+1\\
=cos(60*1991)+cos(60*1991)+i[sin(60*1991)-sin(60*1991)]+1\\
=2cos(60*1991)+i[0]+1\\
=2cos(60*1991)+1\\
=2cos(60*[6*331+5])+1\\
=2cos(60*6*331+60*5])+1\\
=2cos(360*331+60*5)+1\\
=2cos(60*5)+1\\
=2cos(300)+1\\
=2\times\frac{1}{2}+1\\
=1+1\\$$

$$\\=2\\
\ne 1\\
$The equation is not true$$$

So I agree with Alan's answer.    

Alan's answer is here.

Thank you very much Alan.

I whited the question because it was repeated on the next page (this page) and also I had re-thought my question.

Gino has answered here

I don't understand the question

What is 

4*exp(-5x^2)