Melody

100032 apples on a tree!

That is one very productive apple tree.
If I get paid by the tree to pick the apples, I am going to pick a different tree!     LOL

@@ End of Day Wrap            Sun 19/10/14                 Sydney, Australia Time               10:50pm ♬

Hi all,

I had fun today.  It is weekend so we weren't completely swamped but there were some really good questions. 

Our fantastic answerers were CPhill, Geno3141, Bugzy, Srinu, SevenUP and Alan.    Thanks  so much.  

Here is a few interest posts:

1) Integration by parts - It was fun :)    Melody

Mon 20/10/14

1) These inequalities look interesting.    Thanks Chris and Alan.

highlight yellow question.

Yes I took a leap there.  I shall show you.   :)

Consider

$$cos^2(tan^{-1}x)\\\\$$

Remember that $$tan^{-1}x$$  is an angle

So I used this triangle

$$cos^2(tan^{-1}x)=\left(\frac{1}{\sqrt{x^2+1}}\right)^2=\frac{1}{x^2+1}$$

I derive this every time because I have a really bad memory but I think that you are supposed to memorise

$$\boxed{\frac{d}{dx}tan^{-1}x=\frac{1}{1+x^2}}$$

I wrote that     LOL

It is slang.

"Beats me"  means    "I do not know"

I was just attracting Alan's attention because I don't know  the answer.    

Beats me - Ask Alan   LOL  

2nd one

$$\\\displaystyle\lim_{x\rightarrow1}\;\frac{x-1}{lnx}\\\\
$on first inspection this is 0/0 so you can use L'Hopital's Rule$\\\\\\
=\displaystyle\lim_{x\rightarrow1}\;\frac{1}{1/x}\\\\
=\frac{1}{1/1}\\\\
=1$$

I think that the 3rd one is  

$$\displaystyle\lim_{x\rightarrow0} \;\frac{tan^{-1}x}{x}$$                   Is this correct?

Anyway we have a situation of 0/0  so we can use L'Hopital's Rule

Let

 $$\begin{array}{rll}
\alpha &=& tan^{-1}x\\
x &=& tan\alpha\\
\frac{dx}{d\alpha} &=& sec^2\alpha\\
\frac{d\alpha}{dx} &=& cos^2\alpha\\
\frac{d\alpha}{dx} &=& cos^2(tan^{-1}x)\\
\frac{d\alpha}{dx} &=& \frac{1}{1+x^2}\\
\end{array}$$

$$\displaystyle\lim_{x\rightarrow0} \;\frac{tan^{-1}x}{x}\\\\
=\displaystyle\lim_{x\rightarrow0} \;\frac{\left(\frac{1}{1+x^2}\right)}{1}\\\\
=\frac{1}{1+0^2}
=1$$

check

This is the last one - it is really easy

$$\displaystyle \lim_{x\rightarrow -1}\;\;\dfrac{|x+1|}{x^2+1}=
\dfrac{|-1+1|}{(-1)^2+1}=\dfrac{0}{1+1}=0$$ 

this is the graph if you are curious.

the first one looks straight forward.  It is 0/0 so you just use L'Hopital's Rule  

The second and third one we cannot read.   

The last one is really easy - i did it below.