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thats nice .. thank you alan and melody .. 

is my answer here correct ? 

@@ 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

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

Another without L'Hopital's rule (though you need to know the expansion of ln(1-z)):

(x-1)/ln(x)

Let x = 1-z  so  (x-1)/ln(x) becomes -z/ln(1 - z)

As x → 1 , z → 0

Expand ln(1 - z) in an infinite series:  ln(1 - z) = -z - z²/2 - z³/3 - z⁴/4 - ...

so -z/ln(1 - z) = 1/(1 + z/2 + z²/3 + z³/4 + ...)

When z = 0 this becomes 1/(1 + 0 + 0 + ...) = 1

-z/ln(1 - z) tends to 1 as z tends to 0.

Hence (x-1)/ln(x) tends to 1 as x tends to 1

Edited to take the correct limit!

.

Some can be done without L'Hopitals's rule.  For example:

  (x² - 1)/(x² - 2x +1)  =  (x - 1)(x + 1)/(x - 1)²  =  (x + 1)/(x - 1)  As x goes to 1 from above this goes to +∞; as x goes to 1 from below it goes to -∞.

When x is small then tan(x) → x, so tan^-1(x) → x  (x in radians).  Therefore tan^-1(x)/x → 1

.

Also how do you get this ?

which I highlighted

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.    

Thank you alan but what do you mean by beats me ? 

Beats me - Ask Alan   LOL  

The trace of a matrix is the sum of the diagonal elements.  It is a property of the matrix that can sometimes be useful; for example, it is equal to the sum of all the eigenvalues of the characteristic polynomial of the matrix.  

so there are no way to solve these problems without L'Hopital's Rule  right  ? 

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

cos² + sin² = 1 so the left-hand side of your inequality reduces to -4i, where i is the square root of -1

The right-hand side simplifies to -12u so you have

-4i > -12u

This doesn't make a lot of sense if u is a real number, but if it is imaginary, let's say u = iw, where w is real, then:

-4i < -12iw

w < 4/12

w < 1/3

.

sorry for the pictuer  .. I think this is better  

I want the answers for these questions 

and this is the last question you read it wrong .. 

tell me if my answer is correct 

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.

yes its very clear .. 

what about the rest of the questions ? 

43*4.2 metres

That is assuming that the 43m/s is a horizontal speed and that the ball stays in the air for the whole 4.2 seconds.

It also ignores air resistance, wind etc   

sin-1 is the same as arcsine

so you use 

asin

Note that

4 = 2^2  and 8 = 2^3

So

(2^2)^(3/2) = 2^3

(4)^(3/2) = 2^3

4^(3/2) = 8

$$\sqrt3(-16xy^6)\\\\
=-16\sqrt3xy^6\\\\\\
$or do you mean$\\\\
\sqrt{3(-16xy^6)}\\\\
=\sqrt{-3*16xy^6}\\\\
=4y^3\sqrt{-3x}\\\\
=4y^3\sqrt{3x}\;i\\\\
\;$i is the imaginary number $\sqrt{-1}$$

$$log_3\left(\dfrac{1}{3^1\;3^{0.5}}\right)+2\\\\
=log_3\left(\dfrac{1}{3^{1.5}}\right)+2\\\\
=log_3\left(3^{-1.5}\right)+2\\\\
=-1.5log_3\left(3\right)+2\\\\
=-1.5*1+2\\\\
=-1.5+2\\\\
=0.5$$

Let me add someting here.....it IS possible for a graph to cross its horizontal asymptote but it is NOT POSSIBLE for it to cross a vertical asymptote.