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Compare 5/(t+2) with 5/t for various values of t:

t        5/(t+2)                  5/t

1          1.67                       5

10        0.42                       0.5

100     0.049                      0.05

1000   0.00499                  0.005

...

10⁶    4.99999*10^-6           5*10^-6

10¹⁰  4.999999999*10^-10  5*10^-10

You can see that as t gets bigger, 5/(t+2) becomes closer and closer to 5/t, so it's in this sense that t+2 becomes more and more like t as t tends to infinity.

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$$\\3y+5(55)x^3\\\\
=3y+275x^3$$

Is that the question you intended?

$$\frac{4p^5}{2p^2}=\frac{4ppppp}{2pp}=2p^3$$

Those are certainly huge numbers but I am still not really sure what you are after.

When numbers become that big they become incomprehensible to a normal human brain.

By that big, I mean too big to write meaningfully in scientific notation.

Numbers such as   $$10^{333333333333333^{9999999999999999999}}$$     are meaningless to me.

I mean i know it is a huge number but a full understanding of the emensity is beyond me.      

Sorry.     

ok

I have shown you that 

-2*4   =-8

now if I divide both sides by -2,  I get

$$\\\frac{-2*4}{-2}=\frac{-8}{-2}\\\\
4=\frac{-8}{-2}\\\\
\frac{-8}{-2}=4$$

therefore a negative divided by a negative = a positive

confirm that F(x)= (1/4)x^4lin(x) - (-1/16)x^4+12 is the anti-derivative for F(x)=x^3ln(x) for values of x>0

$$\\F(x)=\frac{x^4ln(x)}{4}-\;\frac{-1x^4}{16}+12\\\\
F(x)=\frac{x^4ln(x)}{4}\;+\;\frac{x^4}{16}+12\\\\
F'(x)=\frac{4x^3ln(x)+\frac{x^4}{x}}{4}\;+\;\frac{4x^3}{16}\\\\
F'(x)=\frac{4x^3ln(x)+x^3}{4}\;+\;\frac{1x^3}{4}\\\\
F'(x)=\frac{4x^3ln(x)+2x^3}{4}\\\\
F'(x)=\frac{2x^3ln(x)+x^3}{2}\\\\
F'(x)=x^3ln(x)+\frac{x^3}{2}$$

If the original question had (+1/16) instead of (-1/16) it would have worked.

Then again, maybe I did something wrong  

Yes, Melody. Neat. But the question was about division! 

Thank you Alan and Chris 

Nice thank you Alan.  

But can you please explain to me your method 

First what do you mean by

" As t tend to infinity t+2 tends to t " 

@@ End of Day Wrap       Thurs 27/11/14       Sydney,   Australia        Time 9:00 pm          ♬

Hi everyone,

Today was a little quieter, perhaps because the 27th Nov  is the Thanksgiving Holiday in the US.

There were ofcourse many great answers.  

These were supplied by Baballama, CPhill, Alan, afroninja5959, Anonymous4338, SelenaPubiT and gliu1.

Gino3141 has final taken possession of an icon.  I love your happy rhino Gino.   

I only collected one interest post:

1) Another limit to memorise      Thanks Alan

Fri 28/11/14

1) Fractional powers of negative numbers.    Thanks Alan

This question does not make any sense to me.  

this site might be useful too :)

If you need to solve by hand then use the quadratic formula

Use the calc on the home page:)

I am not a all sure what you want.

Can't you gather at lease one of the examples and we can talk about it.    

2*x^2-(9*x)+4

$$\\2x^2-9x+4\\\\$$

You need to find 2 numbers that multiply to 2*4=8  and add togeter to give -9

That is easy  -8 and -1

Now split -9x into  -8x - 1x

like this

$$\\2x^2-8x-1x+4\\\\$$

now group them in pairs like this

$$\\2x(x-4)-1(x-4)\\\\$$

Now you have 2x lots of (x-4) take away 1 lot of (x-4)

that is 

$$\\(2x-1)(x-4)\\\\$$

Hi Alan,

Yes I did not think that L'Hopital's rule helped either.

I was just pointing out that the question requested it and it was not used.  That is all.   :)

2x+3=31  subtract 3 from both sides

2x = 28    divide both sides by 2

x = 14

√125 = √(5*25) =  5√5

.

x^2+2x-1=9     subtract 9 from both sides

x^2 + 2x - 10 = 0      this can't be factored......let's just use the onsite calculator to solve

$${{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{10}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\sqrt{{\mathtt{11}}}}{\mathtt{\,-\,}}{\mathtt{1}}\\
{\mathtt{x}} = {\sqrt{{\mathtt{11}}}}{\mathtt{\,-\,}}{\mathtt{1}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = -{\mathtt{4.316\: \!624\: \!790\: \!355\: \!399\: \!8}}\\
{\mathtt{x}} = {\mathtt{2.316\: \!624\: \!790\: \!355\: \!399\: \!8}}\\
\end{array} \right\}$$

Write this as x² +2x - 10 = 0 and use the quadratic formula.

$${{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{10}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\sqrt{{\mathtt{11}}}}{\mathtt{\,-\,}}{\mathtt{1}}\\
{\mathtt{x}} = {\sqrt{{\mathtt{11}}}}{\mathtt{\,-\,}}{\mathtt{1}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = -{\mathtt{4.316\: \!624\: \!790\: \!355\: \!399\: \!8}}\\
{\mathtt{x}} = {\mathtt{2.316\: \!624\: \!790\: \!355\: \!399\: \!8}}\\
\end{array} \right\}$$

.

Melody....I think the difference in our answers lies in the fact that you and I have assumed that the base of the pillar lies on the line y = -2  while heureka has assumed that it lies on the x axis......either orientation could be correct...