Melody

Thanks Chris.

I reallly like that way. It is so short and simple.

Thanks guest.:)

Hi Chris, I expect your logic is excellent but I am having trouble following it.

You have just done the same (sort of) as Tertre and me haven't you?

I mean you have developed a pattern and used the pattern (odd and even powers) to answer the question.

I am just a little confused because you have not shown the pattern but I am assuming that is what etc means.

\lim_{x\longrightarrow\infty}\frac{\sqrt{x^4-10x}-\sqrt{x^4-5x^2+7}}{5}

\(\displaystyle\lim_{x\rightarrow\infty}\;\frac{\sqrt{x^4-10x}-\sqrt{x^4-5x^2+7}}{5}\\ =\displaystyle\lim_{x\rightarrow\infty}\;\frac{ \left(\sqrt{x^4-10x}-\sqrt{x^4-5x^2+7}\right) \left(\sqrt{x^4-10x}+\sqrt{x^4-5x^2+7}\right)}{5 \left(\sqrt{x^4-10x}+\sqrt{x^4-5x^2+7}\right)}\\ =\displaystyle\lim_{x\rightarrow\infty}\;\frac{ x^4-10x-(x^4-5x^2+7) }{5 \left(\sqrt{x^4-10x}+\sqrt{x^4-5x^2+7}\right)}\\ =\displaystyle\lim_{x\rightarrow\infty}\;\frac{ 5x^2 -10x-7 }{5 \left(\sqrt{x^4-10x}+\sqrt{x^4-5x^2+7}\right)}\\ =\displaystyle\lim_{x\rightarrow\infty}\;\frac{ 5x^2 -10x-7 }{5 \left(\sqrt{x^4-10x}\right)\left(1+\frac{\sqrt{x^4-5x^2+7}}{\sqrt{x^4-10x}}\right)}\\ \)

\( =\displaystyle\frac{1}{5}\times\displaystyle\lim_{x\rightarrow\infty}\;\frac{     5x^2  -10x-7                           }{ \left(\sqrt{x^4-10x}\right)} \times\;\;\displaystyle\lim_{x\rightarrow\infty}\;\frac{  1}{\left(1+\frac{\sqrt{x^4-5x^2+7}}{\sqrt{x^4-10x}}\right)}\\ \)

NOW I will look at each of these limits seperately.

\(\displaystyle\lim_{x\rightarrow\infty}\;\frac{     5x^2  -10x-7    }{ \left(\sqrt{x^4-10x}\right)}\\ =\displaystyle\lim_{x\rightarrow\infty}\;\sqrt{\frac{     (5x^2  -10x-7 )^2   }{ x^4-10x}}\\ \text{expanding gives}\\ =\displaystyle\lim_{x\rightarrow\infty}\;\sqrt{\frac{   25x^4-100x^3+30x^2+140x+49   }{ x^4-10x}}\\ \text{Dividing top and bottom by x^4 we get}\\ =\displaystyle\lim_{x\rightarrow\infty}\;\sqrt{\frac{   25-\frac{100}{x}+\frac{30}{x^2}+\frac{140}{x^3}+\frac{49}{x^4}   }{ 1-\frac{10}{x^3}}}\\ =\sqrt{25}\\ =5\\ \text{-------------------------------------------------------}\)

\(\displaystyle\lim_{x\rightarrow\infty}\;\frac{  1}{\left(1+\frac{\sqrt{x^4-5x^2+7}}{\sqrt{x^4-10x}}\right)}\\~\\ =\frac{  1}{\displaystyle\lim_{x\rightarrow\infty}\;\left(1+\frac{\sqrt{x^4-5x^2+7}}{\sqrt{x^4-10x}}\right)}\\ =\frac{  1}{\displaystyle\lim_{x\rightarrow\infty}\;\left(\frac{\sqrt{x^4-5x^2+7}}{\sqrt{x^4-10x}}\right)+1}\\ =\frac{  1}{\displaystyle\lim_{x\rightarrow\infty}\;\sqrt{\left(\frac{x^4-5x^2+7}{x^4-10x}\right)}+1}\\ =\frac{  1}{\displaystyle\lim_{x\rightarrow\infty}\;\sqrt{\left(\frac{1-\frac{5}{x^2}+\frac{7}{x^4}}{1-\frac{10}{x^3}}\right)}+1}\\ =\frac{  1}{\displaystyle\lim_{x\rightarrow\infty}\;\sqrt{\left(\frac{1-\frac{5}{x^2}+\frac{7}{x^4}}{1-\frac{10}{x^3}}\right)}+1}\\ =\displaystyle \frac{1}{2}\)

-------------------------------------------------------------

SO what do we have now.

\(\displaystyle\lim_{x\rightarrow\infty}\;\frac{\sqrt{x^4-10x}-\sqrt{x^4-5x^2+7}}{5}= \frac{1}{5} \times5\times \frac{1}{2}=\boxed{\frac{1}{2}}\)

Yes, CPhill is right, I entered the wrong function.   Sorry.

1    5,   

2    25,

3    125,           odd

4    625,             even

5    3125,           odd

6    15625           even

7     78125             odd

8    390625           even

ok I have a pattern

5^137 = 1000n+ 125 = 125*8*n  +   17*8     +  1  =  8(1000n+17) +1       n has a specific integer value

Yes I made a small stupid numeric error that i am quite sure hatchet288 was smart enough to correct.

This last line should have been

5^137 = 1000n+ 125 = 125*8*n  +   15*8     + 5  =  8(1000n+15) +5       n has a specific integer value

so

\(5^{137}\div 8 \)        has a remainder of 5

It was no big deal, any question asker should have been able to pick up that little error for themsleves.

BUT Good effort Tertre, it was a good catch.

this is the question.

(2,40)   is most likely a number plane co-ordinates

It means start at the point (0,0), go left 2 places (positive x direction) then go up 40 places (positive y direction) put a cross and that marks the point (2,40)

This is the question,

What process could you use to solve equations of the form (x + a)(x + b) = 0? In this case there are only two factors (values being multiplied) that are being set equal to 0. How would you solve the equation if there were more than two factors? Explain.

What you have here is 

(x + a) * (x + b) = 0

The ONLY way that two things can multiply together to give zero is if ONE or BOTH of them IS ZERO.

so

x+a=0            or                 x+b=0

Now you can 'digest' that properly and then finish the question.