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\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.

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

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

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SO what do we have now.

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

why did you switch it to an alternating series?

\(a=\dfrac 1 2\\ b = \dfrac{\sqrt{35}}{2}\)

In the figure, angle A = 28 degrees, angle B = 74 degrees and angle C = 26 degrees. 

If x and  y are the measures of the angles in which they are shown, what is the value of x+y?

\(\begin{array}{|rcll|} \hline x-A-B+y &=& C \\ x+y &=& A+B+C \\ x+y &=& 28^{\circ} + 74^{\circ} + 26^{\circ} \\ \mathbf{x+y} & \mathbf{=} & \mathbf{128^{\circ}} \\ \hline \end{array}\)

Find the sum of the base-2 geometric series
\(0.1_2-0.01_2+0.001_2-0.0001_2+0.00001_2\ldots\);
give your answer as a fraction in which the numerator and denominator are both expressed in base 10.

\(\begin{array}{|rcll|} \hline 0.1_{2} = 1\cdot 2^{-1} &=& \frac{1}{2} \\ 0.01_{2} = 1\cdot 2^{-2} &=& \frac{1}{4} \\ 0.001_{2} = 1\cdot 2^{-3} &=& \frac{1}{8} \\ 0.0001_{2} = 1\cdot 2^{-4} &=& \frac{1}{16} \\ 0.00001_{2} = 1\cdot 2^{-5} &=& \frac{1}{32} \\ \ldots \\ \hline \end{array}\)

Infinite Geometric Series:

\(\begin{array}{|rcll|} \hline s &=& \dfrac{1}{2} - \dfrac{1}{4} + \dfrac{1}{8} - \dfrac{1}{16} + \dfrac{1}{32} \pm \ldots \quad | \quad a=\frac{1}{2},~ r = -\frac{1}{2} \\\\ s &=& \dfrac{a}{1-r} \quad | \quad |r| < 1. \\\\ s &=& \dfrac{\frac{1}{2}}{1- \left(-\frac{1}{2} \right)} \\\\ s &=& \dfrac{\frac{1}{2}}{1+\frac{1}{2}} \\\\ s &=& \dfrac{\frac{1}{2}}{\frac{3}{2}} \\\\ s &=& \dfrac{1}{2} \cdot \dfrac{2}{3} \\\\ \mathbf{s} & \mathbf{=} & \mathbf{\dfrac{1}{3}} \\ \hline \end{array} \)

xxxx

What is the smallest integer n, greater than 1, such that
\(n^{-1} \pmod{130}\)
and
\(n^{-1} \pmod{231} \)
are both defined?

\(\text{$n$ must be coprime to $130$ and $231$ } \\ \text{respectively $\gcd(n,130) = \gcd(n,231) = 1$ }\)

Factorisation:

\(\begin{array}{|rcll|} \hline 130 &=& 2\times 5 \times 13 \\ 231 &=& 3\times 7 \times 11 \\ \hline \end{array}\)

\(\text{The next prime number after $2,3,5,7,11,13$ is ${\color{red}17}$ }\)

So 17 is the smallest integer, greater than 1.

\(17^{-1} \pmod {130} = 23 \\ 17^{-1} \pmod {231} = 68 \)

It means area

It means area

Correct, tertre....!!!

We wil have   

1    -16      remaining

Which translates to     1x - 16  =   x -  16

The guest keyed in the incorrect function.....the limit = 1/2

How to solve this algebraically....I don't know....[maybe a series expansion ???]

Here's graphical proof :   https://www.desmos.com/calculator/hqlnsz5lrt

9.8 is the largest decimal that can be formed given the constraints  and 2.1 is the smallest....so....their difference will be the greatest

Freq  =  1  / Period

Freq =  1 / 0.00024   =   4166.66 Hz  = 4167 Hz

A reflection across the y axis just changes the sign on the x coordinate....the y coordinate remains the same

So 

A = (2, 5)   and   A'  =  (-2, 5)

And

B = ( 6, 3)  and  B' = (-6, 3)

how???????

Sorry...I mis-read that as a multiplication....I think you wanted decimal values

The largest possible difference is

9.8  -  1.2   = 

8.6

its not 70.....

D. frequency

a*b will be largest  when  a = 9  and b = 8  [or  vice-versa]

c*d  will be smallest when  c =2 and d =1  [ or vice-versa] 

So

The largest possible difference is

9 * 8  - 2*1  =

72  -  2  =

70

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

B. rotation

Using the sum of an infinite series, we have :

a1  /  [ 1 - r ]  = the sum

Where a1 is the first term  and  r is the common difference.....so we have...

a1  / [ 1 - (-1/2) ] =  45     simplify

a1 /  [ 3/2]  =  45

a1 =  45 (3/2)

a1  =  67.5

OK.....maybe someone else can shed some light on this....!!!

Sorry, that isn't correct

Thanks, Rom!