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unimatix
 one year ago
Compute the standard parts:
(sqrt(H+1))/(sqrt(2H)+sqrt(H1)), H positive infinite
unimatix
 one year ago
Compute the standard parts: (sqrt(H+1))/(sqrt(2H)+sqrt(H1)), H positive infinite

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SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0\(\large\color{blue}{\displaystyle\lim_{x \rightarrow ~\infty}\frac{\sqrt{x+1}}{\sqrt{2x}+\sqrt{x1}}}\) can I interpret your question like this?

unimatix
 one year ago
Best ResponseYou've already chosen the best response.0I believe that would be the same. Kind of trying to teach myself. The answer is given as 1 / ( 1 + sqrt(2) ) OR sqrt(2)  1. I don't know how to solve it though.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.0rationalize the denominator

triciaal
 one year ago
Best ResponseYou've already chosen the best response.0dw:1441555515323:dw

triciaal
 one year ago
Best ResponseYou've already chosen the best response.0dw:1441555908335:dw

triciaal
 one year ago
Best ResponseYou've already chosen the best response.0sorry this looks a bit messy. all three responses are the same . Rationalize the denominator. can you simplify the above drawing?

unimatix
 one year ago
Best ResponseYou've already chosen the best response.0It's fine. I'm trying to rationalize the denominator. You can simplify the denominator to X1

unimatix
 one year ago
Best ResponseYou've already chosen the best response.0Yeah it would be x +1.

unimatix
 one year ago
Best ResponseYou've already chosen the best response.0In the numerator I have sqrt( 2x^2+2x )  sqrt (x^21)

triciaal
 one year ago
Best ResponseYou've already chosen the best response.0dw:1441556580501:dw

unimatix
 one year ago
Best ResponseYou've already chosen the best response.0I guess you could combine those to a single sqrt?

triciaal
 one year ago
Best ResponseYou've already chosen the best response.0dw:1441556731148:dw

triciaal
 one year ago
Best ResponseYou've already chosen the best response.0trying to work on the top

unimatix
 one year ago
Best ResponseYou've already chosen the best response.0not really sure how you got to what is at the bottom in black

triciaal
 one year ago
Best ResponseYou've already chosen the best response.0work in progress however (x1)^2 = x^2 2 x + 1 thinking about difference of 2 squares

unimatix
 one year ago
Best ResponseYou've already chosen the best response.0I don't really understand how in the third step you swapped the place sqrt(2x) with the sqrt(x+1). Wouldn't that give you a numerator not equal to the one before?

triciaal
 one year ago
Best ResponseYou've already chosen the best response.0@SolomonZelman that was very impressive can you simplify my numerator think difference of 2 squares then divide by (x+1)

freckles
 one year ago
Best ResponseYou've already chosen the best response.2Just divide top and bottom by sqrt(2H)

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[\frac{\frac{H}{2H}+\frac{1}{2H}}{\sqrt{\frac{2H}{H}}+\sqrt{\frac{H}{2H}\frac{1}{H}}}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2oops forgot to right a square root thingy on top

freckles
 one year ago
Best ResponseYou've already chosen the best response.2but you should get the idea

triciaal
 one year ago
Best ResponseYou've already chosen the best response.0dw:1441558647746:dw

triciaal
 one year ago
Best ResponseYou've already chosen the best response.0@freckles can you please simplify this dw:1441559138774:dw

unimatix
 one year ago
Best ResponseYou've already chosen the best response.0@triciaal I've been following what you are doing and am stuck at the same place.

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[\frac{\sqrt{H+1}}{\sqrt{2H}+\sqrt{H1}} \cdot \frac{\frac{1}{\sqrt{2H}}}{\frac{1}{\sqrt{2H}}} \\ =\frac{\frac{\sqrt{H+1}}{\sqrt{2H}}}{\frac{\sqrt{2H}+\sqrt{H1}}{\sqrt{2H}}} \\ =\frac{\sqrt{\frac{H}{2H}+\frac{1}{2H}}}{\frac{\sqrt{2H}}{\sqrt{2H}}+\sqrt{\frac{H}{2H}\frac{1}{2H}}}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2now just take the limit of each term as H goes to infinity

triciaal
 one year ago
Best ResponseYou've already chosen the best response.0dw:1441559208589:dw

freckles
 one year ago
Best ResponseYou've already chosen the best response.2since everybody seems really confused on to do this... we have H tends to infinity \[\frac{\sqrt{\frac{H}{2H}+\frac{1}{2H}}}{\frac{\sqrt{2H}}{\sqrt{2H}}+\sqrt{\frac{H}{2H}\frac{1}{2H}}}=\frac{\sqrt{\frac{1}{2}+0}}{1+\sqrt{\frac{1}{2}0}}\] you can play with this answer to make it prettier

triciaal
 one year ago
Best ResponseYou've already chosen the best response.0dw:1441559573245:dw

triciaal
 one year ago
Best ResponseYou've already chosen the best response.0in the limit set equal square each side cancel common factor

freckles
 one year ago
Best ResponseYou've already chosen the best response.2you just have to divide top and bottom sqrt(2H) that is the key here

freckles
 one year ago
Best ResponseYou've already chosen the best response.2the hint was that H tends to infinity

freckles
 one year ago
Best ResponseYou've already chosen the best response.2@unimatix do you understand what I did above?

unimatix
 one year ago
Best ResponseYou've already chosen the best response.0@ freckles Your answer makes a lot of sense. What you made you decide to multiply by 1/sqrt(2H) though?

freckles
 one year ago
Best ResponseYou've already chosen the best response.2this is similar to having fractions where both numerator and denominator are polynomials when you have a polynomial over a polynomial you divide by the term on bottom that has highest exponent for the numerator and denominator that is of course when you are doing the whole x tends to positive or negative infinity this question is very similar except you do not have polynomials on top and bottom in a sense sqrt(H) was the term we are talking about I chose to divide by sqrt(2H) but sqrt(H) would work just as well or faster

freckles
 one year ago
Best ResponseYou've already chosen the best response.2and what I mean by faster is that you will see your answer more clearly because in the previous answer I gave you have to multiply top nad bottom by sqrt(2) to see you have the same answer

triciaal
 one year ago
Best ResponseYou've already chosen the best response.0dw:1441560259041:dw

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[\frac{\sqrt{\frac{1}{2}}}{1+\sqrt{\frac{1}{2}}} =\frac{\frac{1}{\sqrt{2}}}{1+\frac{1}{\sqrt{2}}} \\=\frac{\frac{1}{\sqrt{2}}}{1+\frac{1}{\sqrt{2}}} \cdot \frac{\sqrt{2}}{\sqrt{2}} \\ =\frac{\frac{\sqrt{2}}{\sqrt{2}}}{\sqrt{2}+\frac{\sqrt{2}}{\sqrt{2}}}=\frac{1}{\sqrt{2}+1}\]

triciaal
 one year ago
Best ResponseYou've already chosen the best response.0dw:1441560392712:dw

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[\lim_{x \rightarrow \infty} \frac{\sqrt{x+1}}{\sqrt{2x}+\sqrt{x1}} \text{ instead we could have divided \top and bottom by } \sqrt{x} \\ \lim_{x \rightarrow \infty}\frac{\frac{\sqrt{x+1}}{\sqrt{x}}}{\frac{\sqrt{2x}}{\sqrt{x}}+\frac{\sqrt{x1}}{\sqrt{x}}}=\lim_{x \rightarrow \infty} \frac{\sqrt{\frac{x}{x}+\frac{1}{x}}}{\sqrt{\frac{2x}{x}}+\sqrt{\frac{x}{x}\frac{1}{x}}}=\frac{\sqrt{1+0}}{\sqrt{2}+\sqrt{10}}\]

unimatix
 one year ago
Best ResponseYou've already chosen the best response.0Thanks freckles, I was really getting frustrated! And thanks to triciaal and Solomon as well!

freckles
 one year ago
Best ResponseYou've already chosen the best response.2similar problem: \[\lim_{x \rightarrow \infty}\frac{2x^2+x+1}{3x^2+1} \\ \lim_{x \rightarrow \infty}\frac{2x^2+x+1}{3x^2+1} \cdot \frac{\frac{1}{x^2}}{\frac{1}{x^2}} \\ \lim_{x \rightarrow \infty}\frac{\frac{2x^2}{x^2}+\frac{x}{x^2}+\frac{1}{x^2}}{\frac{3x^2}{x^2}+\frac{1}{x^2}}=\frac{2+0+0}{3+0}=\frac{2}{3}\]

unimatix
 one year ago
Best ResponseYou've already chosen the best response.0WAHOO! I feel like I understand it now. Just have to remember to divide by polynomial with the highest power in the denominator.

freckles
 one year ago
Best ResponseYou've already chosen the best response.2another similar problem: \[\lim_{x \rightarrow \infty}\frac{\sqrt{2x^2+x+1}}{\sqrt{3x^2+1}} \\ \\ \\\ \lim_{x \rightarrow \infty} \frac{\sqrt{2x^2+x+1}}{\sqrt{3x^2+1}} \cdot \frac{\frac{1}{\sqrt{x^2}}}{\frac{1}{\sqrt{x^2}}} \\ \lim_{x \rightarrow \infty} \frac{\sqrt{\frac{2x^2}{x^2}+\frac{x}{x^2}+\frac{1}{x^2}}}{\sqrt{\frac{3x^2}{x^2}+\frac{1}{x^2}}} \\ =\frac{\sqrt{2+0+0}}{\sqrt{3+0}}=\frac{\sqrt{2}}{\sqrt{3}}\]

unimatix
 one year ago
Best ResponseYou've already chosen the best response.0I was able to work through it on my own and got the same answer!

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[\lim_{x \rightarrow \infty} \frac{\sqrt{2x+1}1}{x} \\ \text{ here we have } x>0 \text{ so } \sqrt{x^2}=x \\ \text{ divide \top and bottom by } x \\ \text{ and we will come back \in use this one fact } \\ \lim_{x \rightarrow \infty} \frac{\frac{\sqrt{2x+1}}{x}\frac{1}{x}}{\frac{x}{x}} \\ \text{ now for the } \sqrt{x^2}=x \\ \text{ we can rewrite that one } x \\ \lim_{x \rightarrow \infty} \frac{\frac{\sqrt{2x+1}}{\sqrt{x^2}}\frac{1}{x}}{\frac{x}{x}} \\ \lim_{x \rightarrow \infty} \frac{\sqrt{\frac{2x+1}{x^2}}\frac{1}{x}}{\frac{x}{x}} \\ \lim_{x \rightarrow \infty} \frac{\sqrt{\frac{2x}{x^2}\frac{1}{x^2}}\frac{1}{x}}{\frac{x}{x}} =\frac{\sqrt{00}0}{1}=\frac{0}{1}=0\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2oops forgot to label that another example

freckles
 one year ago
Best ResponseYou've already chosen the best response.2these can get tricky when dividing top and bottom by sqrt(x^2) because sqrt(x^2)=x when x>0 and sqrt(x^2)=x when x<0 just something to remember when you come face to face with it I'm sure you will

freckles
 one year ago
Best ResponseYou've already chosen the best response.2some awesome notes to read: http://tutorial.math.lamar.edu/Classes/CalcI/LimitsAtInfinityI.aspx

freckles
 one year ago
Best ResponseYou've already chosen the best response.2example 3 is kind of example of what I'm talking about

unimatix
 one year ago
Best ResponseYou've already chosen the best response.0Alright I'll read over it. Thanks so much!
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