Compute the standard parts: (sqrt(H+1))/(sqrt(2H)+sqrt(H-1)), H positive infinite

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Compute the standard parts: (sqrt(H+1))/(sqrt(2H)+sqrt(H-1)), H positive infinite

Calculus1
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\(\large\color{blue}{\displaystyle\lim_{x \rightarrow ~\infty}\frac{\sqrt{x+1}}{\sqrt{2x}+\sqrt{x-1}}}\) can I interpret your question like this?
I 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.
rationalize the denominator

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Other answers:

yes
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sorry this looks a bit messy. all three responses are the same . Rationalize the denominator. can you simplify the above drawing?
It's fine. I'm trying to rationalize the denominator. You can simplify the denominator to X-1
x + 1
Yeah it would be x +1.
In the numerator I have sqrt( 2x^2+2x ) - sqrt (x^2-1)
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needs more
I guess you could combine those to a single sqrt?
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trying to work on the top
not really sure how you got to what is at the bottom in black
work in progress however (x-1)^2 = x^2 -2 x + 1 thinking about difference of 2 squares
I 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?
nevermind that haha
@SolomonZelman that was very impressive can you simplify my numerator think difference of 2 squares then divide by (x+1)
Just divide top and bottom by sqrt(2H)
\[\frac{\frac{H}{2H}+\frac{1}{2H}}{\sqrt{\frac{2H}{H}}+\sqrt{\frac{H}{2H}-\frac{1}{H}}}\]
look at each term
oops forgot to right a square root thingy on top
but you should get the idea
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@freckles can you please simplify this |dw:1441559138774:dw|
@triciaal I've been following what you are doing and am stuck at the same place.
\[\frac{\sqrt{H+1}}{\sqrt{2H}+\sqrt{H-1}} \cdot \frac{\frac{1}{\sqrt{2H}}}{\frac{1}{\sqrt{2H}}} \\ =\frac{\frac{\sqrt{H+1}}{\sqrt{2H}}}{\frac{\sqrt{2H}+\sqrt{H-1}}{\sqrt{2H}}} \\ =\frac{\sqrt{\frac{H}{2H}+\frac{1}{2H}}}{\frac{\sqrt{2H}}{\sqrt{2H}}+\sqrt{\frac{H}{2H}-\frac{1}{2H}}}\]
now just take the limit of each term as H goes to infinity
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since 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
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in the limit set equal square each side cancel common factor
you just have to divide top and bottom sqrt(2H) that is the key here
the hint was that H tends to infinity
@unimatix do you understand what I did above?
@ freckles Your answer makes a lot of sense. What you made you decide to multiply by 1/sqrt(2H) though?
this 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
and 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
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\[\frac{\sqrt{\frac{1}{2}}}{1+\sqrt{\frac{1}{2}}} =\frac{\frac{1}{\sqrt{2}}}{1+\frac{1}{\sqrt{2}}} \\=\frac{\frac{1}{\sqrt{2}}}{1+\frac{1}{\sqrt{2}}} \cdot \frac{\sqrt{2}}{\sqrt{2}} \\ =\frac{\frac{\sqrt{2}}{\sqrt{2}}}{\sqrt{2}+\frac{\sqrt{2}}{\sqrt{2}}}=\frac{1}{\sqrt{2}+1}\]
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\[\lim_{x \rightarrow \infty} \frac{\sqrt{x+1}}{\sqrt{2x}+\sqrt{x-1}} \text{ instead we could have divided \top and bottom by } \sqrt{x} \\ \lim_{x \rightarrow \infty}\frac{\frac{\sqrt{x+1}}{\sqrt{x}}}{\frac{\sqrt{2x}}{\sqrt{x}}+\frac{\sqrt{x-1}}{\sqrt{x}}}=\lim_{x \rightarrow \infty} \frac{\sqrt{\frac{x}{x}+\frac{1}{x}}}{\sqrt{\frac{2x}{x}}+\sqrt{\frac{x}{x}-\frac{1}{x}}}=\frac{\sqrt{1+0}}{\sqrt{2}+\sqrt{1-0}}\]
Thanks freckles, I was really getting frustrated! And thanks to triciaal and Solomon as well!
similar problem: \[\lim_{x \rightarrow \infty}\frac{2x^2+x+1}{3x^2+1} \\ \lim_{x \rightarrow \infty}\frac{2x^2+x+1}{3x^2+1} \cdot \frac{\frac{1}{x^2}}{\frac{1}{x^2}} \\ \lim_{x \rightarrow \infty}\frac{\frac{2x^2}{x^2}+\frac{x}{x^2}+\frac{1}{x^2}}{\frac{3x^2}{x^2}+\frac{1}{x^2}}=\frac{2+0+0}{3+0}=\frac{2}{3}\]
WAHOO! I feel like I understand it now. Just have to remember to divide by polynomial with the highest power in the denominator.
another similar problem: \[\lim_{x \rightarrow \infty}\frac{\sqrt{2x^2+x+1}}{\sqrt{3x^2+1}} \\ \\ \\\ \lim_{x \rightarrow \infty} \frac{\sqrt{2x^2+x+1}}{\sqrt{3x^2+1}} \cdot \frac{\frac{1}{\sqrt{x^2}}}{\frac{1}{\sqrt{x^2}}} \\ \lim_{x \rightarrow \infty} \frac{\sqrt{\frac{2x^2}{x^2}+\frac{x}{x^2}+\frac{1}{x^2}}}{\sqrt{\frac{3x^2}{x^2}+\frac{1}{x^2}}} \\ =\frac{\sqrt{2+0+0}}{\sqrt{3+0}}=\frac{\sqrt{2}}{\sqrt{3}}\]
I was able to work through it on my own and got the same answer!
\[\lim_{x \rightarrow \infty} \frac{\sqrt{2x+1}-1}{x} \\ \text{ here we have } x>0 \text{ so } \sqrt{x^2}=x \\ \text{ divide \top and bottom by } x \\ \text{ and we will come back \in use this one fact } \\ \lim_{x \rightarrow \infty} \frac{\frac{\sqrt{2x+1}}{x}-\frac{1}{x}}{\frac{x}{x}} \\ \text{ now for the } \sqrt{x^2}=x \\ \text{ we can rewrite that one } x \\ \lim_{x \rightarrow \infty} \frac{\frac{\sqrt{2x+1}}{\sqrt{x^2}}-\frac{1}{x}}{\frac{x}{x}} \\ \lim_{x \rightarrow \infty} \frac{\sqrt{\frac{2x+1}{x^2}}-\frac{1}{x}}{\frac{x}{x}} \\ \lim_{x \rightarrow \infty} \frac{\sqrt{\frac{2x}{x^2}-\frac{1}{x^2}}-\frac{1}{x}}{\frac{x}{x}} =\frac{\sqrt{0-0}-0}{1}=\frac{0}{1}=0\]
oops forgot to label that another example
and also fantastic
these 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
some awesome notes to read: http://tutorial.math.lamar.edu/Classes/CalcI/LimitsAtInfinityI.aspx
example 3 is kind of example of what I'm talking about
Alright I'll read over it. Thanks so much!

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