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

1. SolomonZelman

$$\large\color{blue}{\displaystyle\lim_{x \rightarrow ~\infty}\frac{\sqrt{x+1}}{\sqrt{2x}+\sqrt{x-1}}}$$ can I interpret your question like this?

2. unimatix

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.

3. SolomonZelman

rationalize the denominator

4. welshfella

yes

5. triciaal

|dw:1441555515323:dw|

6. triciaal

|dw:1441555908335:dw|

7. triciaal

sorry this looks a bit messy. all three responses are the same . Rationalize the denominator. can you simplify the above drawing?

8. unimatix

It's fine. I'm trying to rationalize the denominator. You can simplify the denominator to X-1

9. triciaal

x + 1

10. unimatix

Yeah it would be x +1.

11. unimatix

In the numerator I have sqrt( 2x^2+2x ) - sqrt (x^2-1)

12. triciaal

|dw:1441556580501:dw|

13. triciaal

needs more

14. unimatix

I guess you could combine those to a single sqrt?

15. triciaal

|dw:1441556731148:dw|

16. triciaal

trying to work on the top

17. unimatix

not really sure how you got to what is at the bottom in black

18. triciaal

work in progress however (x-1)^2 = x^2 -2 x + 1 thinking about difference of 2 squares

19. unimatix

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?

20. unimatix

nevermind that haha

21. triciaal

@SolomonZelman that was very impressive can you simplify my numerator think difference of 2 squares then divide by (x+1)

22. freckles

Just divide top and bottom by sqrt(2H)

23. freckles

$\frac{\frac{H}{2H}+\frac{1}{2H}}{\sqrt{\frac{2H}{H}}+\sqrt{\frac{H}{2H}-\frac{1}{H}}}$

24. freckles

look at each term

25. freckles

oops forgot to right a square root thingy on top

26. freckles

but you should get the idea

27. triciaal

|dw:1441558647746:dw|

28. triciaal

@freckles can you please simplify this |dw:1441559138774:dw|

29. unimatix

@triciaal I've been following what you are doing and am stuck at the same place.

30. freckles

$\frac{\sqrt{H+1}}{\sqrt{2H}+\sqrt{H-1}} \cdot \frac{\frac{1}{\sqrt{2H}}}{\frac{1}{\sqrt{2H}}} \\ =\frac{\frac{\sqrt{H+1}}{\sqrt{2H}}}{\frac{\sqrt{2H}+\sqrt{H-1}}{\sqrt{2H}}} \\ =\frac{\sqrt{\frac{H}{2H}+\frac{1}{2H}}}{\frac{\sqrt{2H}}{\sqrt{2H}}+\sqrt{\frac{H}{2H}-\frac{1}{2H}}}$

31. freckles

now just take the limit of each term as H goes to infinity

32. triciaal

|dw:1441559208589:dw|

33. freckles

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

34. triciaal

|dw:1441559573245:dw|

35. triciaal

in the limit set equal square each side cancel common factor

36. freckles

you just have to divide top and bottom sqrt(2H) that is the key here

37. freckles

the hint was that H tends to infinity

38. freckles

@unimatix do you understand what I did above?

39. unimatix

@ freckles Your answer makes a lot of sense. What you made you decide to multiply by 1/sqrt(2H) though?

40. freckles

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

41. freckles

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

42. triciaal

|dw:1441560259041:dw|

43. freckles

$\frac{\sqrt{\frac{1}{2}}}{1+\sqrt{\frac{1}{2}}} =\frac{\frac{1}{\sqrt{2}}}{1+\frac{1}{\sqrt{2}}} \\=\frac{\frac{1}{\sqrt{2}}}{1+\frac{1}{\sqrt{2}}} \cdot \frac{\sqrt{2}}{\sqrt{2}} \\ =\frac{\frac{\sqrt{2}}{\sqrt{2}}}{\sqrt{2}+\frac{\sqrt{2}}{\sqrt{2}}}=\frac{1}{\sqrt{2}+1}$

44. triciaal

|dw:1441560392712:dw|

45. freckles

$\lim_{x \rightarrow \infty} \frac{\sqrt{x+1}}{\sqrt{2x}+\sqrt{x-1}} \text{ instead we could have divided \top and bottom by } \sqrt{x} \\ \lim_{x \rightarrow \infty}\frac{\frac{\sqrt{x+1}}{\sqrt{x}}}{\frac{\sqrt{2x}}{\sqrt{x}}+\frac{\sqrt{x-1}}{\sqrt{x}}}=\lim_{x \rightarrow \infty} \frac{\sqrt{\frac{x}{x}+\frac{1}{x}}}{\sqrt{\frac{2x}{x}}+\sqrt{\frac{x}{x}-\frac{1}{x}}}=\frac{\sqrt{1+0}}{\sqrt{2}+\sqrt{1-0}}$

46. unimatix

Thanks freckles, I was really getting frustrated! And thanks to triciaal and Solomon as well!

47. freckles

similar problem: $\lim_{x \rightarrow \infty}\frac{2x^2+x+1}{3x^2+1} \\ \lim_{x \rightarrow \infty}\frac{2x^2+x+1}{3x^2+1} \cdot \frac{\frac{1}{x^2}}{\frac{1}{x^2}} \\ \lim_{x \rightarrow \infty}\frac{\frac{2x^2}{x^2}+\frac{x}{x^2}+\frac{1}{x^2}}{\frac{3x^2}{x^2}+\frac{1}{x^2}}=\frac{2+0+0}{3+0}=\frac{2}{3}$

48. unimatix

WAHOO! I feel like I understand it now. Just have to remember to divide by polynomial with the highest power in the denominator.

49. freckles

another similar problem: $\lim_{x \rightarrow \infty}\frac{\sqrt{2x^2+x+1}}{\sqrt{3x^2+1}} \\ \\ \\\ \lim_{x \rightarrow \infty} \frac{\sqrt{2x^2+x+1}}{\sqrt{3x^2+1}} \cdot \frac{\frac{1}{\sqrt{x^2}}}{\frac{1}{\sqrt{x^2}}} \\ \lim_{x \rightarrow \infty} \frac{\sqrt{\frac{2x^2}{x^2}+\frac{x}{x^2}+\frac{1}{x^2}}}{\sqrt{\frac{3x^2}{x^2}+\frac{1}{x^2}}} \\ =\frac{\sqrt{2+0+0}}{\sqrt{3+0}}=\frac{\sqrt{2}}{\sqrt{3}}$

50. unimatix

I was able to work through it on my own and got the same answer!

51. freckles

$\lim_{x \rightarrow \infty} \frac{\sqrt{2x+1}-1}{x} \\ \text{ here we have } x>0 \text{ so } \sqrt{x^2}=x \\ \text{ divide \top and bottom by } x \\ \text{ and we will come back \in use this one fact } \\ \lim_{x \rightarrow \infty} \frac{\frac{\sqrt{2x+1}}{x}-\frac{1}{x}}{\frac{x}{x}} \\ \text{ now for the } \sqrt{x^2}=x \\ \text{ we can rewrite that one } x \\ \lim_{x \rightarrow \infty} \frac{\frac{\sqrt{2x+1}}{\sqrt{x^2}}-\frac{1}{x}}{\frac{x}{x}} \\ \lim_{x \rightarrow \infty} \frac{\sqrt{\frac{2x+1}{x^2}}-\frac{1}{x}}{\frac{x}{x}} \\ \lim_{x \rightarrow \infty} \frac{\sqrt{\frac{2x}{x^2}-\frac{1}{x^2}}-\frac{1}{x}}{\frac{x}{x}} =\frac{\sqrt{0-0}-0}{1}=\frac{0}{1}=0$

52. freckles

oops forgot to label that another example

53. freckles

and also fantastic

54. freckles

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

55. freckles

some awesome notes to read: http://tutorial.math.lamar.edu/Classes/CalcI/LimitsAtInfinityI.aspx

56. freckles

example 3 is kind of example of what I'm talking about

57. unimatix

Alright I'll read over it. Thanks so much!