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anonymous

  • 5 years ago

lim x approaches infinity of (Square root of x^2 +2)/ x-1

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  1. anonymous
    • 5 years ago
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    found a good video that helped with this problem http://www.youtube.com/watch?v=GCJxNkw7-Ec&feature=related

  2. amistre64
    • 5 years ago
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    lim x-->inf sqrt(x^2 +2)/x-1 Since x^2 +2 cannot be factored into "real" numbers it is a prime. Since nothing can be canceled out from top to bottom, there are no "holes" in the graph. sqrt(x^2 +2)/x-1 As x reaches infinity we get an indeterminite value: so I am wondering if we can apply L-Hopitals rule: lim x-->inf (f(x)/g(x)) = lim x-->inf (f'(x)/g'(x)).

  3. anonymous
    • 5 years ago
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    i just divided the top and bottoome byt x and got square root of x^2/x which is equal to absolute value of x divided by x and got 1 as the answer.

  4. anonymous
    • 5 years ago
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    *bottom

  5. amistre64
    • 5 years ago
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    That is the algebra way. The calculus way is also good to know :)

  6. anonymous
    • 5 years ago
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    what is it?

  7. amistre64
    • 5 years ago
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    It is called the L-Hopital rule and you find the derivative of the top and the derivative of the bottom. The derivative of the bottom (x-1) = 1 so that gets rid of that. But the top derives to (x/sqrt(x^2 +2)) which doesnt seem to help us out ... then again, I am still trying to learn it good :)

  8. anonymous
    • 5 years ago
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    that seems pretty simple. thanks for the explanation

  9. anonymous
    • 5 years ago
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    i have another problem that is simple, but i'm not sure how to solve it lim as x approaches -infinity of x^2+2/x-1

  10. anonymous
    • 5 years ago
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    and i've got to solve it algebraically

  11. amistre64
    • 5 years ago
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    algebraicaly you want to multiply this equation by "1" or rather some form of "1" that is useful. in order to do that; you take the highest power of "x" from the bottom and use it like this: (1/x)/(1/x) and this equals "1" When we multiply the top we get: (x^2)/x + (2/x) = x+(2/x). And the bottom becomes: (x/x) + (1/x) = 1+(1/x) As the value of "x" gets very big; the fractions with an "x" in the bottom get very very tiny so that they basically have no effect in the equations so we can throw them away. We are left with x/1. As x--> infinity; the limit is a diagonal line of (y=x). Some people call it an oblique asymptote or a slany asymptote.

  12. amistre64
    • 5 years ago
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    slant not "slany"... i got fat fingers lol

  13. anonymous
    • 5 years ago
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    lol, thanks alot u make much more sense than my professor

  14. amistre64
    • 5 years ago
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    thanx :) There is a general rule for finding the limits. Do you know them?

  15. anonymous
    • 5 years ago
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    nope

  16. amistre64
    • 5 years ago
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    lol...thats fine too

  17. anonymous
    • 5 years ago
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    my calc class is based on youtube and this website

  18. amistre64
    • 5 years ago
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    I thought I typed "want to know them" ... my mistake. Yeah, my calc class is based on what I have taught myself as well.

  19. anonymous
    • 5 years ago
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    no problem, sure, i'd like to know em'

  20. amistre64
    • 5 years ago
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    here they are: In the case that the highest exponent of x is in the bottom, for example: x+1/x^2 -2 When we multiply by the highest exponent of "x" from teh bottom we get: (1/x^2)/(1/x^2) the only numbers we care about here will be the stuff that does not end up with an x in the bottom of it. the top: (x+1)(1/x^2) = (x/x^2)+(1/x^2) = (1/x)+(1/x^2) all these have an "x" in the bottom of the fraction; so they all go to 0 right?. The bottom part: (x^2 -2)(1/x^2) = (x^2/x^2)-(2/x^2) = (1)-(2/x^2). The only number we have here that matters is the (1) because anything with an "x" in the bottom goes to "0". What we end up with is 0/1. So, when the highest exponent of "x" is in the bottom part of the original equation; the limit will always go to: 0/1 = 0

  21. amistre64
    • 5 years ago
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    If the highest exponent of "x" is the same on the top as it is on the bottom: everything goes away except for the coeeficients of the highest exponent of x. Example: (5x^2 +1)/(2x^2 -9x +12)... all we care about is (5x^2)/(2x^2) = 5/2 is your limit

  22. anonymous
    • 5 years ago
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    awesome, thanks alot. this will defintely help me out.

  23. amistre64
    • 5 years ago
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    the other case is this: the highest eponent of "x" is in the top part of the fraction. Example: (x^3 + (other things which dont matter)) / (x + (more stuff that doesnt matter). = (x^2)/x = x. Your limit as x--> infinity will equal the graph of y=x.

  24. anonymous
    • 5 years ago
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    for the second question, i asked you, wouldn't the limit just be infinity, since it equals x and x is approaching infinity.

  25. amistre64
    • 5 years ago
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    that should be (x^2 + (other things....))

  26. amistre64
    • 5 years ago
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    If we were only concerned about straight the horizontal asymptote, then YES. But, we can still predict what the graph will act like. And it acts like the graph for y=x; or some other curve. This type of curved limit is called an oblique asymptote. And it simply means that the limit is inifity but we know how the graph will act near infinity. Makes sense?

  27. anonymous
    • 5 years ago
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    somewhat, lol

  28. anonymous
    • 5 years ago
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    i think i got the jist of it though

  29. amistre64
    • 5 years ago
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    As you do more of them, you will do just fine :)

  30. anonymous
    • 5 years ago
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    thanks for all the help, i really appreciate it

  31. anonymous
    • 5 years ago
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    :-)

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