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hpfan101

  • one year ago

\[\lim_{x \rightarrow \infty}(\sqrt{x^2-ax}-\sqrt{x^2+bx}\]

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  1. jim_thompson5910
    • one year ago
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    Hint: Multiply top and bottom by the conjugate \[\large \lim_{x \rightarrow \infty}\left(\sqrt{x^2-ax}-\sqrt{x^2+bx}\right)\] \[\large \lim_{x \rightarrow \infty}\left[\left(\sqrt{x^2-ax}-\sqrt{x^2+bx}\right) \color{red}{\times \frac{\sqrt{x^2-ax}+\sqrt{x^2+bx}}{\sqrt{x^2-ax}+\sqrt{x^2+bx}}}\right]\]

  2. anonymous
    • one year ago
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    wow am i wrong!

  3. hpfan101
    • one year ago
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    Alright so i multiplied by the conjugate and when I simplified I got \[\frac{ ax-bx }{ x \sqrt{ax}+x \sqrt{bx} }\]

  4. hpfan101
    • one year ago
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    Apparently the answer is \[\frac{ 1 }{ 2 }(a-b)\]

  5. jim_thompson5910
    • one year ago
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    you should get `-ax - bx` in the numerator when you expand things out

  6. hpfan101
    • one year ago
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    Though I don't know how to get to that

  7. hpfan101
    • one year ago
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    Oh ok, I'll change that

  8. jim_thompson5910
    • one year ago
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    do you see how I got `-ax` ?

  9. hpfan101
    • one year ago
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    I'm assuming when you multiplied everything out by the conjugate. Though in my work, I don't see that. I must have made an error

  10. hpfan101
    • one year ago
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    Oh never mind I see what I did wrong

  11. hpfan101
    • one year ago
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    Can the equation be simplified further?

  12. DanJS
    • one year ago
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    Calc 1 is 5% Calc shortcuts and 95% ALgebraaaa

  13. jim_thompson5910
    • one year ago
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    I would factor out -x. Then multiply top and bottom by 1/x \[\Large \frac{-ax-bx}{\sqrt{x^2-ax}+\sqrt{x^2+bx}}\] \[\Large \frac{-x(a+b)}{\sqrt{x^2-ax}+\sqrt{x^2+bx}}\] \[\Large \color{red}{\frac{1/x}{1/x}\times}\frac{-x(a+b)}{\sqrt{x^2-ax}+\sqrt{x^2+bx}}\] \[\Large \frac{(1/x)*-x(a+b)}{(1/x)*\left(\sqrt{x^2-ax}+\sqrt{x^2+bx}\right)}\] \[\Large \frac{-1(a+b)}{(1/x)*\sqrt{x^2-ax}+(1/x)*\sqrt{x^2+bx}}\] Do you see how to finish up?

  14. hpfan101
    • one year ago
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    Yes I do. Thank you!

  15. jim_thompson5910
    • one year ago
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    ok great

  16. DanJS
    • one year ago
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  17. DanJS
    • one year ago
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    you doing the Delta-epsilon definition thing yet? I just reviewed that

  18. hpfan101
    • one year ago
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    Not yet. I just learned in class about the Intermediate Value Theorem. But thanks for the cheat sheet!

  19. IrishBoy123
    • one year ago
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    are you sure that is the right answer? \(\lim_{x \rightarrow \infty}(\sqrt{x^2-ax}-\sqrt{x^2+bx})\) \(=\lim_{x \rightarrow \infty}x(\sqrt{1-\frac{a}{x}}-\sqrt{1+\frac{b}{x}})\) using binomial expansion \(\sqrt{1+x} = 1 + \frac{1}{2}x - \frac{1}{8}x^2 + \frac{1}{16}x^3 - \frac{5}{128}x^4 + \frac{7}{256}x^5 - \cdots\)\ \(=lim_{x \rightarrow \infty}x( \{ 1 +\frac{1}{2}(-\frac{a}{x}) - \frac{1}{8}(-\frac{a}{x})^2 + \cdots \} - \{1 +\frac{1}{2}(\frac{b}{x}) - \frac{1}{8}(\frac{b}{x})^2\})\) \(= -\frac{1}{2}(a+b)\)

  20. IrishBoy123
    • one year ago
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    i mean this |dw:1443186003890:dw|

  21. DanJS
    • one year ago
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    It has the negative in front, -(a+b)/2 From here \[\Large \frac{-x(a+b)}{\sqrt{x^2-ax}+\sqrt{x^2+bx}}\] take an x out of both roots in bottom and cancel it with the top to get \[\Large \frac{-(a+b)}{\sqrt{1-\frac{ a }{ x }}+\sqrt{1+\frac{ b }{ x }} }\] As x goes to infinity, those fractions go to zero a/x and b/x Left with the limit at infinity as \[\Large \frac{-(a+b)}{\sqrt{1}+\sqrt{1}} = \frac{-(a+b)}{2}\]

  22. DanJS
    • one year ago
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    ** \[\Large \lim_{x \rightarrow \infty } \frac{ 1 }{ x } = 0\]

  23. IrishBoy123
    • one year ago
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    yes @DanJS , just a typo i expect:-)

  24. IrishBoy123
    • one year ago
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    thx

  25. DanJS
    • one year ago
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    what did you do, that looks like the taylor series expansion of that roots

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