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anonymous

  • 5 years ago

how can I delete indetermination in the following expression. Limx(x->0)((sqrt(x+2)-sqrt(2))/x)

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  1. anonymous
    • 5 years ago
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    \[limit _{x->0}(\sqrt{x+2}-\sqrt{2})/2\] this is the equation

  2. anonymous
    • 5 years ago
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    conjugation

  3. anonymous
    • 5 years ago
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    limitx−>0(x+2−−−−−√−2−−√)/x Sorry

  4. anonymous
    • 5 years ago
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    you mean by multiplyting the numerator and the denominator by \[\sqrt{x+2}+\sqrt{2}\]

  5. anonymous
    • 5 years ago
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    Top and bottom

  6. anonymous
    • 5 years ago
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    ok let me try

  7. anonymous
    • 5 years ago
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    Changuanas, I actually get rid of indetermination. However when I apply the limit x->0, the answer is not the corrrect. The answer int he book is 1/2sqrt(2).

  8. anonymous
    • 5 years ago
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    Sorry. I was wrong. I already got the answer! Can you help me with another exercise? Just to check

  9. anonymous
    • 5 years ago
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    just post question, if fall asleep someone else would help

  10. anonymous
    • 5 years ago
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    \[Limit _{x->infinity} \sqrt{3x ^{4}-2x+4}+x\]

  11. anonymous
    • 5 years ago
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    I don't remember what happens to infinity under square root. Does it go to infinity?

  12. anonymous
    • 5 years ago
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    According to the graph, it increases with no bound

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  13. anonymous
    • 5 years ago
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    Ignore all others and consider only highest exponent of x. What is highest exp?

  14. anonymous
    • 5 years ago
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    I'm sorry. Did you already have an answer: infinity? or it needs more work?

  15. anonymous
    • 5 years ago
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    yes, more work :)

  16. anonymous
    • 5 years ago
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    I think the answer is 1

  17. anonymous
    • 5 years ago
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    what is highest exponent x in your problem?

  18. anonymous
    • 5 years ago
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    inside the root? its 2

  19. anonymous
    • 5 years ago
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    Ignore all others divide sq rt 3x^4 by sq rt x^4

  20. anonymous
    • 5 years ago
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    Wouldn't that one have to be just infinity?

  21. anonymous
    • 5 years ago
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    I think he is in an early class. The class before you study infinity, right Emun?

  22. anonymous
    • 5 years ago
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    yes

  23. anonymous
    • 5 years ago
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    In that case, don't ignore everything else, that is what you do in higher math;keep everything and divide each term by x^2

  24. anonymous
    • 5 years ago
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    It's funny being all done finals but still here during my break lol. why am I here?

  25. anonymous
    • 5 years ago
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    except that expression isn't anything you would need to do that for. maybe if you were dividing by a polynomial

  26. anonymous
    • 5 years ago
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    You here to help Emun I'm going to sleep.

  27. anonymous
    • 5 years ago
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    guys the equation was wrong. The equation is the following: \[\sqrt{x ^{2}-2x+4}+x\] I factored the expression inside the root out and I got \[\sqrt{(x-2)^{2}}+x\] so finally I got \[x-2+x\] as I apply the limit as x=infinity,the answer I got is 2

  28. anonymous
    • 5 years ago
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    Thanks changaunas :)

  29. anonymous
    • 5 years ago
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    The answer would be infinity, wouldn't it?

  30. anonymous
    • 5 years ago
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    wouldn't you get 2x-2 , which as x becomes large.. your expression does too? infinity!

  31. anonymous
    • 5 years ago
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    Yeah and this is the answer: Thanks gusy

  32. anonymous
    • 5 years ago
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    thanks guys :)

  33. anonymous
    • 5 years ago
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    \[\lim_{x \rightarrow 0} (\sqrt{x+2}-\sqrt{2})/x \times (\sqrt{x+2}+\sqrt{2})/(\sqrt{x+2}+\sqrt{2})\] \[\lim_{x \rightarrow 0} (x+2-2)/(x(\sqrt{x+2}+\sqrt{2}))\] \[\lim_{x \rightarrow 0} x/(x(\sqrt{x+2}+\sqrt{2}))\] \[\lim_{x \rightarrow 0} 1/(\sqrt{x+2}+\sqrt{2})\] \[1/(\sqrt{0+2}+\sqrt{2})\] \[1/(\sqrt{2}+\sqrt{2})\] \[1/(2\sqrt{2})\]

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