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Math Help - Limits and Continuity Help

  1. #1
    Junior Member
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    Limits and Continuity Help

    Can anyone help me out with these 3 calculus problems?

    The first one:
    f(x) = [1-cos(x)]/(2 sin^2(x))
    lim(x-->0)f(x)
    f(x) * x/x = [1-cos(x)]/x * x/(2 sin^2(x))
    lim(x-->0)[1-cos(x)]/x = 0
    lim(x-->0) x/sin x = 1
    lim(x-->0) x/sin x = 1
    lim(x-->0) 1/(2 sin(x)) = DNE

    My teacher told us that the answer is 1/4, but I can't imagine how she got to this point.

    For both the 2nd and 3rd- Find the points of discontinuity and label them as removable or nonremovable. I am looking for confirmation, and possibly correction if I am wrong.
    The second one:
    lim(x--> neg infinity)(5x^4-9x)/(5x^3+9)
    Apply direct substitution to get (inf-(neg inf))/(-inf) = -infinity

    The third one:
    lim(x-->neg inf)sqrt(9x^2+2)/(4x+3)
    Apply direct substitution to get (infinity)/(-infinity) = -infinity

    Thanks in advance for everyone's help!
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  2. #2
    Banned
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    Quote Originally Posted by macosxnerd101 View Post
    Can anyone help me out with these 3 calculus problems?

    The first one:
    f(x) = [1-cos(x)]/(2 sin^2(x))
    lim(x-->0)f(x)
    f(x) * x/x = [1-cos(x)]/x * x/(2 sin^2(x))
    lim(x-->0)[1-cos(x)]/x = 0
    lim(x-->0) x/sin x = 1
    lim(x-->0) x/sin x = 1
    lim(x-->0) 1/(2 sin(x)) = DNE

    My teacher told us that the answer is 1/4, but I can't imagine how she got to this point.


    \color{red}\mbox {You can use L'Hospital's Rule and it follows at once, or else use a}
    \color{red}\mbox{little trigonometry:}

    \color{blue}\sin^2x = 1-cos^2x\Longrightarrow\frac{1-cosx}{2sin^2x}=\frac{1}{2(1+cosx)}


    For both the 2nd and 3rd- Find the points of discontinuity and label them as removable or nonremovable. I am looking for confirmation, and possibly correction if I am wrong.
    The second one:
    lim(x--> neg infinity)(5x^4-9x)/(5x^3+9)
    Apply direct substitution to get (inf-(neg inf))/(-inf) = -infinity



    \color{red}\lim_{x\rightarrow-\infty}\frac{5x^2-9x}{5x^3+9}=\frac{x-\frac{9}{5x^2}}{1+\frac{9}{5x^3}}
    \color{red}\mbox{Take it from here (pay attention to the fact that we factored out}\,\, 5x^3
    \color{red}\mbox{above both in the denominator and in the numerator}
    \color{red}\mbox{Do the next exercise in a simmilar way}

    \color{blue}Tonio

    The third one:
    lim(x-->neg inf)sqrt(9x^2+2)/(4x+3)
    Apply direct substitution to get (infinity)/(-infinity) = -infinity

    Thanks in advance for everyone's help!
    ..
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  3. #3
    Junior Member
    Joined
    Sep 2009
    Posts
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    Thanks for your help. I'll give #3 a try:

    sqrt(9x^2+2)/(4x+3)
    [3x/x + sqrt(2)/x] //take 3x out of the radical
    divided by (4x/x + 3/x) =
    3/4.

    Can you confirm this as the right answer?
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  4. #4
    Banned
    Joined
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    Quote Originally Posted by macosxnerd101 View Post
    Thanks for your help. I'll give #3 a try:

    sqrt(9x^2+2)/(4x+3)
    [3x/x + sqrt(2)/x] //take 3x out of the radical
    divided by (4x/x + 3/x) =
    3/4.

    Can you confirm this as the right answer?

    Yup, this is right!

    Tonio
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