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Math Help - [SOLVED] Limit Problem

  1. #1
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    [SOLVED] Limit Problem

    Hi I was wondering if someone could help me with this limit problem:

    Find the lim x->a-, lim x->a+ and lim x->a
    for the f(x)=(x^2-16)/(squareroot(x^2-8x+16)) a=4

    I've tried to simplify the problem but after that i dont know wut to do =(
    This is wut I've done:
    y=(x^2-16)/(squareroot(x-4)^2)
    y=(x^2-16)/abs(x-4)

    any help would be appreciated.
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  2. #2
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    Hello,

    Quote Originally Posted by chukie View Post
    Hi I was wondering if someone could help me with this limit problem:

    Find the lim x->a-, lim x->a+ and lim x->a
    for the f(x)=(x^2-16)/(squareroot(x^2-8x+16)) a=4

    I've tried to simplify the problem but after that i dont know wut to do =(
    This is wut I've done:
    y=(x^2-16)/(squareroot(x-4)^2)
    y=(x^2-16)/abs(x-4)

    any help would be appreciated.
    When x->4-, x-4 is negative. Therefore abs(x-4)=-(x-4).
    Then, simplify with x-16=(x-4)(x+4)

    When x->4+, x-4 > 0. Therefore abs(x-4)=(x-4).
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  3. #3
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    Quote Originally Posted by chukie View Post
    Hi I was wondering if someone could help me with this limit problem:

    Find the lim x->a-, lim x->a+ and lim x->a
    for the f(x)=(x^2-16)/(squareroot(x^2-8x+16)) a=4

    I've tried to simplify the problem but after that i dont know wut to do =(
    This is wut I've done:
    y=(x^2-16)/(squareroot(x-4)^2)
    y=(x^2-16)/abs(x-4)

    any help would be appreciated.
    The numerator is a difference of squares, so you should be able to factor it:

    \frac{x^2-16}{\sqrt{x^2 - 8x + 16}}

    =\frac{\left(x + 4\right)\left(x - 4\right)}{\sqrt{\left(x - 4\right)^2}}

    =\left(x + 4\right)\,\frac{x-4}{|x - 4|}

    Now, when you take the limit, notice that x - 4 is always negative when x<4 (which will be the case for the left-handed limit) and nonnegative when x\geq4 (which will be the case for the right-handed limit). So, get rid of the absolute value bars (negating when appropriate) and simplify.
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  4. #4
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    thanks for ur help in the previous question. i was also wondering if the absolute value function was in the numerator, wut sign would i need to take for the lim->a-, lim ->a+?
    Example:
    f(x)=abs(x^2+5x+6)/(x^2-9)

    i did the same thing as u guys hv instructed me for the other question by taking a negative value of the absolute value of the numerator wen lim->a- and positive wen lim->a+. then i was able to simplify. i was able to get a number for limits from the left and from the right. but wen i graph the function on my calculator i see that the answer is supposed to be negative infinity for the left and positive infinity for the right. im a bit confused
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  5. #5
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    Quote Originally Posted by chukie View Post
    thanks for ur help in the previous question. i was also wondering if the absolute value function was in the numerator, wut sign would i need to take for the lim->a-, lim ->a+?
    Example:
    f(x)=abs(x^2+5x+6)/(x^2-9)
    \frac{|x^2+5x+6|}{x^2-9}

    = \frac{|(x + 2)(x + 3)|}{(x + 3)(x - 3)}

    = \frac{|x + 2||x + 3|}{(x + 3)(x - 3)}

    Now, the absolute value will depend on the values that x is taking. As x\to a-, is x + 3 positive or negative? What about x + 2 and x - 3? How about as x\to a+? This will all, of course, depend on the value of a.

    When the expression is positive, you can remove the absolute bars, and when it is negative you can remove the bars as long as you negate the expression.
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  6. #6
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    Quote Originally Posted by Reckoner View Post
    \frac{|x^2+5x+6|}{x^2-9}

    = \frac{|(x + 2)(x + 3)|}{(x + 3)(x - 3)}

    = \frac{|x + 2||x + 3|}{(x + 3)(x - 3)}

    Now, the absolute value will depend on the values that x is taking. As x\to a-, is x + 3 positive or negative? What about x + 2 and x - 3? How about as x\to a+? This will all, of course, depend on the value of a.

    When the expression is positive, you can remove the absolute bars, and when it is negative you can remove the bars as long as you negate the expression.
    so if a=3
    then from wut u mean wen the lim->3-, the equation will become ((x+2)(x+3))/((x+3)(x-3)) since the absolute value expression will be positive? then simplifying it will become (x+2)/(x-3)?
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    Quote Originally Posted by chukie View Post
    so if a=3
    then from wut u mean wen the lim->3-, the equation will become ((x+2)(x+3))/((x+3)(x-3)) since the absolute value expression will be positive? then simplifying it will become (x+2)/(x-3)?
    That is correct.
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