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Math Help - Limit of a function as x tends to real number

  1. #1
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    Limit of a function as x tends to real number

    a) lim ( x tends to 2 ) 1 / ( x - 2 )
    b) lim (x tends to 2) 1/ (x-2)^2

    I would be grateful if you can help me with this, providing clear explanations

    Thanks in advance!
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  2. #2
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    Quote Originally Posted by ose90 View Post
    a) lim ( x tends to 2 ) 1 / ( x - 2 )
    b) lim (x tends to 2) 1/ (x-2)^2

    I would be grateful if you can help me with this, providing clear explanations

    Thanks in advance!
    As x tends to 2, x - 2 tends to 0 (obviously), and so will (x - 2)^2. In other words, the denominator gets extremely small, and as the denominator gets smaller, \frac{1}{denominator} gets bigger.

    So the answer to both is infinity.
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  3. #3
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    Good explanation , thanks!

    However, the answer shown in my notes for a) is no limit, is it obviously wrong?
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  4. #4
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    Quote Originally Posted by ose90 View Post
    Good explanation , thanks!

    However, the answer shown in my notes for a) is no limit, is it obviously wrong?
    Yes.
    Consider x to 2, with x>2 (from the right). Then x-2>0.
    So it will be positive infinity.

    Consider x to 2, with x<2 (from the left). Then x-2<0.
    So it will be negative infinity.

    Thus \lim_{\substack{x \to 2 \\ x>2}} \quad \frac{1}{x-2} \neq \lim_{\substack{x \to 2 \\ x<2}} \quad \frac{1}{x-2}

    This means the limit doesn't exist, because the limit is not the same from the right and from the left.
    Last edited by Moo; September 3rd 2008 at 01:23 AM.
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  5. #5
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    Quote Originally Posted by Prove It View Post
    As x tends to 2, x - 2 tends to 0 (obviously), and so will (x - 2)^2. In other words, the denominator gets extremely small, and as the denominator gets smaller, \frac{1}{denominator} gets bigger.

    So the answer to both is infinity.
    It's essential that left hand and right and right hand limits are always considered.

    It's not good enough to say x - 2 --> 0 etc.

    You need to consider x \rightarrow 2^+ and x \rightarrow 2^-, whether the limiting value of x - 2 etc. is 0^+ or 0^- and consequently whether the left and right handed limits are +oo or -oo ....
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  6. #6
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    Thanks everyone, I have understood.
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