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Thread: Limits

  1. #1
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    Limits

    Evaluate lim x->0 [√(1+x) -1]/x

    when I sub in 0 i get 0/0

    I am trying to simply the expression or cancel terms but I am having no luck. the answer at the back of the book is 1/2.
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  2. #2
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    Re: Limits

    Quote Originally Posted by edwardkiely View Post
    Evaluate lim x->0 [√(1+x) -1]/x

    when I sub in 0 i get 0/0

    I am trying to simply the expression or cancel terms but I am having no luck. the answer at the back of the book is 1/2.
    Rationalize the numerator ...

    $\dfrac{\sqrt{1+x}-1}{x} \cdot \dfrac{\sqrt{1+x}+1}{\sqrt{1+x}+1}$

    ... then determine the limit.
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  3. #3
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    Re: Limits

    when I sub 0 into the simplified version I get 0
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    Re: Limits

    Quote Originally Posted by skeeter View Post
    Rationalize the numerator ...

    $\dfrac{\sqrt{1+x}-1}{x} \cdot \dfrac{\sqrt{1+x}+1}{\sqrt{1+x}+1}$

    ... then determine the limit.
    $\dfrac{\sqrt{1+x}-1}{x} \cdot \dfrac{\sqrt{1+x}+1}{\sqrt{1+x}+1} = \dfrac{(1+x)-1}{x(\sqrt{1+x}+1} = \dfrac{1}{\sqrt{1+x}+1}$

    $\lim \limits_{x\to 0} \dfrac{1}{\sqrt{1+x}+1} = \dfrac{1}{\sqrt{1+0}+1} = \dfrac 1 2$
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  5. #5
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    Re: Limits

    \dfrac{\sqrt{1+x}-1}{x} \cdot \dfrac{\sqrt{1+x}+1}{\sqrt{1+x}+1} = \frac{ (1+x)-1}{x\left(\sqrt{1+x}+1\right)} = \frac {x}{x\left(\sqrt{1+x}+1\right)} = \frac {1}{\sqrt{1+x}+1}
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