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Math Help - Finding the limit

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    Finding the limit

    Im stuck trying to find the limit of (sqrt(x^2 + 12)-2*x)/(x-2) as x approaches 2
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    Re: Finding the limit

    Quote Originally Posted by dnftp View Post
    Im stuck trying to find the limit of (sqrt(x^2 + 12)-2*x)/(x-2) as x approaches 2
    \frac{\sqrt{x^2+12}-2x}{x-2}=\frac{x^2+12-4x^2}{(x-2)(\sqrt{x^2+12}+2x)}=\frac{12-3x^2}{(x-2)(\sqrt{x^2+12}+2x)}=\frac{3(4-x^2)}{(x-2)\sqrt{x^2+12}+2x}=...

    Can you proceed now?
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  3. #3
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    Re: Finding the limit

    Quote Originally Posted by dnftp View Post
    Im stuck trying to find the limit of (sqrt(x^2 + 12)-2*x)/(x-2) as x approaches 2
    Alternatively, L'Hospital's Rule works nicely in this case since it is of the indeterminate form \displaystyle \frac{0}{0}.

    \displaystyle \begin{align*} \lim_{x \to 2}\frac{\sqrt{x^2 + 12} - 2x}{x - 2} &= \lim_{x \to 2}\frac{\frac{d}{dx}\left(\sqrt{x^2 + 12} - 2x\right)}{\frac{d}{dx}\left(x - 2\right)} \\ &= \lim_{x \to 2}\frac{\frac{x}{\sqrt{x^2 + 12}}-2}{1} \\ &= \lim_{ x \to 2}\frac{x}{\sqrt{x^2 + 12}} - 2\end{align*}
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