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Calculate $\lim\limits_{x\to\infty}\left(\dfrac{x-2}{x+2}\right)^{3x}$.

I need to solve this without using L'Hôpital's rule. I tried setting $y= x+2$ to get the limit to the form: $(1-\frac{2}{y})^{3(y-2)}$ and trying to relate it to the definition of $e$, but I'm unsure where to go from there.

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$$\left[\lim_{x\to\infty}\left(1+\frac{-4}{x+2}\right)^{\dfrac{x+2}{-4}}\right]^{\lim_{x\to\infty}\dfrac{-4\cdot3x}{x+2}}$$

Now for the inner limit use $\lim_{m\to\infty}\left(1+\dfrac1m\right)^m=e$

For the exponent $\lim_{x\to\infty}\dfrac{-4\cdot3x}{x+2}=-12\cdot\lim_{x\to\infty}\dfrac1{1+2/x}=\cdots$

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  • $\begingroup$ Thanks! I understood it now. One question tho, how do you justify "distributing" the limit? Is there a property I can quote? $\endgroup$ – YoTengoUnLCD Mar 23 '15 at 18:19
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    $\begingroup$ The property is continuity. Limits distribute over continuous functions. $\endgroup$ – Alan Apr 4 '15 at 6:42
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$$\left(\frac{x-2}{x+2}\right)^{3x}=\left[\left(1-\frac4{x+2}\right)^{x+2}\right]^3\cdot\left(1-\frac4{x+2}\right)^{-6}\xrightarrow[x\to\infty]{}\left(e^{-4}\right)^3\cdot1=e^{-12}$$

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