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$$\begin{align}\lim x → ∞\end{align}$$

$$\begin{align} f(x) = {\frac{2^{x+1}+{3^{x+1}}}{2^x + 3^x}} \\ \end{align}$$

I tried using L Hopitable but that gives the same expression. Also tried using substitution but I didn't get anywhere. Help would be appreciated.

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I think it would be good to see an argument based on general big-$O$ principles, but I'm not remembering details well enough to write one right now. – dfeuer Oct 7 '13 at 18:55
up vote -2 down vote accepted

To begin with $$f(x)=\frac{2\cdot(\frac23)^x+3}{(\frac23)^x+1}$$ it follows that $$\lim_{x\to+\infty}f(x)=3.$$ However $$f(x)=\frac{2+3\cdot(\frac32)^x}{1+(\frac32)^x}$$ in such a case, we get that $$\lim_{x\to-\infty}f(x)=2.$$

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I think you have a serious mistake in evaluating the limit the second time. Check it ( the limit is only when $\;x\to \infty\;$ . That $\;-\infty\;$ has no business there) – DonAntonio Oct 7 '13 at 17:37
It is the second one, always has been. In complex analysis it can be as the first option, and even then it usually is defined and agreed upon. – DonAntonio Oct 7 '13 at 18:30
@ace , it is done so because then in each case you get fractions that when taking the limit become less than one and thus converge to zero: $$\lim_{x\to -\infty}\left(\frac32\right)^x=0$$ – DonAntonio Oct 7 '13 at 18:32
@ace This is not a solution. The limit DOES exist. – J. W. Perry Oct 7 '13 at 19:07
@ziangchen In what world or language has $x \rightarrow \infty$ ever meant anything other than positive infinity? – J. W. Perry Oct 7 '13 at 19:21

Here is another idea. Try writing, \begin{align*} \frac{2^{x+1}+3^{x+1}}{2^x+3^x} &= \frac{2 \cdot 2^x+3 \cdot 3^x}{2^x+3^x} \\ &=\frac{3\cdot 2^x - 2^x+3 \cdot 3^x}{2^x+3^x} \\ &=\frac{3\left( 2^x+3^x \right)-2^x}{2^x+3^x} \\ &=3-\frac{2^x}{2^x+3^x} \\ &=3-\frac{\frac{2^x}{2^x}}{\frac{2^x}{2^x}+\frac{3^x}{2^x}} \\ &=3-\frac{1}{1+\left( \frac{3}{2} \right)^x}. \end{align*} Now most people would just look at this and say $$\lim_{x \rightarrow \infty} \left( 3-\frac{1}{1+\left( \frac{3}{2} \right)^x} \right)=3.$$

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What "more rigour" is needed there? It is simply arithmetic of limits. – DonAntonio Oct 7 '13 at 18:33

Hint: For the formal argument, it is useful to divide top and bottom by $3^x$. Before doing that, it is a good idea to think about the numbers, and decide what the limit will be.

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Divide the numerator and the denominator by $3^{x+1}$.

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