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$$\lim _{x\to \infty \:}\left(\frac{2^x-3^x}{3^x+4^x}\right)$$

When substituting, it has the form

$$\frac {\infty-\infty}{\infty+\infty}$$

I think it has to do with L'Hopital rule, But I couldn't put in an acceptable form $\dfrac{\infty}{\infty}$ or $\dfrac{ 0}0$

I think also, since the numerator is a difference and the denominator is addition, the limit will go to zero? Is this "rough way" useful?

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  • $\begingroup$ your rough way isn't useful . For example if you had $2^x-4^x$ in the numerator, and $3^x+4^x$ in the denominator, the the limit would be $-1$ and not $0$. $\endgroup$
    – voldemort
    Dec 19 '14 at 23:33
  • $\begingroup$ The more useful "rough way" is that the $4^x$ in the denominator will dominate as $x$ grows. $\endgroup$
    – Arthur
    Dec 19 '14 at 23:34
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    $\begingroup$ $2^x - 3^x \operatorname*{\sim}_{x\to\infty} -3^x$, $3^x + 4^x \operatorname*{\sim}_{x\to\infty} 4^x$, so $\frac{2^x - 3^x}{3^x + 4^x}\operatorname*{\sim}_{x\to\infty} \frac{- 3^x}{4^x} = -\left(\frac{3}{4}\right)^x$. $\endgroup$
    – Clement C.
    Dec 19 '14 at 23:34
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The general principle that I recommend is to divide top and bottom by the dominant term of the denominator. In this case, $4^x$: $$\frac{2^x-3^x}{3^x+4^x}=\frac{(\frac{2}{4})^x-(\frac{3}{4})^x}{(\frac{3}{4})^x-1}$$ Now, $\lim_{x\to\infty}(\frac{2}{4})^x=0$, $\lim_{x\to\infty}(\frac{3}{4})^x=0$ so the limit is $$\frac{0-0}{0-1}=0$$

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    $\begingroup$ Very elegant!$\!\!\!\!$ $\endgroup$
    – flawr
    Dec 19 '14 at 23:40
  • $\begingroup$ @flawr: This is essentially same as my answer- which I had posted earlier, without making it so explicit, so that the OP has to think a bit. $\endgroup$
    – voldemort
    Dec 19 '14 at 23:43
  • $\begingroup$ @vadim123: It's clear that you don't understand what I wrote. It's ok that you are bounty hunting. My method is essentially same, and I wrote it befor you. Plus, you also have to evaluate two limits. $\endgroup$
    – voldemort
    Dec 19 '14 at 23:48
  • $\begingroup$ @vadim123: and now you have resorted to ugly downvoting. Sure- just shows your "class". Happy bounty hunting. $\endgroup$
    – voldemort
    Dec 19 '14 at 23:51
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$$\lim_{x\to \infty} \frac{2^x-3^x}{3^x+4^x}$$ $$=\lim_{x\to \infty} \frac{(2/3)^x-1}{1+(4/3)^x}$$ $$=\frac{-1}{\infty}$$ $$=0$$

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The best way to do this is to factor out $3^x$ from the numerator, and $4^x$ from the denominator. Then $\bigg(\dfrac{3}{4}\bigg)^x \rightarrow 0$, and the rest of the fraction goes to $-1.$

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  • $\begingroup$ Can the downvoter please leave a comment as to why the oldest answer was downvoted? $\endgroup$
    – voldemort
    Dec 19 '14 at 23:52
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Hint: $$\frac{2^x-3^x}{3^x+4^x}=\frac{2^x}{3^x+4^x}-\frac{3^x}{3^x+4^x}$$ and consider each term separately.

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