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I'm searching for another way to solve $\lim\limits_{x \to 0} \frac{9^x - 4^x}{2^x - 3^x}$.

I used L'Hospital's rule making it $\frac{\ln9(9^x) - \ln4(4^x)}{\ln2(2^x) - \ln3(3^x)}$

This gave me an answer of -2. I'm searching for another way I can do this problem. I tried multiplying by the conjugate of both the top and bottom but it always becomes 0. Does anyone have any suggestions on another way I can do this? Am I missing something fundamental and obvious about limits?

Martin Sleziak
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Daniel Kunkler
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1 Answers1

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$$ \begin{align} \lim_{x\to 0}\frac{9^x-4^x}{2^x-3^x}&=\lim_{x\to 0}\frac{(3^x-2^x)(3^x+2^x)}{2^x-3^x}\\ &=\lim_{x\to 0}\frac{-(2^x-3^x)(3^x+2^x)}{2^x-3^x}\\ &=\lim_{x\to 0} -(3^x+2^x)\\ &=-2\\ \end{align}$$

Hamada Al
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Franklin Pezzuti Dyer
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  • Does it depend on the specific choice of the numbers? What if one replaces 9 with, say, 10? – Vim Nov 08 '17 at 01:27
  • 10 stills has a square root. It just isn't an integer (or even rational). – Nicolas Miari Nov 08 '17 at 03:34
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    ...But of course, for the cancellation to work you would need to also change the 3 in the denominator to sqrt(10). – Nicolas Miari Nov 08 '17 at 03:37
  • @Vim Nilknarf has noticed that the numerator can be factored out into terms that include the denominator. This allows cancelling the denominator. If the numerator had different values, it might not be possible to factor into terms that include the denominator. If not, then the cancellation can't be performed, and you'll need a different approach. – jpmc26 Nov 08 '17 at 05:43