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Could anyone tell me if this is correct?

$$ \lim_{x\to 0} \frac{e^{2x}-\pi^{x}}{sin(3x)} = \lim_{x\to 0} \frac{2e^{2x}-\pi^{x} ln(\pi)}{3cos(3x)} = \frac{2e^{2(0)}-\pi^{(0)} ln(\pi)}{3cos(3(0))} = \frac{2\cdot 1-0}{3\cdot1} = \frac{2}{3} $$

I'm getting a bit of a different result on Wolfram|Alpha

If it's incorrect, mind telling me where I went wrong? Thanks!

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2 Answers

up vote 3 down vote accepted

Note that $\log(\pi) \neq 0$. You need not resort to L'Hospital's rule for this limit. You can get it from first principles. $$\dfrac{e^{2x} - \pi^x}{\sin(3x)} = \dfrac{3x}{\sin(3x)} \times \dfrac{(e^{2x} - 1) - (\pi^x-1)}{3x}\\ = \dfrac{3x}{\sin(3x)} \times \left(\dfrac{e^{2x} - 1}{2x} \times \dfrac23 - \dfrac{e^{x\log(\pi)}-1}{x \log(\pi)} \times \dfrac{\log(\pi)}{3} \right)$$ Now recall the following limits $$\lim_{y \to 0} \dfrac{\sin(y)}{y} = 1$$ $$\lim_{y \to 0} \dfrac{e^{y}-1}{y} = 1$$ Hence, you get the limit as $$1 \times \left( 1 \times \dfrac23 - 1 \times \dfrac{\log(\pi)}3\right) = \dfrac{2 - \log(\pi)}3$$

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Ouch..I don't know why I assumed \pi^0 would be zero. Thanks for the thorough explanation I learnt from it. :) –  eee3 Oct 27 '12 at 19:15
    
It would seem that L'hopital would be easier and a lot more straightforward. –  Mike Oct 27 '12 at 21:13
    
@Mike For L'Hopital, you need additional machinery like derivatives etc, which in turn rely on the limits from first principles. If a problem is easier from first principles, then I prefer that way as opposed to L'Hopital. –  user17762 Oct 27 '12 at 21:15
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Your error is here: $$\pi^0\ln(\pi)\neq0$$

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