# Limits with trig, log functions and variable exponents

Would someone mind verifying this?

$$\lim_{x\to \infty} \frac{2 \cdot 3^{5x} + 5}{3^{5x} + 2^{5x}} = \lim_{x\to \infty} \frac{3^{5x}(2 + \frac{5}{3^{5x}})}{3^{5x}(1 + (\frac{2}{3})^{5x})} = \lim_{x\to \infty} \frac{2 + \frac{5}{3^{5x}}}{1 + (\frac{2}{3})^{5x}} = \frac{2 + \frac{5}{3^{5(\infty)}}}{1 + (\frac{2}{3})^{5(\infty)}} = \frac{2 + 0}{1 + 0} = 2$$

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

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In the first, I would mark down the "substitution" of "$\infty$." You should instead observe that $\lim_{x\to\infty} \left(2+\frac{5}{e^{5x}} \right)=2$, with similar observation for the bottom. –  André Nicolas Nov 10 '12 at 3:03

$$\lim_{x\to\infty}\frac{2\cdot 3^{5x}+5}{3^{5x}+2^{5x}}=\lim_{x\to\infty}\frac{2+5\cdot3^{-5x}}{1+\left(3/2\right)^{-5x}}=2$$ and $$\lim_{x\to0}\frac{e^{2x}-e^{x\ln\pi}}{\sin 3x}=\lim_{x\to0}\frac{(2-\ln\pi)x+O(x^2)}{3x}=\frac{1}{3}(2-\ln\pi).$$ In the second one, you can use that $e^u=1+u+O(u^2)$.