Limits with trig, log functions and variable exponents Would someone mind verifying this?
$$ \lim_{x\to \infty} \frac{2 \cdot 3^{5x} + 5}{3^{5x} + 2^{5x}}
= \lim_{x\to \infty} \frac{3^{5x}(2 + \frac{5}{3^{5x}})}{3^{5x}(1 + (\frac{2}{3})^{5x})}
= \lim_{x\to \infty} \frac{2 + \frac{5}{3^{5x}}}{1 + (\frac{2}{3})^{5x}}
= \frac{2 + \frac{5}{3^{5(\infty)}}}{1 + (\frac{2}{3})^{5(\infty)}}
= \frac{2 + 0}{1 + 0}
= 2 $$
$$ \lim_{x\to 0} \frac{e^{2x}-\pi^{x}}{sin(3x)}
= \lim_{x\to 0} \frac{2e^{2x}-\pi^{x} ln(\pi)}{3cos(3x)}
= \frac{2e^{2(0)}-\pi^{(0)} ln(\pi)}{3cos(3(0))}
= \frac{2\cdot 1 - 1\cdot ln(\pi)}{3\cdot1}
= \frac{2-ln(\pi)}{3} $$

Yes those are both right.
$$\lim_{x\to\infty}\frac{2\cdot 3^{5x}+5}{3^{5x}+2^{5x}}=\lim_{x\to\infty}\frac{2+5\cdot3^{-5x}}{1+\left(3/2\right)^{-5x}}=2$$
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 $e^u=1+u+O(u^2)$.
