Two limits without L’Hopital $$\lim_{x \to \infty}\frac{\ln(x^3-5x+3)}{\ln(x^5-6x^2-6)}$$
$$\lim_{x\to \infty}\frac{\ln(1+e^{3x})}{x}$$
For the first function I intended using $\lim_{x\to \infty}\frac{\ln x}{x}=0$ but I couldn’t figure it out. For the second limit I have no idea. Please help me solve them.
 A: Hints:
$$\frac{\ln(x^3-5x+3)}{\ln(x^5-6x^2-6)}=\frac{3\ln(x)+\ln(1-5x^{-2}+3x^{-3})}{5\ln(x)+\ln(1-6x^{-3}-6x^{-5})}.$$
$$\lim_{x\to \infty}\frac{\ln(1+e^{3x})}{x}=\lim_{t\to \infty}\frac{\ln(1+t^3)}{\ln(t)}.$$
A: Note
$$\lim_{x \to \infty}\frac{\ln(x^3-5x+3)}{\ln(x^5-6x^2-6)}
=\lim_{x \to \infty}\frac{\ln x^3+ \ln(1-\frac5{x^2}+\frac3{x^3})}{\ln x^5+ \ln(1-\frac6{x^3}+\frac6{x^5})}=\lim_{x \to \infty}\frac{3\ln x }{5\ln x }= \frac35
$$
$$\lim_{x\to \infty}\frac{\ln(1+e^{3x})}{x}
=\lim_{x\to \infty}\frac{\ln e^{3x}+\ln(1+e^{-3x})}{x}
= \lim_{x\to \infty}\frac{{3x}}{x}=3
$$
A: For the first, divide the numerator and denominator by $\ln x$, giving$$\lim_{x\to\infty}\frac{3+\tfrac{\ln\left(1-\frac{5}{x^2}+\frac{3}{x^3}\right)}{\ln x}}{5+\tfrac{\ln\left(1-\frac{6}{x^3}-\frac{6}{x^5}\right)}{\ln x}}=\frac{3+0/\infty}{5+0/\infty}=\frac35.$$For the second, subtract out $3$, giving$$3+\lim_{x\to\infty}\frac{\ln(1+e^{-3x})}{x}=3+\frac{0}{\infty}=3.$$
A: i) just factor $x^3$ in the argument of the logarithm of the numerator and then use log properties to break it into a sum.
In the denominator do the same but factoring $x^5$. The result should follow easily.
ii) just factor $e^{3x}$ and then break the logarithm of the product as the sum of the logarithms. The result also here follows.
