Two limits without L’Hopital

Question

$$\lim_{x \to \infty}\frac{\ln(x^3-5x+3)}{\ln(x^5-6x^2-6)}$$

$$\lim_{x\to \infty}\frac{\ln(1+e^{3x})}{x}$$

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.

Answer 1

Hints:

$$\frac{\ln(x^3-5x+3)}{\ln(x^5-6x^2-6)}=\frac{3\ln(x)+\ln(1-5x^{-2}+3x^{-3})}{5\ln(x)+\ln(1-6x^{-3}-6x^{-5})}.$$

$$\lim_{x\to \infty}\frac{\ln(1+e^{3x})}{x}=\lim_{t\to \infty}\frac{\ln(1+t^3)}{\ln(t)}.$$

Answer 2

Note

$$\lim_{x \to \infty}\frac{\ln(x^3-5x+3)}{\ln(x^5-6x^2-6)} =\lim_{x \to \infty}\frac{\ln x^3+ \ln(1-\frac5{x^2}+\frac3{x^3})}{\ln x^5+ \ln(1-\frac6{x^3}+\frac6{x^5})}=\lim_{x \to \infty}\frac{3\ln x }{5\ln x }= \frac35 $$

$$\lim_{x\to \infty}\frac{\ln(1+e^{3x})}{x} =\lim_{x\to \infty}\frac{\ln e^{3x}+\ln(1+e^{-3x})}{x} = \lim_{x\to \infty}\frac{{3x}}{x}=3 $$

Answer 3

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$$3+\lim_{x\to\infty}\frac{\ln(1+e^{-3x})}{x}=3+\frac{0}{\infty}=3.$$

Answer 4

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.