How can I evaluate $\lim_{x\to\infty}\frac{\ln(3e^{2x}+5e^x-2)}{\ln(27e^{3x}-1)}$?

Question

How can I evaluate this limit

$$\lim_{x\to\infty}\dfrac{\ln(3e^{2x}+5e^x-2)}{\ln(27e^{3x}-1)}?$$

My attempt:

$$\lim_{x\to\infty}\dfrac{\ln(3e^{2x}+5e^x-2)}{\ln(27e^{3x}-1)} =\lim_{x\to\infty}\dfrac{\ln(3e^{x}-1)(e^x-2)}{\ln(27e^{3x}-1)} $$ $$=\lim_{x\to\infty}\dfrac{\ln(3e^{x}-1)+\ln(e^x-2)}{\ln(27e^{3x}-1)} $$

Above limit leads to form $\dfrac{\infty}{\infty}$, I used L'Hospital's rule $$=\lim_{x\to\infty}\dfrac{\dfrac{3e^{x}}{3e^{x}-1}+ \dfrac{e^{x}}{e^{x}-2} }{\dfrac{81e^{3x}}{27e^{3x}-1} } $$ $$=\lim_{x\to\infty}\dfrac{\dfrac{3}{3-e^{-x}}+ \dfrac{1}{1-2e^{-x}} }{\dfrac{81}{27-e^{-3x}}} $$ $$=\dfrac{\dfrac{3}{3-0}+\dfrac{1}{1-0} }{\dfrac{81}{27-0} }$$ $$=\dfrac{2}{3}$$

So my answer is $\dfrac23$. Can I evaluate this limit without L'Hospital's rule?

Please help me solve this limit without L'Hospital's rule. Thank in advance.

Answer 1

Yes, you can easily evaluate it without L'Hospital's rule as follows $$\lim_{x\to\infty}\dfrac{\ln(3e^{2x}+5e^x-2)}{\ln(27e^{3x}-1)}$$ $$=\lim_{x\to\infty}\dfrac{\ln(e^{2x}(3+5e^{-x}-2e^{-2x}))}{\ln(e^{3x}(27-e^{-3x}))}$$ $$=\lim_{x\to\infty}\dfrac{\ln(e^{2x})+\ln(3+5e^{-x}-2e^{-2x})}{\ln(e^{3x})+\ln(27-e^{-3x})}$$

$$=\lim_{x\to\infty}\dfrac{2x+\ln(3+5e^{-x}-2e^{-2x})}{3x+\ln(27-e^{-3x})}$$ $$=\lim_{x\to\infty}\dfrac{2+\frac1x\ln(3+5e^{-x}-2e^{-2x})}{3+\frac1x\ln(27-e^{-3x})}$$ $$=\frac{2+0}{3+0}$$ $$=\color{blue}{\frac23}$$

Answer 2

The fastest method uses asymptotic equivalence:

As a polynomial is asymptotically equivalent to its leading term, we have $$3\mathrm e^{2x}+5\mathrm e^x-2\sim_\infty3\mathrm e^{2x}, \qquad27\mathrm e^{3x}-1\sim_\infty 27\mathrm e^{3x}, $$ $$\text{whence }\qquad \begin{cases} \ln\bigl(3\mathrm e^{2x}+5\mathrm e^x-2\bigr)\sim_\infty\ln\bigl(3\mathrm e^{2x}\bigr)=\ln 3+2x\sim_\infty 2x, \\ \text{and similarly}\\[0.5ex] \ln\bigl(27e^{3x}-1\bigr)\sim_\infty \ln27+3x \sim_\infty 3x, \end{cases} $$ so that $$\frac{\ln\bigl(3\mathrm e^{2x}+5\mathrm e^x-2\bigr)}{\ln\bigl(27e^{3x}-1\bigr)}\sim_\infty \frac{2\not x}{3\not x}= \frac23.$$