Limit of a ratio of logarithms at infinity
Evaluate the limit: $$ L = \lim_{x \to +\infty} \frac{\ln\left(2+e^{3x}\right)}{\ln\left(3+e^{2x}\right)}. $$
Explanation of Method
As $x\to+\infty$, both $e^{3x}$ and $e^{2x}$ dominate the additive constants $2$ and $3$ inside each logarithm. Factoring out the exponential from inside each $\ln$ splits it into the dominant linear term ($3x$ or $2x$) plus a correction term that vanishes as $x\to+\infty$. This makes the asymptotic behavior of both numerator and denominator explicit and rigorous, rather than relying only on an informal "$\approx$".
- Factor the exponential out of each logarithm's argument.
- Split each logarithm into a linear term plus a vanishing correction using $\ln(ab) = \ln a + \ln b$.
- Divide numerator by denominator and take the limit, showing the correction terms vanish.
Step 1 · Factor the exponential out of each logarithm
Write the numerator's argument as:
$$ 2 + e^{3x} = e^{3x}\left(1 + 2e^{-3x}\right), $$
and the denominator's argument as:
$$ 3 + e^{2x} = e^{2x}\left(1 + 3e^{-2x}\right). $$
Step 2 · Split each logarithm
Using $\ln(ab) = \ln a + \ln b$:
$$ \ln\left(2+e^{3x}\right) = \ln\left(e^{3x}\right) + \ln\left(1+2e^{-3x}\right) = 3x + \ln\left(1+2e^{-3x}\right), $$
$$ \ln\left(3+e^{2x}\right) = \ln\left(e^{2x}\right) + \ln\left(1+3e^{-2x}\right) = 2x + \ln\left(1+3e^{-2x}\right). $$
As $x \to +\infty$, $e^{-3x} \to 0$ and $e^{-2x} \to 0$, so both correction terms $\ln\left(1+2e^{-3x}\right)$ and $\ln\left(1+3e^{-2x}\right)$ tend to $\ln(1) = 0$.
Step 3 · Divide and take the limit
The ratio becomes:
$$ L = \lim_{x \to +\infty} \frac{3x + \ln\left(1+2e^{-3x}\right)}{2x + \ln\left(1+3e^{-2x}\right)}. $$
Dividing numerator and denominator by $x$:
$$ L = \lim_{x \to +\infty} \frac{3 + \dfrac{\ln\left(1+2e^{-3x}\right)}{x}}{2 + \dfrac{\ln\left(1+3e^{-2x}\right)}{x}}. $$
Since the correction terms $\ln\left(1+2e^{-3x}\right)$ and $\ln\left(1+3e^{-2x}\right)$ both tend to $0$, dividing them by $x\to\infty$ makes each fraction tend to $0$ as well. So:
$$ L = \frac{3+0}{2+0} = \frac{3}{2}. $$
Summary Table
| Expression | Exact split | Behavior as $x \to +\infty$ |
|---|---|---|
| $\ln\left(2+e^{3x}\right)$ | $3x+\ln\left(1+2e^{-3x}\right)$ | $\sim 3x$ |
| $\ln\left(3+e^{2x}\right)$ | $2x+\ln\left(1+3e^{-2x}\right)$ | $\sim 2x$ |
| Ratio | — | $\to \dfrac{3}{2}$ |
Additional Notes
- Factoring out the dominant exponential before taking the logarithm is the rigorous version of the informal shortcut "$\ln(\text{const}+e^{kx}) \approx kx$ for large $x$" — it makes explicit that the neglected term vanishes rather than merely asserting it.
- The general pattern $\lim_{x\to+\infty}\dfrac{\ln(a+e^{px})}{\ln(b+e^{qx})} = \dfrac{p}{q}$ holds for any positive constants $a,b$ and any $p,q>0$, since the additive constants never affect the leading linear growth rate of the logarithm.
- This is analogous to comparing the "growth rates" of two functions at infinity: what matters is the coefficient of $x$ inside each exponential, not the constants added alongside it.
Result
The value of the limit is: