Limit of ln(1+2^x) times ln(1+3/x)
Evaluate the limit: $$ L = \lim_{x \to \infty} \ln\left(1+2^x\right)\cdot\ln\left(1+\frac3x\right). $$
Explanation of Method
As $x\to\infty$, the first factor $\ln(1+2^x)\to\infty$ (since $2^x$ dominates the added $1$), while the second factor $\ln\left(1+\frac3x\right)\to0$ (since $\frac3x\to0$), giving the indeterminate form $\infty\cdot0$. Approximating each factor by its dominant behavior — the first by $x\ln2$ (dropping the negligible $+1$ inside a large exponential), the second by $\frac3x$ (via $\ln(1+u)\approx u$) — turns the product into $(x\ln2)\cdot\frac3x$, where the factors of $x$ cancel exactly, leaving a finite constant.
- Approximate $\ln(1+2^x)$ for large $x$ by factoring out the dominant exponential.
- Approximate $\ln\left(1+\frac3x\right)$ for large $x$ using $\ln(1+u)\approx u$.
- Multiply the two approximations and simplify.
Step 1 · Approximate $\ln(1+2^x)$
Factor out the dominant exponential: $1+2^x = 2^x\left(1+2^{-x}\right)$, so:
$$ \ln(1+2^x) = \ln(2^x) + \ln\left(1+2^{-x}\right) = x\ln2 + \ln\left(1+2^{-x}\right). $$
As $x\to\infty$, $2^{-x}\to0$, so $\ln(1+2^{-x})\to0$, and:
$$ \ln(1+2^x) \sim x\ln2 \qquad \text{as } x \to \infty. $$
Step 2 · Approximate $\ln\left(1+\dfrac3x\right)$
As $x\to\infty$, $\dfrac3x\to0$, so using $\ln(1+u)\approx u$ for small $u$:
$$ \ln\left(1+\frac3x\right) \sim \frac3x \qquad \text{as } x \to \infty. $$
Step 3 · Multiply and simplify
Substituting both approximations:
$$ L = \lim_{x\to\infty} \ln(1+2^x)\cdot\ln\left(1+\frac3x\right) \sim \lim_{x\to\infty} (x\ln2)\cdot\frac3x = \lim_{x\to\infty} 3\ln2 = 3\ln2. $$
Summary Table
| Factor | Leading behavior as $x\to\infty$ |
|---|---|
| $\ln(1+2^x)$ | $\sim x\ln2 \ (\to\infty)$ |
| $\ln\left(1+\dfrac3x\right)$ | $\sim\dfrac3x \ (\to0)$ |
| Product | $\to3\ln2$ |
Additional Notes
- This is a typical $\infty\cdot0$ resolution: one factor's rate of divergence (linear in $x$, via $x\ln2$) is exactly matched by the other factor's rate of vanishing (proportional to $\frac1x$), so their product settles to a finite constant rather than diverging or vanishing.
- The specific base $2$ only enters the final answer through $\ln2$; replacing $2^x$ with $c^x$ for another constant $c>1$ would give $\ln(1+c^x)\sim x\ln c$, and the final answer would become $3\ln c$ instead.
- Likewise, the constant $3$ in $\frac3x$ appears directly as a multiplicative factor in the final answer — replacing it with a general constant $k$ would give $\lim_{x\to\infty}\ln(1+2^x)\ln\left(1+\frac kx\right) = k\ln2$.
Result
The value of the limit is: