Evaluate the limit of ln(1 + 2^x) times ln(1 + 2/x) as x approaches positive infinity.

Step-by-step solution

1. 1

   Identify the indeterminate form. As $x\to+\infty$ the first factor $\ln(1+2^x)\to+\infty$ while the second factor $\ln\left(1+\tfrac{2}{x}\right)\to\ln 1=0$. The product is of type $\infty\cdot 0$, so it may converge to anything; the two factors have to be measured against each other.

2. 2

   Replace the growing factor by its dominant part. Factor $2^x$ out inside the logarithm:

   $\ln(1+2^x)=\ln\left[2^x\left(1+2^{-x}\right)\right]=x\ln 2+\ln\left(1+2^{-x}\right)$

   The correction $\ln(1+2^{-x})\to 0$, so

   $\frac{\ln(1+2^x)}{x\ln 2}\to 1,\qquad\text{i.e.}\qquad \ln(1+2^x)\sim x\ln 2$

   Writing the exact identity rather than waving at it is what makes the equivalence rigorous.

3. 3

   Replace the vanishing factor by its linear approximation. With $t=\tfrac{2}{x}\to 0$, the standard equivalence $\ln(1+t)\sim t$ gives

   $\ln\left(1+\frac{2}{x}\right)\sim\frac{2}{x}$

   This comes straight from $\lim_{t\to 0}\frac{\ln(1+t)}{t}=1$, the derivative of $\ln(1+t)$ at $t=0$.

4. 4

   Multiply the two equivalences. Asymptotic equivalence is preserved under multiplication, so

   $\ln(1+2^x)\ln\left(1+\frac{2}{x}\right)\sim (x\ln 2)\cdot\frac{2}{x}=2\ln 2$

   The $x$ cancels exactly — the logarithm grows linearly and the other factor decays like $1/x$, so the product settles at a finite constant.

5. 5

   Conclude and evaluate.

   $\boxed{\lim_{x\to+\infty}\ln(1+2^x)\ln\left(1+\frac{2}{x}\right)=2\ln 2=\ln 4\approx 1.3863}$

6. 6

   Check the convergence numerically. At $x=100$ the product is $1.37261$, at $x=1000$ it is $1.38491$, and at $x=10^4$ it is $1.386156$ — creeping up on $2\ln 2=1.386294$. The approach is slow (the error is of order $1/x$), which is exactly what the discarded terms predict.