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

Step-by-step solution

1. 1

   Do not split the logarithm of a sum. $\ln(e^x+x)$ is not $\ln e^x+\ln x$. The legitimate move is to factor out the dominant term inside the logarithm, then use $\ln(ab)=\ln a+\ln b$ on the product that results.

2. 2

   Factor the dominant term $e^{x}$ inside the log. For large $x$, $e^x$ dwarfs $x$, so

   $e^{x}+x=e^{x}\left(1+xe^{-x}\right)$

   and therefore

   $\ln\left(e^{x}+x\right)=\ln\left(e^{x}\right)+\ln\left(1+xe^{-x}\right)=x+\ln\left(1+xe^{-x}\right)$

3. 3

   Show the correction term is negligible. Since exponentials beat polynomials, $xe^{-x}\to 0$, hence

   $\ln\left(1+xe^{-x}\right)\to\ln 1=0$

   So the numerator is $x$ plus something that vanishes: it grows exactly like $x$, no faster.

4. 4

   Compare the two growth rates. The quotient becomes

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

   A degree-1 numerator over a degree-2 denominator always tends to zero.

5. 5

   State the limit and confirm numerically.

   $\lim_{x\to+\infty}\frac{\ln\left(e^{x}+x\right)}{x^{2}+1}=0$

   At $x=100$ the value is $0.009999$ and at $x=500$ it is $0.002000$ — halving as $x$ doubles, precisely the $1/x$ decay predicted.