First we will try to guess the result informally, and then we will formally prove it using the squeeze theorem. For large values of \(x\), the value \(1\) is not very significant relative to \(2^x\), and so the given function is close to the function \(\displaystyle \frac{\log(2^x)}{x}=\frac{x \log 2}{x}=\log 2. \)
Let \( f(x)=\frac{\log(2^x)}{x}=\log 2, \\ h(x)=\frac{\log(1+2^x)}{x}\text{ and }\\ g(x)=\frac{\log(2^x+2^x)}{x}=\frac{\log(2^{x+1})}{x}=\log 2\frac{x+1}{x}. \)
Evidently for \(x\geq 0\) we have \( f(x)\leq h(x)\leq g(x) \).
This statement plus the simple limits \(\displaystyle \lim_{x\to\infty}f(x)=\log 2=\lim_{x\to\infty}g(x) \) guarantee that the limit of the given function satisfies \(\displaystyle \lim_{x\to \infty}\frac{\log(1+2^x)}{x}=\lim_{x\to \infty}h(x)=\log 2. \)
