1. **State the problem:** We want to find the limit $$\lim_{x \to +\infty} \frac{e^x + \ln x}{e^x - 1}$$ 2. **Recall the behavior of functions as $x \to +\infty$:** - $e^x$ grows exponentially and dominates most functions. - $\ln x$ grows slowly compared to $e^x$. 3. **Rewrite the expression to analyze dominant terms:** $$\frac{e^x + \ln x}{e^x - 1} = \frac{e^x(1 + \frac{\ln x}{e^x})}{e^x(1 - \frac{1}{e^x})}$$ 4. **Simplify by canceling $e^x$ in numerator and denominator:** $$= \frac{\cancel{e^x}(1 + \frac{\ln x}{e^x})}{\cancel{e^x}(1 - \frac{1}{e^x})} = \frac{1 + \frac{\ln x}{e^x}}{1 - \frac{1}{e^x}}$$ 5. **Evaluate the limit of each fraction as $x \to +\infty$:** - Since $e^x$ grows faster than $\ln x$, $\frac{\ln x}{e^x} \to 0$. - Also, $\frac{1}{e^x} \to 0$. 6. **Substitute these limits back:** $$\lim_{x \to +\infty} \frac{1 + 0}{1 - 0} = \frac{1}{1} = 1$$ **Final answer:** $$\boxed{1}$$