Question

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$lim_{x to +infty}frac{ln(1+e^x)}{x}$

Answer 1

$displaystylelim_{x to +infty} frac{ln(1+e^x)}{x}=lim_{x to +infty}frac{ln(e^x(1+e^{-x}))}{x}=lim_{x to +infty}frac{ln e^x+ln(1+e^{-x})}{x}=\ \ \ =lim_{x to +infty}frac{x}{x}+lim_{x to +infty}frac{ln(1+e^{-x})}{x}=1+lim_{x to +infty}e^{-x}=1+0=1$