Hey! I'm not quite sure how to proceed in calculating this limit. If you could look at it and give me some feedback I would appreciate it. $\displaystyle \lim_{x \to +\infty} \frac{xln(e^x+e^{-x})+\sqrt{x^3+1}}{x^2+1}$ Thanks in advance.
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these might help you $\displaystyle xln(e^x+e^{-x})$ =$\displaystyle x[ln(1+e^{-2x})+x]$
After using that, divide both numerator and denominator by $\displaystyle x^2$.
Thank you!
Here's a hand-waving argument, for large $\displaystyle x$ $\displaystyle \frac{x \ln(e^x+e^{-x})+\sqrt{x^3+1}}{x^2+1} \approx \frac{x \ln(e^x)}{x^2} = \frac{x^2}{x^2} = 1$
Originally Posted by mei Hey! I'm not quite sure how to proceed in calculating this limit. If you could look at it and give me some feedback I would appreciate it. $\displaystyle \lim_{x \to +\infty} \frac{xln(e^x+e^{-x})+\sqrt{x^3+1}}{x^2+1}$ Thanks in advance. You can divide all three terms by $\displaystyle x^2$ and concentrate on $\displaystyle {\ln(e^x+e^{-x})\over x}$
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