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#1

$Old$ July 4th, 2009, 04:50 AM

mei $mei is offline$

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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.

$\lim_{x \to +\infty} \frac{xln(e^x+e^{-x})+\sqrt{x^3+1}}{x^2+1}$

Thanks in advance.

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#2

$Old$ July 4th, 2009, 05:02 AM

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these might help you

$xln(e^x+e^{-x})$

= $x[ln(1+e^{-2x})+x]$

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#3

$Old$ July 4th, 2009, 05:48 AM

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After using that, divide both numerator and denominator by $x^2$ .

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$Old$ July 4th, 2009, 08:24 AM

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Thank you!

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#5

$Old$ July 4th, 2009, 08:35 AM

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Here's a hand-waving argument, for large $x$

$\frac{x \ln(e^x+e^{-x})+\sqrt{x^3+1}}{x^2+1} \approx \frac{x \ln(e^x)}{x^2} = \frac{x^2}{x^2} = 1$

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$Old$ July 4th, 2009, 08:56 AM

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Quote:

Originally Posted by mei $View Post$

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.

$\lim_{x \to +\infty} \frac{xln(e^x+e^{-x})+\sqrt{x^3+1}}{x^2+1}$

Thanks in advance.

You can divide all three terms by $x^2$

and concentrate on ${\ln(e^x+e^{-x})\over x}$

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