Question #e487f

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Answer 1 · 2017-06-02T01:06:35

You do not really need l'Hospital:

$lim_(x->oo) (e^x+sinx)/(e^x+cosx) = lim_(x->oo) (e^x( 1+e^(-x)sinx))/(e^x(1+e^(-x)cosx))$

$lim_(x->oo) (e^x+sinx)/(e^x+cosx) = lim_(x->oo) ( 1+e^(-x)sinx)/(1+e^(-x)cosx) = 1$

Answer 2 · 2017-06-02T03:45:01

$lim_(x->oo) frac{e^x + sinx}{e^x + cosx} = 1$

$lim_(x->oo) frac{e^x + sinx}{e^x + cosx}$

Multiply the fraction by $frac{e^(-x)}{e^(-x)}$ :

$= lim_(x->oo) frac{1+ sinx/(e^x) }{ 1+ cosx/(e^x) }$

By direct substitution, this is equal to:
$= frac{1+0}{1+0} = 1$

Check by looking at the graph:
graph{(e^x + sinx)/(e^x + cosx) [-2, 18, -3.5, 5.5]}

Note that L'Hospital's rule was not needed because after factoring the fraction, there was no indeterminate case for which the rule would be needed.