How do you find the limit of $\frac{ln (3 x + 5 e^{x})}{ln (7 x + 3 e^{2 x})}$ $ln ( 3x + 5e^x )/ ln ( 7x + 3e^{2x})$ as x approaches infinity?

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Answer 1 · 2016-07-03T18:54:30

$lim_{x->oo}log_e ( 3x + 5e^x )/ log_e (7x+3e^{2x}) =1/2$

$log_e ( 3x + 5e^x )/ log_e ( 7x + 3e^{2x})$

For large $x$ follows that
$log_e ( 3x + 5e^x ) approx log_e (5e^x )$
$log_e ( 7x + 3e^{2x}) approx log_e (3e^{2x})$

so
$lim_{x->oo}log_e ( 3x + 5e^x )/ log_e (7x+3e^{2x}) = lim_{x->oo}log_e (5e^x )/ log_e (3e^{2x})$

but

$lim_{x->oo}log_e (5e^x )/ log_e (3e^{2x}) = lim_{x->oo}(log_e 5 + x)/(log_3 3+2x)= 1/2$

How do you find the limit of ln ( 3x + 5e^x )/ ln ( 7x + 3e^{2x})ln(3x+5ex)ln(7x+3e2x) ln ( 3x + 5e^x )/ ln ( 7x + 3e^{2x}) as x approaches infinity?