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lim x->infinity for ln6x/e^5x
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inf/inf invokes Lhpoital
L'Hopital
\[\lim x \rightarrow \infty \ln6x/e ^{5x}\]
take the derivative of the top and bottom, don't use product rule
0
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to apply L'H rule
ln(6x) = ln(6) + ln(x) by the way
are you asking this \[\lim_{x \rightarrow \infty}\frac{\ln 6x}{e^{5x}}\]
\[e^{5x}>>\ln(6x)\]
\[\frac{1/inf=0}{5e^{5inf}=inf}\] either way you go you get to 0
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also notice that e^5x grows alot faster than ln(6x) so it over powers it
1/1000000000 is a good visual; which is very very tiny
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