How do you find the limit of #(ln (e^x - 3*x)) / x# as x approaches infinity?

1 Answer
Eddie
Jul 4, 2016

1

Explanation:

#lim_{x to oo} (ln (e^x - 3x)) / x#

if we scope this out we know that

#(ln (e^(oo) - 3(oo))) / (oo) approx (ln (e^(oo) )) / (oo)# as #lim_{x to oo} e^(x) > > 3x# and so #Lim = oo/oo = 1#. that's a first order guess but it's pretty straightforward.

but clearly we have the indeterminate form here so we can use L'Hopital's Rule

so

#lim_{x to oo} (ln (e^x - 3x)) / x = lim_{x to oo} (1/ (e^x - 3x)*(e^x-3 ) )/ 1#

# = lim_{x to oo} (e^x-3 )/ (e^x - 3x) #

We can go again with L'Hopital as as #lim_{x to oo} e^(x) > > 3x# so this is still #oo/oo#

# = lim_{x to oo} (e^x )/ (e^x - 3) #

it's starting to becomes obvious but we can go one final time with L'Hopital's Rule

# = lim_{x to oo} (e^x )/ (e^x) = 1#

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