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How do you find the limit of #(e^x - 3*x)^ (1/x) # as x approaches infinity?

Answer 1

Scarlett Allison

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

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

#= exp (lim_{x to oo} color{red}{ ln (e^x - 3x)/x )}#

applying L'Hopital to term in red as it amounts to #oo / oo# because #e^x > > 3x# as #x to oo#

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

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

#= exp (lim_{x to oo} (1 - 3e^{-x})/ (1 - color{green}{3xe^{-x}}) )#

looking at the green term, we can see already that the exponential term is the dominant term but we can apply L'Hopital to see this through

#lim_{x to oo} 3x e^{-x} = lim_{x to oo} (3x)/e^x = lim_{x to oo} 3/e^x = 0#

so the limit is

#exp ( (1 )/ (1 ))#

#= e#

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Answer 2

Charles Anctil

To find the limit of (e^x - 3*x)^(1/x) as x approaches infinity, we can use the properties of exponential and logarithmic functions. By taking the natural logarithm of both sides, we can simplify the expression. Applying the limit properties, we find that the limit is equal to e^(-3).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.