How to evaluate a limit if the function is given by y=[(a^x+b^x)/2]^1/x, a>0, b>0, x-->0 ?

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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We'll create the remarcable limit:

lim(1 + x)^(1/x) = e

We'll add and subtract 1:

lim[1 + (a^x + b^x)/2 - 1]^(1/x) = lim[1+(a^x + b^x - 2)/2]^(1/x) lim{[1+(a^x + b^x - 2)/2]^[2/(a^x+b^x-2)]}^(a^x+b^x-2)/2x = e^lim (a^x+b^x-2)/2x

We'll calculate the limit of the superscript:

lim (a^x+b^x-2)/2x = (a^0+b^0-2)/2*0 = 0/0

We'll apply L'Hospital rule:

lim (a^x+b^x-2)/2x = lim (a^x+b^x-2)'/(2x)' lim (a^x+b^x-2)'/(2x)' = lim(a^x*lna + b^x*lnb)/2

We'll substitute x by 0:

lim(a^x*lna + b^x*lnb)/2 = (a^0*lna + b^0*lnb)/2 lim(a^x*lna + b^x*lnb)/2 = ln(a*b)/2 = ln sqrt(a*b)

The limit of the given function is: lim[(a^x+b^x)/2]^(1/x) = e^ln sqrt(a*b)

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