Asked by Alex
Find the linear approximation of ln(1+x)/x*e^x near x=0.
I keep getting an undefined answer, however this is not correct.
I keep getting an undefined answer, however this is not correct.
Answers
Answered by
oobleck
I assume you mean
y = ln(1+x)/(x e^x)
Clearly, y(0) = 0/0, but the limit is 1, via l'HΓ΄pital's Rule
y' = (x - (1+x^2)ln(1+x))/(x^2(1+x)e^x)
y'(0) = (0 - 1*0)/(0*1*1) = 0/0, but the limit is -1, so the tangent line is
y-1 = -1(x-0)
y = 1-x
y = ln(1+x)/(x e^x)
Clearly, y(0) = 0/0, but the limit is 1, via l'HΓ΄pital's Rule
y' = (x - (1+x^2)ln(1+x))/(x^2(1+x)e^x)
y'(0) = (0 - 1*0)/(0*1*1) = 0/0, but the limit is -1, so the tangent line is
y-1 = -1(x-0)
y = 1-x
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