Asked by Momo
Find the limit:
lim {[1/(2+x)]-(1/2)}/x as x->0
How do you remove the complex fraction???
(or just complex fractions in general, I have trouble with them...)
Thank you!
lim {[1/(2+x)]-(1/2)}/x as x->0
How do you remove the complex fraction???
(or just complex fractions in general, I have trouble with them...)
Thank you!
Answers
Answered by
drwls
That is not a complex fraction in the sense of being a complex number (1.e., containing the square root of -1).
As x-> 0, 1/(2+x) behaves like (1/2)[1/[1 + (x/2)] = (1/2)[1 - (x/2)]
Subtract 1/2 from that, divide by x, and you have
(-x/4)/x = -1/4
You could also use L'Hopital's rule to get the limit, but I don't know if you have learned that yet. It says that the limit is the same as the limit of the ratio of the derivatives of the numerator and denominator, which is
Lim -(1/(2+x))^2 = -1/4
x->0
As x-> 0, 1/(2+x) behaves like (1/2)[1/[1 + (x/2)] = (1/2)[1 - (x/2)]
Subtract 1/2 from that, divide by x, and you have
(-x/4)/x = -1/4
You could also use L'Hopital's rule to get the limit, but I don't know if you have learned that yet. It says that the limit is the same as the limit of the ratio of the derivatives of the numerator and denominator, which is
Lim -(1/(2+x))^2 = -1/4
x->0
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