lim as x approaches 8 of (2 - cube root x)/(8-x)

Question

i know the answer, but i can't figure out how to go about doing it....?

Answer 1

The cube root of 8+h where h is small and positive or negative can be expressed binomially: (8+h)^(1/3)=2(1+(h/8))^(1/3), which expands to 1+(1/3)(h/8)+((1/3)((1/3)-1)/2)h^2/64+... To a first approximation this is just 1+h/24.

So 2-2(1+(h/24)+...)=-h/12-.... Now, x=8+h, so 8-x=-h and if we divide -h/12 by -h we get 1/12. Therefore, the limit is 1/12 as h approaches 0, i.e., x approaches 8.