How would I evaluate the following limit?

Question

lim 3√x-4 / x-64

x->64

How would I answer this?

at first I tried doing the change of variables and I ended up with an extremely large answer. now I’m thinking of rationalizing the numerator? Does that sound right? Thanks.

Answer 1

If you mean (³√x - 4) / (x - 64), then the given limit is the definition of the derivative of f(x) = ³√x when x = 64.

Recall that f'(c) = lim_x→c [(f(x) - f(c)) / (x - c)]

So, since f'(x) = (1/3)x^-2/3, the value of the limit is f'(64) = (1/3)(64)^-2/3 = 1/48

Answer 2

Hey Anne,

In this particular case, x = 64 is a discontinuity that isn't removable by any method. You can graph the function to confirm this. I'm not sure if its [3sqrt(x)-4]/(x-64) or [3sqrt(x-4)]/(x-64) that was intended, but if it's one of those two, you end up with the same conclusion, the limit does not exist.