Limit of (x-16)/(x^(7/4)-128) as x approaches 16?

Question

I don't quiet get how to evaluate

limx→16 (x-16)/(x^(7/4)-128)

Factorization does not work, and I don't see how to apply substitution or the squeeze theorem here. Can anyone help me out?

Answer 1

This is a cool problem, especially if you do not know LHopital's Rule:

As x -> 16, the numerator -> 0

The denominator 16^7/4 - 128 = 2⁷ - 128 = 0, also goes to 0. So the form is indeterminate (0/0). That means L'Hopital's rule can be applied (derivative of numerator / derivative of denominator).

lim_x->16 1/[(7/4)x^3/4] = 1/[(7/4)16^3/4] = 4/[7(8)] = 1/14, which indeed is the limit.

The other approach: let u = x^1/4. As x -> 16, u -> 2 and x = u⁴.

The numerator becomes:

u⁴ - 16, which factors as a difference of squares:

(u - 2)(u+2)(u² + 4)

The denominator becomes:

u⁷ - 128

Now the idea is to get the (u-2) in the numerator to cancel so that when we take the limit the indeterminate form is gone:

You might try synthetic division of u⁷ - 128 by (u -2).

2 | 1 0 0 0 0 0 0 -128

This results in (u-2)(u⁶ + 2u⁵ + 4u⁴ + 8u³ + 16u² + 32u + 64)

lim_u->2 [(u-2)(u+2)(u² + 4)/ (u-2)(the 7 term polynomial shown above)]

The (u-2) factors cancel, so the limit is: 32/448 = 1/14

desmos.com/calculator/i5z1xilppk