Evaluate the function f(h) at the given numbers (correct to six decimal places).

Question

Evaluate the function f(h) at the given numbers (correct to six decimal places).

f(h) = (2 + h)5 − 32

h

,

h = ±0.5, ±0.1, ±0.01, ±0.001, ±0.0001

h	f(h)
0.5
0.1
0.01
0.001
0.0001
	h	f(h)
−0.5
−0.1
−0.01
−0.001
−0.0001

Guess the value of the limit (correct to six decimal places). (If an answer does not exist, enter DNE.)

lim h→0 (2 + h)5 − 32

h

Answer 1

It's a little hard to tell from the formatting in the question, but I'm guessing the problem is

Plugging in these values of "h" we get these associated values of f(h)

0.5 → f(0.5) = [(2 + 0.5)⁵ - 32]/0.5 = 65.65625/0.5 = 131.3125

0.1 → f(0.1) = [(2 + 0.1)⁵ - 32]/0.1 = 8.84101/0.1 = 88.4101

0.01 → f(0.01) = [(2 + 0.01)⁵ - 32]/0.01 = 0.8080401001/0.01 = 80.80401001

0.001 → f(0.001) = [(2 + 0.001)⁵ - 32]/0.001 = 0.08008004/0.001 = 80.08004001

0.0001 → f(0.0001) = [(2 + 0.0001)⁵ - 32]/0.0001 = 0.0080008/0.0001 = 80.0080004

-0.5 → f(-0.5) = [(2 + -0.5)⁵ - 32]/(-0.5) = (-24.40625)/(-0.5) = 48.8125

-0.1 → f(-0.1) = [(2 + -0.1)⁵ - 32]/(-0.1) = (-7.23901)/(-0.1) = 72.3901

-0.01 → f(-0.01) = [(2 + -0.01)⁵ - 32]/(-0.01) = (-0.7920399001)/(-0.01) = 79.20399001

-0.001 → f(-0.001) = [(2 + -0.001)⁵ - 32]/(-0.001) = (-0.07992004)/(-0.001) = 79.92003999

-0.0001 → f(-0.0001) = [(2 + -0.0001)⁵ - 32]/(-0.0001) = (-0.0079992)/(-0.0001) = 79.9920004

The limit as h→0 = 80.000000

To verify this, you can plug in numbers closer to 0 if you like or you can solve this algebraically.

lim h→0 [h⁵ + 10h⁴ + 40h³ + 80h² + 80h +32 - 32]/h

lim h→0 [h⁵ + 10h⁴ + 40h³ + 80h² + 80h]/h

lim h→0 (h⁴ + 10h³ + 40h² + 80h + 80) = 80