(1+h): - 1 A. lim
h>0
h
f(0) =
х
a =
1
B. lim
h>0
cos(1 + h) +1
h
f(3) =
a =
7T
5(4+ h)" - 5120
C. lim
h->0
h
f(x) =
a =
4
o
D. lim
h 0
e4h - 1
h
f(x) =
a =
e
Х
Pls take a look at the picture

Each of these limits correspond to the derivative of some function f(x) at a point x = a.

Recall the limit definition of a function f(x) :

f '(x) = lim [h → 0] ( f(x + h) - f(x) ) / h

Then if x = a, we get

f '(a) = lim [h → 0] ( f(a + h) - f(a) ) / h

From here, it's easy to identify what each function and point should be:

(a) f (a + h) = (1 + h)¹ʹ³   →   f(x) = x ¹ʹ³ and a = 1

(that's a 1/3 in the exponent)

(b) f (a + h) = cos(π + h)   →   f(x) = cos(x) and a = π

(c) f (a + h) = 5 (4 + h)⁵   →   f(x) = 5x ⁵ and a = 4

(d) f (a + h) = exp(4h) = exp(4 (0 + h))   →   f(x) = exp(4x) and a = 0

(where exp(x) = eˣ )

25/4

step-by-step explanation:

call segment cd's length x. because dc and ab are parallel, angle acd and cab, so triangles adc and abc are similar and we can set up the equation x / ac = ac / ab, or x / 5 = 5 / 4 and x = 25/4.

the equivalent expression are always the hardest ones but you look at the gcf also

step-by-step explanation: