…then lim x–>0 (f(x)/g(x)) is…?
Thank you for any help!
4 Answers

Hi Julia,
df(x)/dx = cos(x).
dg(x)/dx = 1.
Integrate:
f(x) = sin(x) + constant1.
g(x) = x + constant2.
Use initial conditions:
f(0) = 0:
f(0) = sin(0) + constant1 = 0 + constant1 = 0.
So, constant1 = 0.
Thus, f(x) = sin(x).
g(0) = 0:
g(0) = 0 + constant2 = 0.
So, constant2 = 0.
Thus, g(x) = x.
Substitute in functions:
lim [x –> 0] (f(x) / g(x)) = lim [x –> 0] (sin(x) / x)
This is 0 / 0. Therefore, you can use L’Hopital’s rule:
= lim [x –> 0] (f'(x) / g'(x)) = lim [x –> 0] (cos(x) / 1) = lim [x –> 0] cos(x) = cos(0) = 1.
Solution:
1.

“If f'(x) = cos x”
So f(x) = sin(x) + C, where C = some constant.
“and g'(x) = 1”
So g(x) = x + D where D = some other constant.
“f(0) = g(0) = 0”
Plug in x = 0. That tells you what C and D need to be.
Then take the limit as x>0.

The antiderivative of cosx = sinx which is f(x)
The antiderivative of g'(x) = 1 is x
So f(0): sin(0) = 0
and g(0) = 0
Therefore lim x>0 of f(x)/g(x) = 0/0 which is indeterminate

im taking calculas next semester, thanks for scaring me now lol