# Help with Calculus?? If f'(x) = cos x and g'(x) = 1 for all x, and if f(0) = g(0) = 0….?

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…then lim x–>0 (f(x)/g(x)) is…?

Thank you for any help!

• 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 anti-derivative of cosx = sinx which is f(x)

The anti-derivative 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

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