Let f be a continuous, periodic and bounded function with period 3 such that ∫ 0 3 f ( t ) d t = 6 .
If g ′ ( x ) = f ( x ) , such that g ( 0 ) = 0 , find the value of x → 0 lim x g ( x 1 ) .
If you think the limit doesn't exists, enter 5 5 5 .
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tried to do in a little rigorous procedure
Nice solution!
First, let's get some key insights on the function f ( x ) and g ( x ) .
∫ 0 3 f ( t ) d t = 6 d x d g = f ( x )
Using the definition of an integral, we glean that:
g ( 3 ) − g ( 0 ) = 6 g ( 0 ) = 0 g ( 3 ) = 6
Now, let's think practically. Sure, there are some sine and cosine waves that work with these formulas, so if you want to use these, be my guest:
Ugly Sine Waves
I'm going to use the practical way. The line that passes through the points ( 0 , 0 ) and ( 3 , 6 ) is g ( x ) = 2 x . d x d g = 2 = f ( x ) . Surprisingly, this function satisfies the rest of the requirements! The integral will be just the area of a rectangle with side lengths 2 and 3 , which will be 6 units 2 . And because it's a constant function, f ( x ) = f ( x + 3 ) , so this function technically speaking has a period of 3 .
The rest is computation:
x → 0 lim x g ( x 1 ) = x → 0 lim x x 2 = x → 0 lim 2 = 2
Our answer is 2 .
Your solution is okay for getting an answer. But just taking an example of a function is not a complete solution. You have to show that the limit is same for all functions satisfying the requirements.
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