Calculating Limit of Integrals Involving Cos and Exponential Functions

[Metal]

Ok, i have this problem:

Calculate: Limit x->0 of ((integral of cos t² dt between 0 and x²) / (integral of e^(-t²) dt))

So, the limit of both integrals is 0 since the interval between both integrals tends to 0. I used L'Hopital then, so:

Limit x->0 F'(x)/G'(x) = (cos (x^4) * 2x)/(e^(-x²)), F(x) and G(x) being both integrals.

I know that this is right, i just don't remember why F'(x) = cos (x^4) * 2x and not F'(x) = cos (x^4)... I know it have something to do with the fact that 2x is the derivative of x² but there's something missing in this explanation.

Tks in advance.

 

[AKG]

Define

A(x) = \int _0 ^x \cos t^2\, dt

B(x) = x^2

then

F = A o B, so F'(x) = B'(x)A'(B(x)) by chain rule, giving:

F'(x) = 2xA'(x²) = 2xcos(x⁴) by FTC

 

[Metal]

Tks for that... but what i really wanted to know is why F'(x) = B'(x)A'(B(x))... IOW a proof of the chain rule. I guess was naturally inclined to think that F'(x) would equal A'(B(x)).

 

[StatusX]

Here's a sketch of a proof of the chain rule:

\frac{d}{dx}(f(g(x)))= \lim_{h\rightarrow 0} \frac{f(g(x+h))-f(g(x))}{h}

Now define:

k=g(x+h)-g(x)

Then as h goes to zero, k goes to zero, and:

\lim_{h \rightarrow 0} \frac{k}{h} = g'(x)

Then we have:

\lim_{h\rightarrow 0} \frac{f(g(x+h))-f(g(x))}{h} =\lim_{k\rightarrow 0} \frac{f(g(x)+k)-f(g(x))}{h}

And as k gets very small, h can be replaced by k/g'(x), and:

\lim_{k\rightarrow 0} \frac{f(g(x)+k)-f(g(x))}{h}= \lim_{k\rightarrow 0} \frac{f(g(x)+k)-f(g(x))}{k/g'(x)} = \lim_{k\rightarrow 0} \frac{f(g(x)+k)-f(g(x))}{k}g'(x) =f'(g(x)) g'(x)


There are some details to fill in if you want to be rigorous.

 

[Metal]

Great, Tks...