Jameson said:
If you differentiate $f(a)$ or $a$ then that is 0 since those are both constants.
So [math]\frac{d}{dx} \sin(x)-\sin(a)=\cos(x)-0=\cos(x)[/math] and [math]\frac{d}{dx} x-a=1-0=1[/math]
Thus the derivative of the numerator divided by the derivative of the denominator is [math]\frac{\cos(x)-0}{1-0}=\cos(x)[/math] which intuitively makes sense because we know that [math]\frac{d}{dx}\sin(x)=\cos(x)[/math]
Hi there, can you explain to me something? I just don't understand the ONE part(the last part) of this problem. Actually, this is a solution to the question on finding the limit of 1/(sin(x) - sin(a)) - 1/((x - a)cos(a)).
The solution is:
Clearly we can't use l'Hopital's rule yet, so let's use the hint. We need a Taylor series for sin(x) centred at x = a. We know the successive derivatives of sin(x), so this should be fairly simple. Moreover, since the derivatives are all still everywhere bounded by -1 and 1, Taylor's theorem will still prove that the series will converge to sin(x) for all x. So, we get that, for all x:
sin(x) = (sin(a)/0!) + (cos(a)/1!)(x - a) - (sin(a)/2!)(x - a)^2 - (cos(a)/3!)(x - a)^3 + (sin(a)/4!)(x - a)^4 + ...
sin(x) - sin(a) = (x - a)(cos(a)/1! - (sin(a)/2!)(x - a) - (cos(a)/3!)(x - a)^2 + (sin(a)/4!)(x - a)^3 + ...)
Let f(x) = cos(a)/1! - (sin(a)/2!)(x - a) - (cos(a)/3!)(x - a)^2 + (sin(a)/4!)(x - a)^3 + ...
i.e. (sin(x) - sin(a)) / (x - a) if x is not equal to a, or cos(a) if it is. The important thing to realize is that, as a function defined by a power series, it is infinitely differentiable everywhere on the interior of its domain (in this case, everywhere), and therefore continuous and differentiable everywhere. Then:
1 / (sin(x) - sin(a)) - 1 / ((x - a)cos(a))
= 1 / ((x - a)f(x)) - 1 / ((x - a)cos(a))
= (1 / (x - a))(1 / f(x) - 1 / cos(a))
= -(f(x) - cos(a)) / ((x - a)cos(a)f(x))
-(f(x) - cos(a)) / ((x - a)cos(a)f(x))
Here's where we could use l'Hopital's rule, but I think it would be redundant. We can separate the factors like so:
-1 / (cos(a)f(x)) * (f(x) - cos(a)) / (x - a)
The first factor is continuous so long as the limit of f(x) as x approaches a is not 0. But, f(x) is continuous, so the limit is f(a), which is clearly equal to cos(a). Since cos(a) appears in the denominator, we already presuppose that cos(a) is non-zero, so the limit of the first factor is -1 / cos^2(a).
The second factor can be rewritten as such:
(f(x) - f(a)) / (x - a)
The limit as x approaches a could be determined by differentiating top and bottom, or by simply noticing that this is a definition of the derivative of f at a point a. We know it exists because f is defined as a power series. We can determine the limit either by differentiating term by term and substituting x = a, or by recalling that power series are their own Taylor expansion around their centre, which means the derivative, divided by 1!, will be the coefficient of (x - a)^1, which is -sin(a)/2!. Thus:
f'(a) = -sin(a) / 2
Therefore, the limit is:
-1 / cos^2(a) * -sin(a) / 2
= sec(a)tan(a) / 2
OK,back to my question. Can you explain this part where he said:
The limit as x approaches a could be determined by differentiating top and bottom, or by simply noticing that this is a definition of the derivative of f at a point a. We know it exists because f is defined as a power series. We can determine the limit either by differentiating term by term and substituting x = a, or by recalling that power series are their own Taylor expansion around their centre, which means the derivative, divided by 1!, will be the coefficient of (x - a)^1, which is -sin(a)/2!.
If I use l'Hopital's rule, I must differentiate the numerator and the denominator,right? The denominator is just x-a so the derivative would be 1 with respect to x,isn't it?What about the numerator?How do I differentiate f(x)-f(a)?