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# L'Hopital rule Help! watch

1. Can anyonw do the question 1 and 2 below
2. i knew there was a reason i stopped doing maths, lol
3. Question two is clearly wrong - e.g take f(x) = x on (1/3,2/3). Rather it should say that |f'(x)| < 1 (I don't think you even need it bounded away from 1) and then apply the mean-value theorem to the fixed point alpha and a second hypothetical fixed point, to get your contradiction.
4. for 1) use lim[f(x)/g(x)] = lim[f'(x)/g'(x)]
5. I still don't know how to do them
6. lim1 cos(pi/2)x / x^2 - 1
= lim1 (-pi/2)sin(pi/2)x / 2x
As x>1 you get (-pi/2)sin(pi/2) /2
= -pi/4.

[Not entirely sure that's right though. I know nothing about L Hopitals rules other than what Fermat stated.]
7. (Original post by superkillball)
I still don't know how to do them
Just apply the formula which Fermat provided. Sorry to be blunt, but Gaz031 managed one fine and he's never met this material.
8. 1a)
If you are going to end up with lim 0/0 then, if lim f'(x)/g'(x) = L then so also does lim (f(x)/g(x)).
let f(x) = cos(pi.x/2) then f'(x)= -(pi/2)sin(pi.x/2) and lim{x->1} f'(x) = -pi/2
let g(x) = x² - 1 then g'(x)= 2x and lim{x->1} g'(x) = 2
lim[f(x)/g(x)] = lim[f'(x)/g'(x)] = -(pi/2)/2 = -pi/4.

1b)
f(x) = ln(4-x) then f'(x) = -1/(4-x) and lim{x->3} = -1
g(x) = √(x² - 9) then g'(x) = x/√(x² - 9) and lim{x->3} g'(x) = ∞
lim[f(x)/g(x)] = lim[f'(x)/g'(x)] = -1/∞ = 0.

1c)
f(x) = (x-π/2)^4
g(x) = cos²x - cot²x

This one takes a bit longer to do. Basically we just extend the original rule, viz
lim[f(x)/g(x)] = lim[f'(x)/g'(x)] = lim[f''(x)/g''(x)] = lim[f'''(x)/g'''(x)] = .....

When we do lim[f'(x)/g'(x)], we end up with 0/0, so we continue and work out lim[f''(x)/g''(x)]
But, lim[f''(x)/g''(x)] also ends up as 0/0.

Eventually we will get f^4(x) = 24 and g^4(x) = -24 (where f^4 is the 4th derivative of f )
(You will also be donig an awful lot of fairly tedious algebra workingout the various derivatives for g(x), but you will get the result!)

So, lim[f(x)/g(x)] = lim[f'(x)/g'(x)] = lim[f''(x)/g''(x)] = lim[f'''(x)/g'''(x)] = lim[f^4(x)/g^4(x)] = 24/-24 = -1.
9. How to do question 2? Thank you.
10. (Original post by superkillball)
How to do question 2? Thank you.
Well it looks like it's wrong!

(Original post by RichE)
Question two is clearly wrong - e.g take f(x) = x on (1/3,2/3). Rather it should say that |f'(x)| < 1 (I don't think you even need it bounded away from 1) and then apply the mean-value theorem to the fixed point alpha and a second hypothetical fixed point, to get your contradiction.

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