# L'Hopital rule Help!

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#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.

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#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.]

= 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.]

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#7

(Original post by

I still don't know how to do them

**superkillball**)I still don't know how to do them

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#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.

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.

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#10

(Original post by

How to do question 2? Thank you.

**superkillball**)How to do question 2? Thank you.

(Original post by

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.

**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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