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    (Original post by Swayum)
    What level of maths are you at? What is your definition of a limit?

    And don't worry, you're not being a pain or anything. Keep em coming.

    Err I did A-level maths and now doing an intro maths module at Uni. Never did limits at a-level and dont understand them at all..

    Er its  f'(x)=lim h -> 0  \frac{(fx+h)-f(x)}{h}
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    (Original post by ukdragon37)
    I suspect by "limit definition" you mean:
     f'(x) = \lim_{h \rightarrow 0} \dfrac{[4+(x+h)^2]^2-(4+x^2)^2}{h}

    The thing to remember is that you can only substitute h with 0 once it is not in the denominator. Since the question gives you a value of x, why don't you put that in and see what happens?

    OKay, when I follow through the above with x=0 I end up with

     \frac{8h^2 +h^4}{h}

    then can cancel one h thorughout...to give 8h +h^3 but no idea what that means...or if its even allowed!
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    (Original post by fn_101)
    OKay, when I follow through the above with x=0 I end up with

     \frac{8h^2 +h^4}{h}

    then can cancel one h thorughout...to give 8h +h^3 but no idea what that means...or if its even allowed!
    You are on the right lines. When you are taking the limit as h approaches 0, h gets arbitrarily small but is not at 0. Therefore it's valid to cancel h from top and bottom. Once h is not in the denominator, you can substitute h for 0 as a limit is what the expression would arrive at if h is allowed to be 0.
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    (Original post by fn_101)
    OKay, when I follow through the above with x=0 I end up with

     \frac{8h^2 +h^4}{h}

    then can cancel one h thorughout...to give 8h +h^3 but no idea what that means...or if its even allowed!
    It's fine to divide through there.

    Let me try and explain what's happening though:

    We're basically saying that "Ok, I know a point (x, f(x)) lies on the graph of my function. Now I want to find the gradient of the tangent to my function at that point. I know that the gradient is kind of change in y/change in x. So let me consider a point close to my original point (x, f(x)). Let's say that its x coordinate is x+h, then its y coordinate must be f(x + h). Now ideally, I want x+h to be as close as possible to x, because (you can see from a sketch) then the gradient will be something meaningful (i.e. it'll be the gradient of the tangent at the point (x, f(x))). So then:

    gradient = change in y/change in x

    = (f(x + h) - f(x))/(x + h - x)"

    Do you see that? Then this idea of getting x + h as close as possible to x is this idea of taking a limit.

    You've worked out that change in y/change in x = 8h + h^3. Now as h becomes smaller and smaller, 8h + h^3 will tend towards 0, right? So the gradient is 0.

    Google "differentiation by first principles" for a better explanation.
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    (Original post by Swayum)
    It's fine to divide through there.

    Let me try and explain what's happening though:

    We're basically saying that "Ok, I know a point (x, f(x)) lies on the graph of my function. Now I want to find the gradient of the tangent to my function at that point. I know that the gradient is kind of change in y/change in x. So let me consider a point close to my original point (x, f(x)). Let's say that its x coordinate is x+h, then its y coordinate must be f(x + h). Now ideally, I want x+h to be as close as possible to x, because (you can see from a sketch) then the gradient will be something meaningful (i.e. it'll be the gradient of the tangent at the point (x, f(x))). So then:

    gradient = change in y/change in x

    = (f(x + h) - f(x))/(x + h - x)"

    Do you see that? Then this idea of getting x + h as close as possible to x is this idea of taking a limit.

    You've worked out that change in y/change in x = 8h + h^3. Now as h becomes smaller and smaller, 8h + h^3 will tend towards 0, right? So the gradient is 0.

    Google "differentiation by first principles" for a better explanation.

    Thanks for the explanation It's making a bit more sense, think I just need to keep reading it through with examples...

    I'm confused on one I'm trying now though where

     f(x) =2+x^2    then f'(1)=2

    I don't really get what I'm supposed to sub into the numerator...is it supposed to be [2+ (x+h)^2 - (2+x)^2 ] and then sub in x =1 throughout?
    Or is it [(2+x+h)^2 - (2+x)^2 ] ?
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    (Original post by fn_101)
    Thanks for the explanation It's making a bit more sense, think I just need to keep reading it through with examples...

    I'm confused on one I'm trying now though where

     f(x) =2+x^2    then f'(1)=2

    I don't really get what I'm supposed to sub into the numerator...is it supposed to be [2+ (x+h)^2 - (2+x)^2 ] and then sub in x =1 throughout?
    Or is it [(2+x+h)^2 - (2+x)^2 ] ?
    It's the first one but it should be -(2+x^2) instead of (2+x)^2. You can tell be replacing x as (x+h) for the f(x+h) part.
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    (Original post by ukdragon37)
    It's the first one but it should be -(2+x^2) instead of (2+x)^2. You can tell be replacing x as (x+h) for the f(x+h) part.
    Sorry I'm confused, why should it be - (2+x^2)...and where did the h just go?
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    (Original post by fn_101)
    Sorry I'm confused, why should it be - (2+x^2)...and where did the h just go?
    I mean it should be [2+ (x+h)^2 - (2+x^2)]. f(x) is (2+x^2) which is different to (2+x)^2 which you said the first time.
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    (Original post by ukdragon37)
    I mean it should be [2+ (x+h)^2 - (2+x^2)]. f(x) is (2+x^2) which is different to (2+x)^2 which you said the first time.
    Ah okay thankyou

    I cannot remember how to do these inequality things at all...I've been looking at some other examples but can't get this one to work...

     \frac{4x^2+2x +4}{x+1}  <= 3x + 2
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    (Original post by fn_101)
    Ah okay thankyou

    I cannot remember how to do these inequality things at all...I've been looking at some other examples but can't get this one to work...

     \frac{4x^2+2x +4}{x+1}  <= 3x + 2
    Do a polynomial long division on the left hand side and then rearrange and simplify. You can solve it as if it's a quadratic equality and then sketch the curve to find the appropriate directions of the signs. Remember that if you multiply by a negative number on both sides (could include x if x is negative) you have to switch the direction of the sign.
 
 
 
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