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Thread: Limit derivative questions

  1. June 9th 2011, 08:44 PM #1
    TeaWithoutMussolini
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    Limit derivative questions

    Use L'opital to find the following limit:

    \lim_{x \to \0}\frac{log(1+x)-x}{cos(2x)-1}

    Second Q (unrelated):
    Use the definition of a derivative (which I know) to show that cos x is the derivative of sin x.

    ^ Ok for this I wrote it out with the h's as per normal. I figured, Change the numerator using sina-sinb = 2sin(x-y/2)cos(x+y/2) but then I'm a little brammed.

    For the first q, the l'Opital, I make it zero but I'm having trouble convincing myself. Both the derivatives are 0 at x=0?
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  2. June 9th 2011, 08:56 PM #2
    sa-ri-ga-ma
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    Taking the derivatives you get { [1/(1+x)] -1}/-2in(2x).
    Now simplify and find the limits.
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  3. June 9th 2011, 09:03 PM #3
    TeaWithoutMussolini
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    Um hum ya, ya, but simplify, how? Indeed? I can differentiate them but?
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  4. June 9th 2011, 09:09 PM #4
    chisigma
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    Quote Originally Posted by TeaWithoutMussolini View Post
    Use L'opital to find the following limit:

    \lim_{x \to \0}\frac{log(1+x)-x}{cos(2x)-1}

    ... the l'Opital, I make it zero but I'm having trouble convincing myself. Both the derivatives are 0 at x=0?
    ... then try to use L'Hopital again!...

    Kind regards

    \chi \sigma
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  5. June 9th 2011, 09:14 PM #5
    TeaWithoutMussolini
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    Is it not ok to stick with what I get? Or does L'Hopital HAVE to result in a none-zero denominator? Is that the only reason to continue? I am convinced that f(x), g(x), f'(x) AND g'(x) are all zero at x=0...

    How about this question@

    Show cos x is the derivative of sin x using the definition of a derivative
    Last edited by TeaWithoutMussolini; June 9th 2011 at 09:27 PM.
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  6. June 9th 2011, 09:27 PM #6
    chisigma
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    Applying L'Hopital rule You obtain first...

    \lim_{x \rightarrow 0} \frac{\ln (1+x)-x}{\cos 2x -1} = \lim_{x \rightarrow 0} \frac{\frac{1}{1+x}-1}{-2\ \sin 2x}

    ... and now, applying L'Hopital rule again...

    \lim_{x \rightarrow 0} \frac{\frac{1}{1+x}-1}{-2\ \sin 2x}= \lim_{x \rightarrow 0} \frac{-\frac{1}{(1+x)^{2}}}{-4\ \cos 2x}

    ... and the goal is realized!...

    Kind regards

    \chi \sigma
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  7. June 9th 2011, 09:32 PM #7
    TeaWithoutMussolini
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    The answer is 1/4, not zero? Oh

    thanks

    please is this the reason? that it must not equal zero in the denominator? In three hours another question like this might come up, I do want to be prepared to know how to tackle it.
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  8. June 9th 2011, 09:43 PM #8
    chisigma
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    Quote Originally Posted by TeaWithoutMussolini View Post
    How about this question@

    Show cos x is the derivative of sin x using the definition of a derivative
    Using the definition of derivative You have...

    \frac{d}{dx} \sin x = \lim_{h \rightarrow 0} \frac{\sin (x+h)-\sin x}{h}= \lim_{h \rightarrow 0} (\sin x\ \frac{\cos h-1}{h} + \cos x\ \frac{\sin h}{h})

    Now using the 'fundamental limits' ...

    \lim_{h \rightarrow 0} \frac{\sin h}{h}=1 (1)

    \lim_{h \rightarrow 0} \frac{\cos h -1}{h}=0 (2)

    ... You prove that...

    \frac{d}{dx} \sin x = \cos x (3)

    Kind regards

    \chi \sigma
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  9. June 9th 2011, 09:46 PM #9
    TeaWithoutMussolini
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    What rules have you used in that first step please?
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  10. June 9th 2011, 09:48 PM #10
    chisigma
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    The definition of derivative of an f(x) is...

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

    Kind regards

    \chi \sigma
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  11. June 9th 2011, 09:52 PM #11
    TeaWithoutMussolini
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    No no no I mean the first step... changing it into cosh-1/h etc?
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  12. June 9th 2011, 10:04 PM #12
    chisigma
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    Quote Originally Posted by chisigma View Post
    Using the definition of derivative You have...

    \frac{d}{dx} \sin x = \lim_{h \rightarrow 0} \frac{\sin (x+h)-\sin x}{h}= \lim_{h \rightarrow 0} (\sin x\ \frac{\cos h-1}{h} + \cos x\ \frac{\sin h}{h})

    Now using the 'fundamental limits' ...

    \lim_{h \rightarrow 0} \frac{\sin h}{h}=1 (1)

    \lim_{h \rightarrow 0} \frac{\cos h -1}{h}=0 (2)

    ... You prove that...

    \frac{d}{dx} \sin x = \cos x (3)

    Kind regards

    \chi \sigma
    An alternative and more 'rigorous' way uses the trigonometric identity...

    \sin p - \sin q = 2\ \sin \frac{p-q}{2}\ \cos \frac{p+q}{2} (1)

    ... so that is...

    \frac{d}{dx} \sin x = \lim_{h \rightarrow 0} \frac{\sin (x+h)-\sin x}{h}= \lim_{h \rightarrow 0} \frac{\sin \frac{h}{2}}{\frac{h}{2}} \ \cos (x+\frac{h}{2}) (2)

    ... and from (2) and the 'fundamental limit' \lim_{h \rightarrow 0} \frac{\sin h}{h}=1 You derive that is...

    \frac{d}{dx} \sin x = \cos x

    Kind regards

    \chi \sigma
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  13. June 9th 2011, 10:31 PM #13
    TeaWithoutMussolini
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    Okay so p is x+h and q is x?

    I can do that.

    limit of sinh/h = 1, is a result that can just be quoted? No proof required?
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  14. June 9th 2011, 10:39 PM #14
    chisigma
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    The 'fundamental limit' \lim_{\theta \rightarrow 0} \frac{\sin \theta}{\theta}=1 has to be prooved 'geometrically' as in...

    Limits Of Trigonometric Functions | Limits Theorems | Tutorvista.com

    Kind regards

    \chi \sigma
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