1. ## 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?

2. Taking the derivatives you get { [1/(1+x)] -1}/-2in(2x).
Now simplify and find the limits.

3. Um hum ya, ya, but simplify, how? Indeed? I can differentiate them but?

4. Originally Posted by TeaWithoutMussolini
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

$\chi$ $\sigma$

5. 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...

Show cos x is the derivative of sin x using the definition of a derivative

6. 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

$\chi$ $\sigma$

7. 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.

8. Originally Posted by TeaWithoutMussolini

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

$\chi$ $\sigma$

9. What rules have you used in that first step please?

10. The definition of derivative of an f(x) is...

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

Kind regards

$\chi$ $\sigma$

11. No no no I mean the first step... changing it into cosh-1/h etc?

12. Originally Posted by chisigma
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

$\chi$ $\sigma$

13. 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?

14. 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

$\chi$ $\sigma$