1. Trig. Limits

a) Calculate the limit value to $\frac{sinx-x}{x(cosx-1)}$ when $x\rightarrow0$

The point is to use Taylor or McLaurin polynomial..?.

B) Make a smilar task that will equal $\frac{21}{5}$

2. Originally Posted by lynch-mob
a) Calculate the limit value to $\frac{sinx-x}{x(cosx-1)}$ when $x\rightarrow0$

The point is to use Taylor or McLaurin polynomial..?.

B) Make a smilar task that will equal $\frac{21}{5}$
Have you tried substituting the Maclaurin Series for sin x and cos x and then simplifying the resulting expression?

3. too hard

Originally Posted by mr fantastic
Have you tried substituting the Maclaurin Series for sin x and cos x and then simplifying the resulting expression?
The Mclaurin series for $sinx$ is something like $\frac{x^{2n-1}}{(2n-1)!}$

and for cos something like $\frac{x^{2n}}{2n!}$

So should I put these straight into the formula replacing sin and cos?

4. Originally Posted by lynch-mob
The Mclaurin series for $sinx$ is something like $\frac{x^{2n-1}}{(2n-1)!}$

and for cos something like $\frac{x^{2n}}{2n!}$

So should I put these straight into the formula replacing sin and cos?
Yes, but only the first couple of terms of each are necessary.

5. Originally Posted by lynch-mob
a) Calculate the limit value to $\frac{sinx-x}{x(cosx-1)}$ when $x\rightarrow0$

The point is to use Taylor or McLaurin polynomial..?.

B) Make a smilar task that will equal $\frac{21}{5}$
$\sin(x)-x=\bigg[x-\frac{x^3}{3!}+\frac{x^5}{5!}-...\bigg]=\frac{-x^3}{3!}+\frac{x^5}{5!}-...\sim\frac{-x^3}{3!}$

and $x(\cos(x)-1)=x\bigg[1-\frac{x^2}{2!}+\frac{x^4}{4!}-...-1\bigg]=x\bigg[\frac{-x^2}{2!}+\frac{x^4}{4!}-...\bigg]=\frac{-x^3}{2!}+\frac{x^5}{4!}-...\sim\frac{-x^3}{2!}$

So we have that $\lim_{x\to{0}}\frac{\sin(x)-x}{x(\cos(x)-1)}\sim\lim_{x\to{0}}\frac{\frac{-x^3}{3!}}{\frac{-x^2}{2!}}=\frac{1}{3}$

6. Originally Posted by Mathstud28
$\sin(x)-x=\bigg[x-\frac{x^3}{3!}+\frac{x^5}{5!}-...\bigg]=\frac{-x^3}{3!}+\frac{x^5}{5!}-...\sim\frac{-x^3}{3!}$

and $x(\cos(x)-1)=x\bigg[1-\frac{x^2}{2!}+\frac{x^4}{4!}-...-1\bigg]=x\bigg[\frac{-x^2}{2!}+\frac{x^4}{4!}-...\bigg]=\frac{-x^3}{2!}+\frac{x^5}{4!}-...\sim\frac{-x^3}{2!}$

So we have that $\lim_{x\to{0}}\frac{\sin(x)-x}{x(\cos(x)-1)}\sim\lim_{x\to{0}}\frac{\frac{-x^3}{3!}}{\frac{-x^2}{2!}}=\frac{1}{3}$
Im sorry but I don´t understand much of this? Isn´t the point when calculating with lim that if $x\rightarrow0$ then you put x=0 in the expression. When i do that i get only zeros.

And what happens to the x and 1 in x(cosx-1)? Same with the -x in sinx - x??

And how do you get that the cosx (mclaurin...) expression is approximately $\frac{-x^3}{2!}$ and the same with the sin expression?

Hope you have time for at least a brief explanation.

7. Originally Posted by lynch-mob
Im sorry but I don´t understand much of this? Isn´t the point when calculating with lim that if $x\rightarrow0$ then you put x=0 in the expression. When i do that i get only zeros.

And what happens to the x and 1 in x(cosx-1)? Same with the -x in sinx - x??

And how do you get that the cosx (mclaurin...) expression is approximately $\frac{-x^3}{2!}$ and the same with the sin expression?

Hope you have time for at least a brief explanation.

I have done some editing to Mathstud28's reply, but you should have realised (and in fact done) these things yourself:

Originally Posted by Mathstud28 and slightly edited by Mr F to save reinventing the wheel

$\sin(x)-x=\bigg[x-\frac{x^3}{3!}+\frac{x^5}{5!}-...\bigg]{\color{red} - x} =\frac{-x^3}{3!}+\frac{x^5}{5!}-...$ Mr F says: The bit in red was missing. Do you understand this part?

and $x(\cos(x)-1)=x\bigg[1-\frac{x^2}{2!}+\frac{x^4}{4!}-...-1\bigg]=x\bigg[\frac{-x^2}{2!}+\frac{x^4}{4!}-...\bigg]=\frac{-x^3}{2!}+\frac{x^5}{4!} -...$ Mr F says: Do you understand this part?

So we have that $\lim_{x\to{0}}\frac{\sin(x)-x}{x(\cos(x)-1)} = \lim_{x\to{0}}\frac{ \frac{-x^3}{3!} + \frac{x^5}{5!} -...}{ \frac{-x^{\color{red}3}}{2!} + \frac{x^5}{5!} -...}$ Mr F says: The bit in red is a correction.
Now divide numerator and denominator by $x^3$:

$= \lim_{x\to{0}}\frac{ \frac{-1}{3!} + \frac{x^2}{5!} -...}{ \frac{-1}{2!} + \frac{x^2}{5!} -...}$

Now "put x = 0 in the expression" as you say it: $= \frac{\frac{-1}{3!}}{\frac{-1}{2!}} = \frac{2!}{3!}$ $= \frac{1}{3}$

8. Reinventing the wheel

Ok now I get it. Sorry guys for being slow.

I missed a few parts which totally jammed me.

I can´t praise this site and these people helping here enough. Thanks!!