a) Calculate the limit value to $\displaystyle \frac{sinx-x}{x(cosx-1)}$ when $\displaystyle x\rightarrow0$
The point is to use Taylor or McLaurin polynomial..?.
B) Make a smilar task that will equal $\displaystyle \frac{21}{5}$
$\displaystyle \sin(x)-x=\bigg[x-\frac{x^3}{3!}+\frac{x^5}{5!}-...\bigg]=\frac{-x^3}{3!}+\frac{x^5}{5!}-...\sim\frac{-x^3}{3!}$
and $\displaystyle x(\cos(x)-1)=x\bigg[1-\frac{x^2}{2!}+\frac{x^4}{4!}-...-1\bigg]=x\bigg[\frac{-x^2}{2!}+\frac{x^4}{4!}-...\bigg]=\frac{-x^3}{2!}+\frac{x^5}{4!}-...\sim\frac{-x^3}{2!}$
So we have that $\displaystyle \lim_{x\to{0}}\frac{\sin(x)-x}{x(\cos(x)-1)}\sim\lim_{x\to{0}}\frac{\frac{-x^3}{3!}}{\frac{-x^2}{2!}}=\frac{1}{3}$
Im sorry but I donīt understand much of this? Isnīt the point when calculating with lim that if $\displaystyle x\rightarrow0$ 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 $\displaystyle \frac{-x^3}{2!}$ 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:
Now divide numerator and denominator by $\displaystyle x^3$:
$\displaystyle = \lim_{x\to{0}}\frac{ \frac{-1}{3!} + \frac{x^2}{5!} -...}{ \frac{-1}{2!} + \frac{x^2}{5!} -...}$
Now "put x = 0 in the expression" as you say it: $\displaystyle = \frac{\frac{-1}{3!}}{\frac{-1}{2!}} = \frac{2!}{3!}$ $\displaystyle = \frac{1}{3}$