Find lim_{x->inf} (x-sin(sinx)) / (x-sinx)
The followup is to find the limit when x -> 0 for the same expression.
I think this problem is related to the Taylor series chapter, I don't know how to apply that stuff here.
As x goes to infinity, x-sin(sin(x)) and x-sin(x) are both between x-1 and x+1, so you can use the squeeze theorem. I think you can use L'Hôpital's rule for the follow-up.
I think you can also expand the numerator and denominator in Taylor series to get the limit as x goes to 0. To get the Taylor series for sin(sin(x)), you'll need to plug the series for sin(x) into itself - just figure out the first few terms.
- Hollywood
-1 < sin(x) < 1 <=> 1 > -sin(x) > 1 <=> x+1 > x-sin(x) > x-1
lim_{x->inf} x+1 = lim_{x->inf} x-1 = inf => x-sin(x) = inf (Squeeze)
Same result for x-sin(sin(x)).
lim_{x->inf} x-sin(sin(x)) = lim_{x->inf} x-sin(x) = inf => lim_{x->inf} (x-sin(sin(x))) / (x-sin(x)) = 1?
I think that's correct, but a little hard to follow. I would have put it like this:
$\displaystyle \lim_{x\rightarrow\infty}\frac{x+1}{x-1}=\lim_{x\rightarrow\infty}\frac{1+\frac{1}{x}}{1-\frac{1}{x}}=1$
$\displaystyle \lim_{x\rightarrow\infty}\frac{x-1}{x+1}=\lim_{x\rightarrow\infty}\frac{1-\frac{1}{x}}{1+\frac{1}{x}}=1$, and
$\displaystyle \frac{x-1}{x+1} \le \frac{x-\sin(\sin{x})}{x-\sin{x}} \le \frac{x+1}{x-1}$
so by the squeeze theorem,
$\displaystyle \lim_{x\rightarrow\infty}\frac{x-\sin(\sin{x})}{x-\sin{x}}=1$.
- Hollywood