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.

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- Oct 30th 2012, 03:21 AMCinnamanLimit problem
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. - Oct 30th 2012, 07:34 AMhollywoodRe: Limit problem
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 - Oct 30th 2012, 08:15 AMCinnamanRe: Limit problem
-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? - Oct 30th 2012, 04:41 PMhollywoodRe: Limit problem
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