Find limx->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'H˘pital'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.
-1 < sin(x) < 1 <=> 1 > -sin(x) > 1 <=> x+1 > x-sin(x) > x-1
limx->inf x+1 = limx->inf x-1 = inf => x-sin(x) = inf (Squeeze)
Same result for x-sin(sin(x)).
limx->inf x-sin(sin(x)) = limx->inf x-sin(x) = inf => limx->inf (x-sin(sin(x))) / (x-sin(x)) = 1?