Problem: Find the following limits if they exist. a) lim x->Infinity (x-sin(x))/x Work done so far: g(x) = x which goes to infinty as x goes to infinity, thus applying L'Hospitals Rule the limit must equal: 1-cos(x) / 1 = 1-cos(x) as x approaches infinity. Obviously... cos(x) oscillates from 1 to -1... not giving us a real limit. Alternative: (x-sin(x))/x = 1 - sin(x)/x = 1 - (1/x)sin(x) as X approachs infinity this limit becomes 1 - 0 = 1. My problem is that the back of the book tells me the answer is 0... I have no clue how they got this and I was wondering if anyone could give me a push in the right direction.
Be careful when you're applying L'hopital's rule. i.e. is (x - sinx) / x of the 0 over 0 type? Infinity over infinity type? ie, what is the limit of (2x+3)/(x-4) when x approaches 5? The limit would be 13/1 = 13. Applying L'Hopital's rule, when it shouldn't be applied, would result in 2/1 = 2 (which is incorrect) for this problem. Also, the limit of sin(x) / x is not equal to 0. And, you can apply L'hopital's rule to this one (or the limit should have been proven earlier in your text by using the squeezing theorem.)
How can I apply L'hopitals rule effectively as x approaches infinity for sin(x)/x seems to be my main concern. sin(x)/x (Allowed to use since x-> Inf.) Thus the limit is equal to cos(x) as x approaches infinity. But cos(x) has no limit as x approaches infinity.
I'm starting the believe the book is just wrong in this case... When I graph (x-sin(x))/x it leads me to believe the limit approaches 1 as x goes to infinity as I keep coming up with. Proof: (x-sin(x))/x = 1-sin(x)/x = 1-(1/x)sin(x) Lim as x-> Infinity = 1 - 0 * sin(x) = 1-0*[-1,1] (range of sin), though since its times 0 it doesnt really matter... thus = 1 - 0 = 1.
The limit of sinx / x as x approaches infinity isn't one of the indeterminant cases though; it's not 0 over 0, nor is it the type infinity over infinity. The value of that part of the limit is zero. Thus, your fraction is equal to x/x - (sinx)/x. The first part is 1, the second part is 0, the value of the limit is 1, as you originally stated in your alternative method (which I believe is correct.) edit edit: all is well in my head again... Your original alternate version is correct. Since you can easily re-write the function such that it isn't an infinity over infinity type, you don't apply L'Hopital's rule.
What did you make the original function into? I would like to see a more direct proof as I'm not totally sure how rigorous it is to say that 0* anything in the range is 0 for the proof.
[tex]x-\sin(x)[/tex] goes to [itex]0[/itex] as [itex]x[/itex] goes to [itex]0[/itex]. as does [itex]x[/itex], so it's a [itex]\frac{0}{0}[/itex]. Now, taking the derivative of the top and bottom, the answer should be pretty clear. ... Dr. Pizza: Please be careful when answering people: [tex]\lim_{x \rightarrow 0} \frac{\sin(x)}{x}=1[/tex]
x isnt going to 0. The problem says x goes to infinity Edit: Dont be mean to Pizza, he made the same assumption as you which I believe the book typoed on this problem. I would assume the book meant to put as x approaches zero as that would make the most sense for the problem (at least when the chapter is proving L'H rule). Though, as it stands the problem has approaching infinity, which allows you to use L'H rule... though the derivative has no clear limit L.
Lot's of cranial methane here... Ah, so it should be infinity - if you don't want to deal with the derivative, you can still squeeze the limit. [tex]\frac{x-1}{x}\leq \frac{x-\sin(x)}{x} \leq \frac{x+1}{x}[/tex] Now, you have two expressions (on the outside) that are easy to apply l'Hopital's rule to.