# (limits) sandwich theorem - lim x approaches 0 of xcos(x)

1. Sep 14, 2008

### ashleyrc

(limits) sandwich theorem -- lim x approaches 0 of xcos(x)

1. The problem statement, all variables and given/known data
Use the sandwich theorem to find:
lim as x approaches 0 of xcos(x).

2. Relevant equations
I know that the range is between -1 and 1
-1<xcos(x)<1

3. The attempt at a solution
-1(x)<xcos(x)<1(x)???
Then what do i do? i think the answer is 0, but i figured that out by graphing, i need to find it analytically. please help!

2. Sep 14, 2008

### Dick

Re: (limits) sandwich theorem -- lim x approaches 0 of xcos(x)

Now take the limits of -1*x and 1*x. Your limit of x*cos(x) is somewhere between those two limits, right?

3. Sep 14, 2008

### ashleyrc

Re: (limits) sandwich theorem -- lim x approaches 0 of xcos(x)

ok, so -x<xcos(x) <x, but how am i going to get a value from that? do i need to use the unit circle?
and would i need to multiply cos(x) by x too? =cos(x)^2???

4. Sep 14, 2008

### ashleyrc

Re: (limits) sandwich theorem -- lim x approaches 0 of xcos(x)

i'm not sure if it helps, but when i graph -x and x, the limits as x approaches 0 for both of them are 0, so i think that should pinpoint the limit for xcos(x), however, i don't know if that is actually the right way to answer the question.

5. Sep 14, 2008

### danago

Re: (limits) sandwich theorem -- lim x approaches 0 of xcos(x)

Pretty much. You shouldnt really need to graph x and -x, since they are polynomials (hence contininuous), you can just sub x=0 into them and find that the limit is zero. Since x cos x must be somewhere in between (or equal) to 0 and 0, the only possibility is that the limit of x cos x is also zero!

6. Sep 14, 2008

### ashleyrc

Re: (limits) sandwich theorem -- lim x approaches 0 of xcos(x)

Thanks! i guess i forgot about subbing in the 0 at the end. i understand it a lot better now!

7. Nov 2, 2009

### kimanampaka

Re: (limits) sandwich theorem -- lim x approaches 0 of xcos(x)

When you solve limit problems first of all try substitution, and other methods come after, like the usage of the calculator, sandwich limit and algebra.

In this case begin by substitution :

When you plug 0 into xcos(x) you got 0*cos(0) (remember cosine of 0 equals 1),
we have now 0*1.
and the answer to your limit problem is 0.

I hope i'ts going to help you
Good luck.

My english is not so good, I only speak french.

8. Nov 2, 2009

### kimanampaka

Re: (limits) sandwich theorem -- lim x approaches 0 of xcos(x)

With the sandwich method :

Because the range of the cosine function is from negative 1 to positive 1, whenever you multiply a number by the sine of anything, the result either stays the same distance from zero or gets closer to zero. Thus, xcos(x) will never get above |x| or below -|x|. So try graphing the functions |x| and -|x| along with xcos(x) to see if |x| and -|x| make adequate bread functions for xcos(x).

let f(x)=|x|, g(x)=xcos(x) and h(x)=-|x|

f(x)>=g(x)>=h(x)

and limit when x apporaches 0 of f(x)=limit when x approaches 0 of h(x), it follows that g(x) must have the same limit.

limit when x approaches 0 of g(x)=0

check the graph here http://www.wolframalpha.com/input/?i=xcosx%2Cabs%28x%29%2C-abs%28x%29

Good luck

Last edited: Nov 2, 2009