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

[ashleyrc]

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

Homework Statement

Use the sandwich theorem to find:
lim as x approaches 0 of xcos(x).

Homework Equations

I know that the range is between -1 and 1
-1<xcos(x)<1

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!

 

[Dick]

Homework Statement

Use the sandwich theorem to find:
lim as x approaches 0 of xcos(x).

Homework Equations

I know that the range is between -1 and 1
-1<xcos(x)<1

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!

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

 

[ashleyrc]

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?

 

[ashleyrc]

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.

 

[danago]

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.

Pretty much. You shouldn't 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!

 

[ashleyrc]

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

 

[kimanampaka]

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.

 

[kimanampaka]

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,abs(x),-abs(x)"

Good luck