Squeeze thereom question

compaqlaptop1

Using the squeeze theorem to find lim xCos(1/x) as x approaches 0 -

-1 <= Cos (1/x) <= 1

multiply across by x

-x <= xCos(1/x) <= x

And as lim -x and lim x = 0, limt xCox(1/x) = 0


OK thats fine, but what I dont get is how you can say
-1 <= Cos (1/x) <= 1

As Cos(1/x) is not defined at x = 0.

So why is the squeeze theorem allowed to be used here?

Michael Collins

compaqlaptop1 wrote: »

Using the squeeze theorem to find lim xCos(1/x) as x approaches 0 -

-1 <= Cos (1/x) <= 1

multiply across by x

-x <= xCos(1/x) <= x

And as lim -x and lim x = 0, limt xCox(1/x) = 0


OK thats fine, but what I dont get is how you can say
-1 <= Cos (1/x) <= 1

As Cos(1/x) is not defined at x = 0.

So why is the squeeze theorem allowed to be used here?

You are correct that cos(1/x) is not defined at x = 0, because 1/x is not defined at x = 0. But when you use limits you're asking what happens as x goes arbitrarily close to 0, but not equal to zero. So in this case as x gets smaller and smaller x.cos(1/x) gets closer and closer to zero. Try it yourself with some small values of x.

It's the same with sin(x)/x, as x gets closer and closer to zero, this ratio gets closer and closer to one. But at x = 0, it is not defined.

LeixlipRed

The fact that cos(1/x) isn't defined at x=0 is the initial issue that you need to overcome to evaluate the limit at x=0.

Why is
-1 <= Cos (1/x) <= 1
true?

Well all this is saying is that the range of values of the function f(x)=Cos(1/x) is between -1 and 1. This is true for the Cosine of any angle, the only possible value of the Cos is between -1 and 1. The function may not be defined at x=0 but it's defined at every other x and the range is still between -1 and 1.