Question

Consider the limit. Use a theorem to find the limit symbolically/algebraically.

Answer 1

\[\lim_{x \rightarrow 0^{+}} xcos(\frac{ 1 }{ x })\]

Answer 2

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

-x <= x cos(1/x) <= x (since x is positive)

lim x->0+ (-x) = 0
lim x->0+ (x) = 0

By $5 foot long subway theorem,
lim x->0+ (x cos(1/x)) = 0

:D

Answer 3

Try using the squeeze theorem. Notice that y=xcos(1/x) is always < y=x and always greater than -x (as cosine is always between -1 and 1. Since both x and -x approach 0 as x approaches 0 and xcos(1/x) is always between -x and x, the limit has to approach 0.

Answer 4

Oh, yes I did try the Squeeze Theorem, but wasn't so sure on how to write my final answer, because leaving it at "x" seemed incomplete... I don't think I'm phrasing my question correctly, though.... In other words, I know the limit is zero, but why is it that x = -1 on one side, and x = 1 on the other side, but when I substitute the values, and I still get zero as the limit for both?... I think it's a matter of understanding the way the Squeeze Theorem is written which is what is confusing me. I arrived at the same answer as sourwing, but the result seems to be hanging--at least, for me....

Answer 5

In other words, why isn't "0" written on the left-hand and right-hand sides of the Squeeze Theorem's result (for this problem)? Or is it just assumed that one reads it as "0" for the limit?

Answer 6

I'm not sure if i exactly get what you're saying... but if you're saying what I think you're saying, then yes, the limit from either the left or the right will still evaluate to 0

Answer 7

Thanks for both your help!