Evaluate: lim x--0 xsin(1/x) by using the Squeeze theorem.

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Although the function `sin ( 1 / x ) ` infinitely oscillates between `1 ` and `- 1 ` as `x ` tends to zero (so it does not have a limit), the given product of `x ` and `sin ( 1 / x ) ` does have a limit, and it is zero. And the Squeeze theorem can be applied here.

This theorem requires two functions that estimate the given function from the below and from the above and both have the same limit at the given point. Then these functions guarantee the same limit for the initial function.

Indeed, sine is bounded in absolute value by `1 ` for any values of the argument, `-1 lt= sin ( 1 / x ) lt= 1 . ` Therefore we can estimate `f ( x ) ` from the above and from the below:

`- | x | lt= | f ( x ) | = | x sin ( 1 / x ) | = | x | * | sin ( 1 / x ) | lt= | x |,` so
`- | x | lt= x sin ( 1 / x ) lt= | x | .`

It is clear that both `- | x | ` and `| x | ` have limit `0 ` as `x ` tends to `0 .`

This way the function `f ( x ) ` is continuous at the entire real line (no problems for `x != 0`). But note that it does not have a derivative at `x = 0 .`

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