7. Prove the function y = 1/x * sin (1/x). ...

Proof: The transcendental proof function has no upper bound in the interval (0, 1).

Because any M>0, the point x0 can always be found in (0, 1], so f (x0) >; M. For example, if x0= 1/(2kπ+π/2) and k∈N, then f(x0)=2kπ+π/2. When k is large enough, f (x0) >: m. So the function is on the upper bound of (0, 1).

It is proved that the function y = f (x) =1/x * sin (1/x) is not the infinity of x→0+.

Because any M>0, δ > 0, you can always find point x0, so 0.

Finally finished. Very sad. Add something.

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