Is the result of multiplication of infinite number and bounded function infinite number

Is the result of multiplication of infinite number and bounded function infinite number

no
For example, Lim X - > 0 (SiNx) × (1 / x)
Obviously, SiNx is a bounded function and 1 / X is infinite when X - > 0
But Lim X - > 0 (SiNx) × (1 / x) = 1

Is the function f (x) = 1 / xsin1 / X bounded on (0,1 / 2)
Is the function f (x) = 1 / X multiplied by sin1 / X bounded on (0,1 / 2)

Let x = 1 / (2k π + π / 2) k = 1, 2, 3 be unbounded, then x tends to 0, so f (x) = (2k π + π / 2) * sin (2k π + π / 2) = (2k π + π / 2) tends to positive infinity, so f (x) has no upper bound, and then let x = 1 / (2k π - π / 2) k = 1, 2, 3

F (x + 13 / 42) + F (x) = f (x + 6) + F (x + 1 / 7) f (x) is a bounded real function. It is proved that f (x) is a periodic function

If the title is wrong, it should be f (x + 13 / 42) + F (x) = f (x + 1 / 6) + F (x + 1 / 7)
Let g (x) = f (x + 1 / 7) - f (x), then G (x + 1 / 6) = f (x + 13 / 42) - f (x + 1 / 6) = f (x + 1 / 7) - f (x) = g (x), that is, 1 / 6 is the period of G (x)
Let H (x) = f (x + 1) - f (x) = (f (x + 1) - f (x + 6 / 7)) +... + (f (x + 1 / 7) - f (x)) = g (x + 6 / 7) +... + G (x)
Then 1 / 6 is also the period of H (x), so 1 is also the period of H (x)
If not, let H (a) = B ≠ 0, then for any integer k, H (a + k) = H (a) = B
Then f (a + k) - f (a) = H (a + k-1) + H (a + K-2) +... + H (a) = KB, the absolute value of F (a + k) - f (a) = H (a + k-1) +... + H (a) = KB, which is contradictory to f (x) boundedness
So we get f (x + 1) - f (x) = H (x) = 0, that is, 1 is the period of F (x)