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why sin1/x is treated as bounded function?
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think about the values c cannot be?
i mean x
now compare it to a normal sinx graph
because even though 1/x is unbounded when approaching 0, the sine function itself is bounded.
hey joemath
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\[\sin(junk)\] will always be bounded @hahd yo :)
@joe come on chat
If you think about sin as the \(y\) component of an arm on a unit circle, regardless of the angle the \(y\) component will always be between \(-1\) and \(1\)
So the function is bounded above by 1 and bounded below by -1.
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