Let be an integrable and uniformly continuous function. Prove that.
Suppose that the limit is not zero. Then let us pick a sequence and an such that foreach to both hold and . Since is uniformly continuous there exists such that
However in the interval it holds that
foreach positive. The last is a consequence of . And of course this contradicts the fact that the function is integrable. Hence the conclusion.
Note: We can’t drop the condition of uniform continuity. If so, then the function is a counterexample. Indeed it is continuous, integrable but the limit at infinity is not .
Let be a continuous function such that
Prove that .
and thus the conclusion.
Evaluate the rational number .
since it is known that for any -th root of unity rather than . Thus:
Thus is there is a at the upper left corner and ‘s along the anti diagonal in the lower right block. Thus:
Evaluate the limit
The fastest way is by making use of probabilities. Let us consider independent Poisson distributions with parameter . Then is Poisson with parameter . From the central limit theorem converges in distribution to the standard normal distribution. In particular, if follows a standard normal distribution then
and the conclusion follows.