arildno
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Looks good to me. 
As long as f is continuous, and the area beneath the curves formed by each member in your sequence equals 1, then this argument really shows that ANY such sequence is good enough to represent the delta function(al).
As long as f is continuous, and the area beneath the curves formed by each member in your sequence equals 1, then this argument really shows that ANY such sequence is good enough to represent the delta function(al).