Using the identity lima→∞∫a0e−xtdt=1x,x>0,
can I get a hint to show that limb→∞∫b0sinxxdx=π2.
Answer
Hint:
limb→∞∫b0sinxxdx=lima,b→∞∫b0∫a0e−xtdtsinxdx=lima,b→∞∫b0dt∫a0e−xteix−e−ix2idx=lima,b→∞∫b0dt∫a0e−(t−i)x−e−(i+t)x2idx
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How to find limh→0sin(ha)h without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...
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