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Question Number 34951 by rahul 19 last updated on 13/May/18
limn→∞(e×an)n=?HereaϵR+
Answered by MJS last updated on 14/May/18
ae=p∈R+limn→∞pnnn=LL=limn→∞pnnn=limn→∞ddn[pn]ddn[nn]=limn→∞ln(p)×pn(1+ln(n))×nn==L×limn→∞ln(p)1+ln(n)=L×0=0
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