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Evaluatelimn{12tanx2+122tanx22+....+12ntanx2n}

a
xtanx2
b
1xcotx2
c
x−cotx2
d
1x−cot x

detailed solution

Correct option is D

limn→∞{12tanx2+122tanx22+....+12ntanx2n}  = limn→∞{−cot x+(cotx+12tanx2+122tanx22+....+12ntan x2n)}limn→∞{−cot x+(12cotx2+122tanx22)+....+12ntan x2n} (∵cot x+12tanx2=12cotx2) Proceeding like this we get =limn→∞{−cotx+12ncotx2n} =−cotx+limn→∞(x/2ntanx2n.1x)=−cotx+1x

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