(a) (i)limn→∞(2n+1−2n)
Rationalising the given expression-
=limn→∞(2n+1−2n)×2n+1+2n2n+1+2n=limn→∞2n+1+2n(2n+1)2−(2n)2=limn→∞2n+1+2n1
Applying limits
= 0
(ii) limn→∞4n(n−1)(n−2)3n3−n+8
Now writing it my simplifying-
=limn→∞4n(n2−2n−n+2)3n3−n+8=limn→∞4n3−12n2+8n3n3−n+8=limn→∞n3(4−n12+n28)n3(3−n21+n38)=4−0+03−0+0=43
(b) Given, sequence
an=(−1)n−2n
a1=−1−2=−3a2=1−4=−3a3=−1−6=−7a4=−7a5=−1−10=−11a6=1−12=−11
⇒<an>=−3,−3,−7,−7,−11,−11...
Here, a1≥a2≥a3≥a4≥a5≥a6...
we know a sequence <an> is monotonic decreasing if an+1≤an∀n∈N
Hence Given sequence is monotone decreasing.
So limn→∞(an)=−∞.
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