Question

Limits

Answer 1

\[\LARGE \lim_{n \rightarrow \infty} \cos^2 \pi(\sqrt[3]{n^3+n^2+2n}-n)=k\]
Then find 8k+3

Answer 2

ratonalize the eqn of n inside cos

Answer 3

tried,no avail,show me please

Answer 4

thn use (a-b)(a^2+ b^2 +ab) = a^3 -b^3

Answer 5

do you know how to take derivative yet?
L'Hopitals Rule?

Answer 6

oh well, here is how i would do it
sub n = 1/u , change limit to u->0
\[\large \lim_{u \rightarrow 0} \sqrt[3]{\frac{1}{u^{3}}+\frac{1}{u^{2}}+\frac{2}{u}} - \frac{1}{u}\]
\[\large = \lim_{u \rightarrow 0} \frac{\sqrt[3]{1+u+2u^{2}}-1}{u}\]
this results in 0/0
apply L'hopitals...differentiate
\[\large \lim_{u \rightarrow 0} \frac{4u+1}{3(1+u+2u^{2})^{2/3}} = \frac{1}{3}\]
\[k = \cos^{2} (\pi*\frac{1}{3}) = \frac{1}{4}\]
\[8k+3 = 8*\frac{1}{4}+3 = 5\]

Answer 7

how would @shubhamsrg do it :)

Answer 8

divu sir ne jo kiya wo bhi kar sakte ho
a= cuberoot(n^3 +n^2 +2n)
b= n
a-b likha hai
a^2 + b^2 + ab se multiply kardo

Answer 9

multiply divide