Limn -3n - -1n-4n - -1n is equal to

The correct option is D 3/4 lim_n →∞ 3n + ( 1)^n4n ( 1)^n =lim_n →∞ 3 + ( 1)^nn4 + ( 1)^nn = 34 ...

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Summation by Sigma Method

limn →∞-3n + ...

Question

$lim n \to \infty \frac{- 3 n + (- 1)^{n}}{4 n - (- 1)^{n}}$ is equal to

\frac{3}{4}

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0

n

is even

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- \frac{3}{4}

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none of these

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6

n - \frac{n (n - 1) (n - 2)}{⌊ 3} \cdot 3 + \frac{n (n - 1) (n - 2) (n - 3) (n - 4)}{⌊ 5} \cdot 3^{2} - . . . . .,

n - \frac{n (n - 1) (n - 2)}{⌊ 3} \cdot \frac{1}{3} + \frac{n (n - 1) (n - 2) (n - 3) (n - 4)}{⌊ 5} \cdot \frac{1}{3^{2}} - . . . . .,

lim n \to \infty {(\frac{(n + 1) (n + 2) . . . . . .3 n}{n^{2 n}})}^{\frac{1}{n}}

\begin{matrix} l i m n \to \infty \end{matrix} (\frac{(n + 1) (n + 2) . . .3 n}{n^{2 n}})^{\frac{1}{n}}

$l i m n \to \infty {(\frac{(n + 1) (n + 2) . . .3 n}{n^{2 n}})}^{\frac{1}{n}}$ is equal to:

∣ ∣ ∣ ∣ ∣ \begin{matrix} {n + 2}_{C_{n}} & n + 3_{C_{n + 1}} & n + 4_{C_{n + 2}} n + 3_{C_{n + 1}} & n + 4_{C_{n + 2}} & n + 5_{C_{n + 3}} n + 4_{C_{n + 2}} & n + 5_{C_{n + 3}} & n + 6_{C_{n + 4}} \end{matrix} ∣ ∣ ∣ ∣ ∣

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