displaystyle lim n arrow ∞(((n/n+1))α+ sin (1/n))n ( where α ∈ Q) is equal to

Question

displaystyle lim n arrow ∞(((n/n+1))α+ sin (1/n))n ( where α ∈ Q) is equal to (A) e -α (B) -α (C) e 1-α (D) e 1+α

Tardigrade Admin · Accepted Answer

displaystyle 1∞ form L = e displaystyle lim n arrow ∞ n(((n/n+1))α+ sin (1/n)-1) = e displaystyle lim n arrow ∞ n sin (1/n)+ displaystyle lim n arrow ∞ n(((n/n+1))α-1) Consider, displaystyle lim n arrow ∞ n(((n/n+1))α-1) = displaystyle lim n arrow ∞ n(((1/1+1 / n))α-1) Put n =(1/ y ) = displaystyle lim y arrow 0 (1/y)(((1/1+y))α-1) = displaystyle lim y arrow 0 (1-(1+y)α/y)=-α (Using binomial) ∴ L = e 1-α

Q. $n \to \infty lim ((\frac{n}{n + 1})^{α} + sin \frac{1}{n})^{n} ($ where $α \in Q)$ is equal to

$e^{- α}$

$- α$

$e^{1 - α}$

$e^{1 + α}$

Q. n→∞lim​((n+1n​)α+sinn1​)n( where α∈Q) is equal to

e−α

−α

e1−α

e1+α

Q. $n \to \infty lim ((\frac{n}{n + 1})^{α} + sin \frac{1}{n})^{n} ($ where $α \in Q)$ is equal to

$e^{- α}$

$- α$

$e^{1 - α}$

$e^{1 + α}$