Question

If $\lim_{n\rightarrow \infty}$($\sqrt{n^2-n-1}$ + nα+β)=0 then 8(α + β) is equal to

• A. 4
• B. -8
• C. -4
• D. 8

Answer 1

Answer: (C)

-4

Answer 2

$\lim_{n\rightarrow \infty}$($n^2−n−1$+nα+β)=0
=$\lim_{n\rightarrow \infty}$ n[$\sqrt{1-\frac{1}{n}-\frac{1}{n^2}}$+α+$\frac{\beta}{n}$]=0
∴ α = –1
Now,
$\lim_{n\rightarrow \infty}$ n[{1−($\frac{1}{n}-\frac{1}{n^2}$)}$^{\frac{1}{2}}$+$\frac{\beta }{n}$−1]=0
$\lim_{n\rightarrow \infty}$ n(1−$\frac{1}{2}$($\frac{1}{n}+\frac{1}{n^2}$)+…)+$\frac{\beta }{n}$−1=0
⇒ β–$\frac{1}{2}$=0
∴β=$\frac{1}{2}$
Now,
8(α+β)=8(-$\frac{1}{2}$)=-4