Question

$\underset{n\to \infty }{\lim }\left(\frac{2^{-n}(n^2+5n+6)}{(n+4)(n+5)}\right)=$

• A. $0$
• B. $1$
• C. $\infty$
• D. $-\infty$

Answer 1

The correct option is A

$0$

Explanation for the correct option:

Finding the value of the given limit:

$$\begin{gathered} \underset{n\to \infty }{\lim }\left(\frac{2^{-n}(n^2+5n+6)}{(n+4)(n+5)}\right) \\ =\underset{n\to \infty }{\lim }\left(2^{-n}\frac{n^2+3n+2n+6}{(n+4)(n+5)}\right) \\ =\underset{n\to \infty }{\lim }\left(2^{-n}\frac{(n+2)(n+3)}{(n+4)(n+5)}\right)[\mathrm{Solving}\mathrm{the}\mathrm{numerator}\mathrm{by}\mathrm{splitting}\mathrm{the}\mathrm{middle}\mathrm{term}] \\ =\underset{n\to \infty }{\lim }\left(\frac{1}{2^n}\times \frac{{\displaystyle \frac{(n+2)(n+3)}{n^2}}}{{\displaystyle \frac{(n+4)(n+5)}{n^2}}}\right) \\ =\underset{n\to \infty }{\lim }\left(\frac{1}{2^n}\times \frac{\left(1+{\displaystyle \frac{2}{n}}\right)\left(1+{\displaystyle \frac{3}{n}}\right)}{\left(1+{\displaystyle \frac{4}{n}}\right)\left(1+{\displaystyle \frac{5}{n}}\right)}\right) \end{gathered}$$

Applying the limits,

$$\begin{gathered} =\frac{1}{2^{\infty }}\left(\frac{\left(1+\frac{2}{\infty }\right)\left(1+\frac{3}{\infty }\right)}{\left(1+\frac{4}{\infty }\right)\left(1+\frac{5}{\infty }\right)}\right) \\ =\frac{1}{\infty }\left(\frac{1\times 1}{1\times 1}\right)\left[\because x^{\infty }=\infty \mathrm{and}\frac{\mathrm{x}}{\infty }=0\right] \\ =\frac{1}{\infty } \\ =0 \end{gathered}$$

Therefore, the correct answer is option (A).