Question

$li{m}_{x\to \infty }\frac{(2x+1{)}^{40}(4x-1{)}^5}{(2x+2{)}^{45}}$ is equal to

• A. 16
• B. 24
• C. 32
• D. 8

Answer 1

The correct option is C 32

Hance,

$\underset{x\to \infty }{lim}\frac{{\left(2x+1\right)}^{40}{\left(4x-1\right)}^5}{{\left(2x+2\right)}^{45}}\left[\frac{\infty }{\infty }for\right]$

$\underset{x\to \infty }{lim}\frac{x^{40}{\left(2+\frac{1}{x}\right)}^4\times 9^5{\left(4-\frac{1}{x^5}\right)}^5}{x^{45}{\left(2+\frac{2}{x}\right)}^{45}}$

$\begin{array}{c}{lim}_{x\to \infty }\frac{x^{45}{\left(2+\frac{1}{5}\right)}^{40}{\left(4-\frac{1}{x^5}\right)}^5}{x^{45}{\left(2+\frac{2}{x}\right)}^{45}}\\ =\frac{{\left(2+0\right)}^{40}{\left(4-0\right)}^5}{{\left(2+0\right)}^{45}}\\ =\frac{2^{40}\times {\left(2^2\right)}^5}{2^{45}}\end{array}$

$\begin{array}{c}=\frac{2^{50}}{2^{45}}\\ =2^5\\ =32\end{array}$

Hence,

Option (c) is correct answer