Question

${lim}_{x\to \infty }{2}^x⟮4ta{n}^{-1}(1+2^{-x})-\pi ⟯$ is equal to

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The correct option is B 2

$li{m}_{x\to \infty }\frac{4{\tan }^{-1}(1+2^{-x})-\pi }{2^{-x}}$

Let $2^{-x}=a$

When $x\to \infty ,a\to 0$

$li{m}_{a\to 0}\frac{4{\tan }^{-1}(1+a)-\pi }{a}$

L-Hopital Rule;

$\frac{4}{1+(1+a{)}^2}$

$\Rightarrow \frac{4}{1+1}=2$