Question

$\text{If}f\left(x\right)=\underset{t\to \infty }{lim}\frac{(1+\cos \frac{\pi x}{2}{)}^t-1}{(1+\cos \frac{\pi x}{2}{)}^t+1},$ then which of the following is/are correct?

• A. $\underset{x\to 1^{+}}{lim}f\left(x\right)=-1$
• B. $\underset{x\to 1^{-}}{lim}f\left(x\right)=1$
• C. $\underset{x\to 2^{+}}{lim}f\left(x\right)=1$
• D. $\underset{x\to 2^{-}}{lim}f\left(x\right)=-1$

Answer 1

Answer: (A), (B), (D)

The correct options are
A $\underset{x\to 1^{+}}{lim}f\left(x\right)=-1$
B $\underset{x\to 1^{-}}{lim}f\left(x\right)=1$
D $\underset{x\to 2^{-}}{lim}f\left(x\right)=-1$

$x\to 1^{+}\Rightarrow \frac{\pi x}{2}>\frac{\pi }{2}\Rightarrow 2\text{nd quadrant}\Rightarrow \cos \left(\frac{{\pi }^{+}}{2}\right)\text{will be}-\text{ve}\therefore \underset{t\to \infty }{lim}\frac{(1+\cos \frac{{\pi }^{+}}{2}{)}^t-1}{(1+\cos \frac{{\pi }^{+}}{2}{)}^t+1}=\frac{0-1}{0+1}=-1$


$x\to 1^{-}\Rightarrow \frac{\pi x}{2}<\frac{\pi }{2}\Rightarrow 1\text{st quadrant}\Rightarrow \cos \left(\frac{{\pi }^{-}}{2}\right)\to +\text{ve}\Rightarrow \underset{t\to \infty }{lim}{[1+\cos (\frac{{\pi }^{-}}{2})]}^t\to \infty \therefore \underset{t\to \infty }{lim}\frac{(1+\cos \frac{{\pi }^{-}}{2}{)}^t-1}{(1+\cos \frac{{\pi }^{-}}{2}{)}^t+1}=\underset{t\to \infty }{lim}\frac{1-\frac{1}{(1+\cos \frac{{\pi }^{-}}{2}{)}^t}}{1+\frac{1}{(1+\cos \frac{{\pi }^{-}}{2}{)}^t}}=\frac{1-0}{1+0}=1$


$x\to 2^{+}\Rightarrow \frac{\pi x}{2}>\pi \Rightarrow 3\text{rd quadrant}\Rightarrow \cos {\pi }^{+}\to -\text{ve}\therefore \underset{t\to \infty }{lim}\frac{(1+\cos {\pi }^{+}{)}^t-1}{(1+\cos {\pi }^{+}{)}^t+1}=\frac{0-1}{0+1}=-1$


$x\to 2^{-}\Rightarrow \frac{\pi x}{2}<\pi \Rightarrow 2\text{nd quadrant}\Rightarrow \cos {\pi }^{-}\to -\text{ve}\therefore \underset{t\to \infty }{lim}\frac{(1+\cos {\pi }^{-}{)}^t-1}{(1+\cos {\pi }^{-}{)}^t+1}=\frac{0-1}{0+1}=-1$