Question

Let $f\left(x\right)={lim}_{n\to \infty }\frac{x^{2n}-1}{x^{2n}+1}$ then

• A. $f\left(x\right)=1$ for $\left|x\right|>1$
• B. $f\left(x\right)=-1$ for $\left|x\right|<1$
• C. $f\left(x\right)$ is not defined for any value of $x$
• D. $f\left(x\right)=1$ for $\left|x\right|=1$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $f\left(x\right)=1$ for $\left|x\right|>1$
B $f\left(x\right)=-1$ for $\left|x\right|<1$
C $f\left(x\right)$ is not defined for any value of $x$

Let $f\left(x\right)={lim}_{n\to \infty }\frac{x^{2n}-1}{x^{2n}+1}$

Option$\left(a\right)$

$\left|x\right|>1$

Then $f\left(x\right)={lim}_{n\to \infty }\frac{x^{2n}(1-\frac{1}{x^{2n}})}{x^{2n}(1+\frac{1}{x^{2n}})}$

$=\frac{1-0}{1+0}=1$

Option$\left(b\right)$

$\left|x\right|<1$

Then $f\left(x\right)={lim}_{n\to \infty }\frac{x^{2n}-1}{x^{2n}+1}$

$=\frac{0-1}{0+1}=-1$

Option$\left(c\right)$

From the alternates $\left(a\right)$ and $\left(b\right)$,$1\ne -1$

$\therefore f\left(x\right)$ is not defined for any value of $x$

Option$\left(d\right)$

$\left|x\right|=1$

Then $f\left(x\right)=0$