Question

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

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

Answer 1

Answer: (A), (B)

$f\left(x\right)=1$ for $|x|>1$
$f\left(x\right)=-1$ for $|x|<1$

Let us take $|x|<1$, then $x=\frac{1}{q}$

$\therefore x^{2n}={\left(\frac{1}{q}\right)}^{2n}$

${\left(\frac{1}{q}\right)}^{2n}\to 0$ when $n\to \infty$

Substituting this value in the limit we get,

$f\left(x\right)=\frac{0-1}{0+1}=-1$

Again, let us take $|x|>1$ then we can write the limit as

$\underset{n⟶\infty }{lim}\frac{x^{2n}}{x^{2n}}\left(\frac{1-\frac{1}{x^{2n}}}{1+\frac{1}{x^{2n}}}\right)$

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

Hence, option A and B are correct.