Question

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

• A. $f\left(1^{+}\right)+f\left(1^{-}\right)=0$
• B. $f\left(1^{+}\right)+f\left(1^{-}\right)+f\left(1\right)=\frac{3}{2}$
• C. $f(-1^{+})+f(-1^{-})=-1$
• D. $f\left(1^{+}\right)+f(-1^{-})=0$

Answer 1

Answer: (B), (C), (D)

$f\left(1^{+}\right)+f\left(1^{-}\right)+f\left(1\right)=\frac{3}{2}$
$f\left(1^{+}\right)+f(-1^{-})=0$
$f(-1^{+})+f(-1^{-})=-1$

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

$f\left(1^{+}\right)=\underset{n\to \infty }{lim}\frac{x}{x^{2n}+1}(x=p;p>1)$

$=\frac{1}{p^{\infty }+1}=\frac{1}{\infty }=0$

$f\left(1^{-}\right)=\underset{n\to \infty }{lim}\frac{1}{{\left(\frac{1}{q}\right)}^{2n}+1}(x=\frac{1}{q};q>1)$

$=\frac{1}{q^{\infty }+1}=\frac{1}{0+1}=1$

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

$f\left({-1}^{+}\right)=\underset{n\to \infty }{lim}\frac{-1}{{\left(\frac{-1}{r}\right)}^{2n}+1}(r>1)$

as $x\to {-1}^{+}$

$-1<x<0$

$x=-\frac{1}{5}$

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

$f\left({-1}^{-}\right)=\underset{n\to \infty }{lim}\frac{-1}{{(-5)}^{2n}+1}(x=-5;5>1)$

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

From above function values

B,C,D are the correct options.