The function f(x) = lim_{n to infty} frac{x^{ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) To evaluate $f(x) = \lim_{n 	o \infty} \frac{x^{2n} - 1}{x^{2n} + 1}$:Case $1$: If $|x| < 1$,then $x^{2n} 	o 0$ as $n 	o \infty$. Thus,$f(x) =

Vedclass · Accepted Answer

$u(x) = 	ext{sgn}(|x| - 1)$

Vedclass · Answer

(C) To evaluate $f(x) = \lim_{n 	o \infty} \frac{x^{2n} - 1}{x^{2n} + 1}$:Case $1$: If $|x| Case $2$: If $|x| > 1$,then divide numerator and denominator by $x^{2n}$,so $f(x) = \lim_{n 	o \infty} \frac{1 - 1/x^{2n}}{1 + 1/x^{2n}} = \frac{1 - 0}{1 + 0} = 1$.Case $3$: If $x = 1$,$f(1) = \frac{1 - 1}{1 + 1} = 0$. If $x = -1$,$f(-1) = \frac{1 - 1}{1 + 1} = 0$.Thus,$f(x) = -1$ for $|x|  1$.Now,consider $u(x) = 	ext{sgn}(|x| - 1)$:If $|x| If $|x| = 1$,$|x| - 1 = 0$,so $	ext{sgn}(|x| - 1) = 0$.If $|x| > 1$,$|x| - 1 > 0$,so $	ext{sgn}(|x| - 1) = 1$.Therefore,$f(x) = u(x)$.

The function $f(x) = \lim_{n \to \infty} \frac{x^{2n} - 1}{x^{2n} + 1}$ is identical to which of the following functions?

Similar Questions

$\lim _{x \rightarrow 2}\left(\frac{5^x+5^{3-x}-30}{5^{3-x}-5^{\frac{x}{2}}}\right)=$

$\lim _{x \rightarrow 0} \frac{\tan (x)+4 \tan (2 x)-3 \tan (3 x)}{x^2 \tan (x)}$ is equal to

$\lim _{x \rightarrow 0} \frac{x \tan 2x - 2x \tan x}{(1 - \cos 2x)^2}$ is

If $|x| < 1$,then $\lim_{n \to \infty} \{(1 + x)(1 + x^2)(1 + x^4) \dots (1 + x^{2^n})\}$ is equal to

The value of $\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 - \cos {x^2}} }}{{1 - \cos x}}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module