For the function f(x) = lim_{n to infty} frac | vedclass

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

Question

(C) Case $1$: If $x \in \mathbb{Z}$,then $\sin(\pi x) = 0$. Thus,$f(x) = \lim_{n 	o \infty} \frac{1}{1 + n(0)} = 1$. Case $2$: If $x 
otin \mathbb{Z

Vedclass · Accepted Answer

$f$ is discontinuous for all $x \in \mathbb{Z}$

Vedclass · Answer

(C) Case $1$: If $x \in \mathbb{Z}$,then $\sin(\pi x) = 0$. Thus,$f(x) = \lim_{n 	o \infty} \frac{1}{1 + n(0)} = 1$. Case $2$: If $x 
otin \mathbb{Z}$,then $\sin^2(\pi x) > 0$. As $n 	o \infty$,$n \sin^2(\pi x) 	o \infty$. Thus,$f(x) = \lim_{n 	o \infty} \frac{1}{1 + n \sin^2(\pi x)} = 0$. Therefore,$f(x) = \begin{cases} 1, & x \in \mathbb{Z} \ 0, & x 
otin \mathbb{Z} \end{cases}$. Since the function jumps between $0$ and $1$ at every integer value,$f$ is discontinuous for all $x \in \mathbb{Z}$.

For the function $f(x) = \lim_{n \to \infty} \frac{1}{1 + n \sin^2(\pi x)}$,which of the following holds?

Similar Questions

$\mathop {\lim }\limits_{x \to a} \frac{{\sqrt {a + 2x} - \sqrt {3x} }}{{\sqrt {3a + x} - 2\sqrt x }} = \dots$ (where $a \ne 0$)

The value of $\lim _{x \rightarrow 0} \frac{x e^{x}-\sin x}{x}$ is equal to:

Let $A = \lim_{x \rightarrow 0^{+}} \left(1 + \tan^2 \sqrt{x}\right)^{\frac{1}{2x}}$,then $\log_{e} A = $

The value of $\lim_{x \rightarrow 1} \frac{(x^{2}-1) \sin^{2}(\pi x)}{x^{4}-2x^{3}+2x-1}$ is equal to

Let $[P]$ denote the greatest integer $\leq P$. If $0 \leq a \leq 2$,then the number of integral values of $a$ such that $\lim _{x \rightarrow a}([x^2]-[x]^2)$ does not exist is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module