If f(x) = begin{cases} sin x, & x neq npi, n | vedclass

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

Question

(D) We need to evaluate $\lim_{x ightarrow 0} g(f(x))$.As $x ightarrow 0$,$x 
eq n\pi$ for $n 
eq 0$. Thus,$f(x) = \sin x$.As $x ightarrow 0$,

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(D) We need to evaluate $\lim_{x ightarrow 0} g(f(x))$.As $x ightarrow 0$,$x 
eq n\pi$ for $n 
eq 0$. Thus,$f(x) = \sin x$.As $x ightarrow 0$,$f(x) = \sin x ightarrow 0$.Since $f(x) ightarrow 0$ but $f(x) 
eq 0$ for $x$ in a deleted neighborhood of $0$,we look at the definition of $g(u)$ as $u ightarrow 0$.For $u 
eq 0, 2$,$g(u) = u^2 + 1$.Therefore,$\lim_{u ightarrow 0} g(u) = \lim_{u ightarrow 0} (u^2 + 1) = 0^2 + 1 = 1$.Thus,$\lim_{x ightarrow 0} g(f(x)) = 1$.

If $f(x) = \begin{cases} \sin x, & x \neq n\pi, n \in I \\ 2, & \text{otherwise} \end{cases}$ and $g(x) = \begin{cases} x^2 + 1, & x \neq 0, 2 \\ 2, & x = 0 \\ 4, & x = 2 \end{cases}$,then find $\lim_{x \rightarrow 0} g(f(x))$.

Similar Questions

If $f: R - \{\frac{3}{7}\} \rightarrow R - \{\frac{3}{7}\}$ is given by $f(x) = \frac{3x+5}{7x-3}$,then the statement which is not true is:

If $f: R \rightarrow R$ and $g: R \rightarrow R$ are defined by $f(x)=|x|$ and $g(x)=[x-3]$ for $x \in R$, then $\{g(f(x)):-\frac{8}{5} < x < \frac{8}{5}\}$ is equal to

If $f(x) = (p - x^n)^{1/n}$, $p > 0$ and $n$ is a positive integer, then $f[f(x)]$ is equal to

If $f : R \rightarrow R$ is defined by $f(x) = x^{2} - 3x + 2$,find $f(f(x))$.

$f(x) = \frac{1}{x}$ and $g(x) = \frac{1}{\sqrt{x}}$,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module