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

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

Class 12 Mathematics Relations and Functions Composition of Functions

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 $\lim_{x \rightarrow 0} g(f(x))$.

A
$3$
B
$2$
C
$4$
D
$1$

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Relations and Functions

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Composition of Functions

$[x]$ represents the greatest integer function. Let $g(x) = 1 + x - [x]$ and $f(x) = \begin{cases} -3, & x < 0 \\ 0, & x = 0 \\ 5, & x > 0 \end{cases}$. Then $f(g(x))$ is:

MediumAP EAMCET 2021

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$f: R \rightarrow R$ and $g: R \rightarrow R$ are two functions such that $f(x)=2x-3$ and $g(x)=x^3+5$. Then,$(f \circ g)^{-1}(-9)$ is

DifficultMHT CET 2023

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If $f:[-6,6] \rightarrow R$ is defined by $f(x)=x^2-3$ for $x \in R$, then $(f \circ f \circ f)(-1)+(f \circ f \circ f)(0)+(f \circ f \circ f)(1)$ is equal to

DifficultTS EAMCET 2008

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For every real number $x \neq -1$, let $f(x) = \frac{x}{x+1}$. Define $f_1(x) = f(x)$ and for $n \geq 2$, $f_n(x) = f(f_{n-1}(x))$. Then the product $f_1(-2) \cdot f_2(-2) \cdot \ldots \cdot f_n(-2)$ is equal to:

MediumWBJEE 2024

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Consider $f: N \rightarrow N$,$g: N \rightarrow N$,and $h: N \rightarrow R$ defined as $f(x) = 2x$,$g(y) = 3y + 4$,and $h(z) = \sin z$,$\forall x, y, z \in N$. Show that $h \circ (g \circ f) = (h \circ g) \circ f$.

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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 $\lim_{x \rightarrow 0} g(f(x))$.

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$[x]$ represents the greatest integer function. Let $g(x) = 1 + x - [x]$ and $f(x) = \begin{cases} -3, & x < 0 \\ 0, & x = 0 \\ 5, & x > 0 \end{cases}$. Then $f(g(x))$ is:

$f: R \rightarrow R$ and $g: R \rightarrow R$ are two functions such that $f(x)=2x-3$ and $g(x)=x^3+5$. Then,$(f \circ g)^{-1}(-9)$ is

If $f:[-6,6] \rightarrow R$ is defined by $f(x)=x^2-3$ for $x \in R$, then $(f \circ f \circ f)(-1)+(f \circ f \circ f)(0)+(f \circ f \circ f)(1)$ is equal to

For every real number $x \neq -1$, let $f(x) = \frac{x}{x+1}$. Define $f_1(x) = f(x)$ and for $n \geq 2$, $f_n(x) = f(f_{n-1}(x))$. Then the product $f_1(-2) \cdot f_2(-2) \cdot \ldots \cdot f_n(-2)$ is equal to:

Consider $f: N \rightarrow N$,$g: N \rightarrow N$,and $h: N \rightarrow R$ defined as $f(x) = 2x$,$g(y) = 3y + 4$,and $h(z) = \sin z$,$\forall x, y, z \in N$. Show that $h \circ (g \circ f) = (h \circ g) \circ f$.

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