If f ( x ) = { s i n x , x ≠ n π , n ϵ I 2 , o t h e r w i s e and g ( x ) = ⎧ ⎪ ⎨ ⎪ ⎩ x 2 + 1 , x ≠ 0 , 2 4 , x = 0 5 , x = 2 , then l i m x → 0 g { f ( x ) } Options: A . &nbsp 5 B . &nbsp 6 C . &nbsp 7 D . &nbsp 1

Answer: Option D : D l i m x → 0 g { f ( x ) } = l i m x → 0 ⎧ ⎪ ⎨ ⎪ ⎩ ( f ( x ) ) 2 + 1 , f ( x ) ≠ 0 , 2 4 , f ( x ) = 0 5 , f ( x ) = 2 = l i m x → 0 ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ s i n 2 x + 1 , s i n x ≠ 0 , 2 a n d x ≠ n π 4 , s i n x = 0 a n d x ≠ n π 5 , s i n x = 2 a n d x ≠ n π 5 , 2 ≠ 0 , 2 a n d x ≠ n π 4 , 2 = 0 a n d x = n π 5 , 2 = 2 a n d x = n π = l i m x → 0 { 1 + s i n 2 x , x ≠ n π 5 , x = n π = 1 + s i n 2 0 = 1 + 0 = 1

If f(x)={sinx,x≠nπ,nϵI2,otherwise and g(x)=⎧⎪

Discussion Forum : Limits Continuity And Differentiability

Question -

If $f (x) = {\begin{matrix} s i n x, & x \neq n π, & n ϵ I 2, & o t h e r w i s e \end{matrix}$ and $g (x) = ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} x^{2} + 1, & x \neq 0, & 2 4, & x = 0 5, & x = 2 \end{matrix}$ , then $l i m x \to 0 g {f (x)}$

Options:

A . 5

B . 6

C . 7

D . 1

Answer: Option D
:
D

l i m x \to 0 g {f (x)} = l i m x \to 0 ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} (f (x))^{2} + 1, & f (x) \neq 0, 2 4, & f (x) = 0 5, & f (x) = 2 \end{matrix} = l i m x \to 0 ⎧ ⎪⎪⎪⎪⎪⎪⎪⎪⎪ ⎨ ⎪⎪⎪⎪⎪⎪⎪⎪⎪ ⎩ \begin{matrix} s i n^{2} x + 1, & s i n x \neq 0, 2 & a n d x \neq n π 4, & s i n x = 0 & a n d x \neq n π 5, & s i n x = 2 & a n d x \neq n π 5, & 2 \neq 0, 2 & a n d x \neq n π 4, & 2 = 0 & a n d x = n π 5, & 2 = 2 & a n d x = n π \end{matrix} = l i m x \to 0 {\begin{matrix} 1 + s i n^{2} x, & x \neq n π 5, & x = n π \end{matrix} = 1 + s i n^{2} 0 = 1 + 0 = 1

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