If f(x) = { sin x x ≠ n π , n = 0 , ± 1 , ± 2... 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 lim x → 0 g { f ( x ) } is Options: A . &nbsp 2 B . &nbsp 0 C . &nbsp 3 D . &nbsp 1

Answer: Option D : D Given that, f(x) = { sin x x ≠ n π , n = 0 , ± 1 , ± 2... 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 lim x → 0 g [ f ( x ) ] = lim x → 9 g ( sin x ) = lim x → 0 ( sin 2 x + 1 ) = 1

If f(x) = {sinxx≠nπ,n=0,±1,±2...2,otherwise and g(x) =

Discussion Forum : Limits Continuity And Differentiability

Question -

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

Options:

A .

2

B .

0

C .

3

D .

1

Answer: Option D
:
D
Given that,
f(x) =

{\begin{matrix} sin x & x \neq n π, n = 0, \pm 1, \pm 2... 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

lim x \to 0 g [f (x)] = lim x \to 9 g (sin x)

= lim x \to 0 ({sin}^{2} x + 1) = 1

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