Let f (x )= sin ⁡x ,g (x )= cos ⁡x ,h (x )=x^2 then lim_x 1 0.05em f (g (h (x ) ) )-f (g (h (1 ) ) )x-1=

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Correct answer: B. (b) Given f (x )= sin ⁡x ,g (x )= cos ⁡x ,h (x )=x^2 limx 1 0.05em f (x (h (x ) ) )-f (g (h (1 ) ) )x-1= limx 1 0.05em sin ⁡ ( cos ⁡x^2 )- sin ⁡ ( cos ⁡1 )x-1 If Apply limit it gives 00 form, then apply L'Hospital rule. ⇒ limx 1 0.05em cos ⁡ ( cos ⁡x^2 )(- sin ⁡x^2) (2x )-01-0 Apply the limit, ⇒-2 (1 ) sin ⁡ (1 ) cos ⁡ ( cos ⁡1 )=-2 sin ⁡ (1 ) cos ⁡ ( cos ⁡1 ) So, option (b) is correct.

$Let f (x) = sin x, g (x) = cos x, h (x) = x^{2} then lim_{x \to 1} \frac{f ( g ( h ( x ) ) ) - f ( g ( h ( 1 ) ) )}{x - 1} =$

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