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(C) We need to evaluate the limit: $\mathop {\lim }\limits_{x 	o 0} \left( {\frac{{	an 3x}}{x} + \cos x} ight)$.Using the property of limits,$\mat

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(C) We need to evaluate the limit: $\mathop {\lim }\limits_{x 	o 0} \left( {\frac{{	an 3x}}{x} + \cos x} ight)$.Using the property of limits,$\mathop {\lim }\limits_{x 	o a} [f(x) + g(x)] = \mathop {\lim }\limits_{x 	o a} f(x) + \mathop {\lim }\limits_{x 	o a} g(x)$,we get:$\mathop {\lim }\limits_{x 	o 0} \frac{{	an 3x}}{x} + \mathop {\lim }\limits_{x 	o 0} \cos x$.For the first term,multiply and divide by $3$:$\mathop {\lim }\limits_{x 	o 0} 3 \cdot \frac{{	an 3x}}{{3x}} + \mathop {\lim }\limits_{x 	o 0} \cos x$.Since $\mathop {\lim }\limits_{	heta 	o 0} \frac{{	an 	heta}}{	heta} = 1$,the first term becomes $3 \cdot 1 = 3$.For the second term,$\mathop {\lim }\limits_{x 	o 0} \cos x = \cos(0) = 1$.Therefore,the total limit is $3 + 1 = 4$.

$\mathop {\lim }\limits_{x \to 0} \left( {\frac{{\tan 3x}}{x} + \cos x} \right) = $

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