mathop {lim }limits_{x to 0} {(cos mx)^{n/{x^ | vedclass

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(B) The given limit is of the form $1^\infty$ as $x 	o 0$.Using the formula $\mathop {\lim }\limits_{x 	o a} f(x)^{g(x)} = e^{\mathop {\lim }\limits

Vedclass · Accepted Answer

$e^{-\frac{m^2n}{2}}$

Vedclass · Answer

(B) The given limit is of the form $1^\infty$ as $x 	o 0$.Using the formula $\mathop {\lim }\limits_{x 	o a} f(x)^{g(x)} = e^{\mathop {\lim }\limits_{x 	o a} (f(x)-1)g(x)}$,we have:$\mathop {\lim }\limits_{x 	o 0} (\cos mx - 1) \cdot \frac{n}{x^2} = \mathop {\lim }\limits_{x 	o 0} - (1 - \cos mx) \cdot \frac{n}{x^2}$Using the standard limit $\mathop {\lim }\limits_{	heta 	o 0} \frac{1 - \cos 	heta}{	heta^2} = \frac{1}{2}$,we substitute $	heta = mx$:$= -n \cdot \mathop {\lim }\limits_{x 	o 0} \frac{1 - \cos mx}{(mx)^2} \cdot m^2 = -n \cdot \frac{1}{2} \cdot m^2 = -\frac{m^2n}{2}$Thus,the limit is $e^{-\frac{m^2n}{2}}$.

$\mathop {\lim }\limits_{x \to 0} {(\cos mx)^{n/{x^2}}}$ equals

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