mathop {lim }limits_{x to 0} frac{sin(mx)}{ta | vedclass

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(B) To evaluate the limit $\mathop {\lim }\limits_{x 	o 0} \frac{\sin(mx)}{	an(nx)}$,we use the standard limits $\mathop {\lim }\limits_{	heta 	o

Vedclass · Accepted Answer

$\frac{m}{n}$

Vedclass · Answer

(B) To evaluate the limit $\mathop {\lim }\limits_{x 	o 0} \frac{\sin(mx)}{	an(nx)}$,we use the standard limits $\mathop {\lim }\limits_{	heta 	o 0} \frac{\sin(	heta)}{	heta} = 1$ and $\mathop {\lim }\limits_{	heta 	o 0} \frac{	an(	heta)}{	heta} = 1$.Divide the numerator and denominator by $x$:$\mathop {\lim }\limits_{x 	o 0} \frac{\frac{\sin(mx)}{x}}{\frac{	an(nx)}{x}}$Multiply and divide by $m$ in the numerator and $n$ in the denominator:$\mathop {\lim }\limits_{x 	o 0} \frac{m \cdot \frac{\sin(mx)}{mx}}{n \cdot \frac{	an(nx)}{nx}}$As $x 	o 0$,$mx 	o 0$ and $nx 	o 0$,so the limit becomes:$\frac{m \cdot 1}{n \cdot 1} = \frac{m}{n}$.

$\mathop {\lim }\limits_{x \to 0} \frac{\sin(mx)}{\tan(nx)} = $

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