The value of mathop {lim }limits_{x to 0} fra | vedclass

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(C) Let $y = \mathop {\lim }\limits_{x 	o 0} \frac{{{e^x} - {e^{ - x}}}}{{\sin x}}$.Using the series expansion for $e^x$ and $e^{-x}$:$e^x = 1 + x +

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$2$

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(C) Let $y = \mathop {\lim }\limits_{x 	o 0} \frac{{{e^x} - {e^{ - x}}}}{{\sin x}}$.Using the series expansion for $e^x$ and $e^{-x}$:$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots$$e^{-x} = 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \dots$Subtracting the two series:$e^x - e^{-x} = 2 \left( x + \frac{x^3}{3!} + \frac{x^5}{5!} + \dots ight)$Substituting this into the limit:$y = \mathop {\lim }\limits_{x 	o 0} \frac{2 \left( x + \frac{x^3}{6} + \dots ight)}{\sin x}$Divide numerator and denominator by $x$:$y = \mathop {\lim }\limits_{x 	o 0} \frac{2 \left( 1 + \frac{x^2}{6} + \dots ight)}{\frac{\sin x}{x}}$Since $\mathop {\lim }\limits_{x 	o 0} \frac{\sin x}{x} = 1$:$y = \frac{2(1 + 0)}{1} = 2$.Alternatively,using $L$-Hospital's rule:$\mathop {\lim }\limits_{x 	o 0} \frac{{{e^x} - {e^{ - x}}}}{{\sin x}} = \mathop {\lim }\limits_{x 	o 0} \frac{\frac{d}{dx}({e^x} - {e^{ - x}})}{\frac{d}{dx}(\sin x)}$$= \mathop {\lim }\limits_{x 	o 0} \frac{{{e^x} + {e^{ - x}}}}{{\cos x}} = \frac{{{e^0} + {e^0}}}{{\cos 0}} = \frac{1 + 1}{1} = 2$.

The value of $\mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - {e^{ - x}}}}{{\sin x}}$ is

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