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

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(C) We know the Taylor series expansion for $e^x$ is $e^x = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \dots$ Substituting this into the limit expression: $\

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$1$

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(C) We know the Taylor series expansion for $e^x$ is $e^x = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \dots$ Substituting this into the limit expression: $\mathop {\lim }\limits_{x 	o 0} \frac{(1 + \frac{x}{1!} + \frac{x^2}{2!} + \dots) - 1}{x}$ $= \mathop {\lim }\limits_{x 	o 0} \frac{\frac{x}{1!} + \frac{x^2}{2!} + \dots}{x}$ $= \mathop {\lim }\limits_{x 	o 0} (1 + \frac{x}{2!} + \dots) = 1$.

The value of $\mathop {\lim }\limits_{x \to 0} \left( \frac{e^x - 1}{x} \right)$ is

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