If I = lim_{x rightarrow 0} sin left( frac{e^ | vedclass

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(C) We evaluate the limit inside the sine function first: $L = \lim_{x \rightarrow 0} \frac{e^{x}-x-1-\frac{x^{2}}{2}}{x^{2}}$ Using the Taylor series

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exists and equals $0$

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(C) We evaluate the limit inside the sine function first: $L = \lim_{x \rightarrow 0} \frac{e^{x}-x-1-\frac{x^{2}}{2}}{x^{2}}$ Using the Taylor series expansion for $e^{x} = 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \dots$, we get: $L = \lim_{x \rightarrow 0} \frac{(1 + x + \frac{x^{2}}{2} + \frac{x^{3}}{6} + \dots) - x - 1 - \frac{x^{2}}{2}}{x^{2}}$ $L = \lim_{x \rightarrow 0} \frac{\frac{x^{3}}{6} + \dots}{x^{2}} = \lim_{x \rightarrow 0} (\frac{x}{6} + \dots) = 0$ Since the function $\sin(u)$ is continuous at $u=0$, we have: $I = \sin(L) = \sin(0) = 0$

If $I = \lim_{x \rightarrow 0} \sin \left( \frac{e^{x}-x-1-\frac{x^{2}}{2}}{x^{2}} \right)$, then the limit

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