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# Find the limit. Use l'Hospital's Rule where appropriate. If there is a more elementary method, consider using it. If l'Hospital's Rule doesn't apply, explain why.$\lim_{x\to 0} \frac{e^x - e^{-x} - 2x}{x - \sin x}$

## $\lim _{x \rightarrow 0} \frac{e^{x}-e^{-x}-2 x}{x-\sin x}=2$

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we know this function is in the form zero over zero. Therefore, we know we can use the hospital's rule. We know this is equivalent to zero over zero. What this means is that we apply the hospital's rule second time. The two is now a constant. Therefore it just completely cancels. This is equivalent to zero over zero We apply The happy tells me we've gone. See that now it's e to the X plus feet of the negative acts over co sign acts Plugin We end up with two over one. This gives us an actual answer of Jim.

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