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

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∻Solution by Steps∻ ‖ step 1 ⋮ To find the limit of the given function as $ x $ approaches 0, we can use L'Hôpital's Rule, which states that if the limit of functions $ f(x) $ and $ g(x) $ as $ x $ approaches a point $ c $ results in an indeterminate form $ \frac{0}{0} $ or $ \frac{\infty}{\infty} $, then the limit of $ \frac{f(x)}{g(x)} $ as $ x $ approaches $ c $ is the same as the limit of $ \frac{f'(x)}{g'(x)} $ as $ x $ approaches $ c $, provided that the latter limit exists. ‖ ‖ step 2 ⋮ The function $ \frac{e^x - e^{\sin(x)}}{x - \sin(x)} $ is in the indeterminate form $ \frac{0}{0} $ as $ x $ approaches 0. Applying L'Hôpital's Rule, we differentiate the numerator and the denominator with respect to $ x $. ‖ ‖ step 3 ⋮ The derivative of the numerator $ e^x - e^{\sin(x)} $ with respect to $ x $ is $ e^x - e^{\sin(x)}\cos(x) $, and the derivative of the denominator $ x - \sin(x) $ with respect to $ x $ is $ 1 - \cos(x) $. ‖ ‖ step 4 ⋮ Taking the limit of the derivatives as $ x $ approaches 0, we get $ \lim_{x \to 0} \frac{e^x - e^{\sin(x)}\cos(x)}{1 - \cos(x)} = \frac{1 - 1}{0} = 0 $. However, this is not the final answer, as we need to apply L'Hôpital's Rule again because we still have an indeterminate form $ \frac{0}{0} $. ‖ ‖ step 5 ⋮ Differentiating the numerator and the denominator again, we get $ \lim_{x \to 0} \frac{e^x - (e^{\sin(x)}\cos(x))'}{(1 - \cos(x))'} $. The derivative of $ e^{\sin(x)}\cos(x) $ is $ e^{\sin(x)}(-\sin(x)\cos(x) + \cos^2(x)) $, and the derivative of $ 1 - \cos(x) $ is $ \sin(x) $. ‖ ‖ step 6 ⋮ Taking the limit of the new derivatives as $ x $ approaches 0, we get $ \lim_{x \to 0} \frac{e^x - e^{\sin(x)}(-\sin(x)\cos(x) + \cos^2(x))}{\sin(x)} $. As $ x $ approaches 0, $ e^x $ and $ e^{\sin(x)} $ both approach 1, $ \sin(x) $ approaches 0, and $ \cos(x) $ approaches 1. ‖ ‖ step 7 ⋮ Simplifying the expression, we get $ \lim_{x \to 0} \frac{1 - (1)(0 + 1)}{0} = \lim_{x \to 0} \frac{1 - 1}{0} = 0 $. However, this is still an indeterminate form, so we apply L'Hôpital's Rule one more time. ‖ ‖ step 8 ⋮ Differentiating the numerator and the denominator one more time, we get $ \lim_{x \to 0} \frac{e^x - (e^{\sin(x)}(-\sin(x)\cos(x) + \cos^2(x)))'}{\sin(x)'} $. The derivative of $ \sin(x) $ is $ \cos(x) $, and the derivative of the numerator simplifies to $ e^x $ as $ x $ approaches 0. ‖ ‖ step 9 ⋮ Taking the limit of the final derivatives as $ x $ approaches 0, we get $ \lim_{x \to 0} \frac{e^x}{\cos(x)} = \frac{1}{1} = 1 $. This is the final result of the limit. ‖ ∻Answer∻ ⚹ $ \lim_{x \to 0} \frac{e^x - e^{\sin(x)}}{x - \sin(x)} = 1 $ ⚹ ∻Key Concept∻ ⚹L'Hôpital's Rule and Taylor Series Expansion⚹ ∻Explanation∻ ⚹The limit was evaluated using L'Hôpital's Rule, which involves taking derivatives of the numerator and denominator until the indeterminate form is resolved. Additionally, the Taylor series expansion of $ e^x $ and $ e^{\sin(x)} $ around $ x = 0 $ confirms that the leading terms cancel out, leaving the constant term 1 as $ x $ approaches 0.⚹◊What is the limit of the given function as $x$ approaches 0?⍭ Generate me a similar question◊

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