Find $\lim _{x \rightarrow 0} \frac{e^{x}-\cos x}{4 \sin x}$.
Choose 1 answer:
(A) $\frac{1}{4}$
(B) $\frac{e-1}{4}$
(C) $-\frac{1}{2}$
(D) The limit doesn't exist.

Question

Find $\lim _{x \rightarrow 0} \frac{e^{x}-\cos x}{4 \sin x}$.
Choose 1 answer:
(A) $\frac{1}{4}$
(B) $\frac{e-1}{4}$
(C) $-\frac{1}{2}$
(D) The limit doesn't exist.

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ Apply L'Hôpital's Rule since the limit is of the indeterminate form $\frac{0}{0}$. ‖ ‖ step 2 ⋮ Differentiate the numerator and denominator separately: $\frac{d}{dx}(e^x - \cos(x)) = e^x + \sin(x)$ and $\frac{d}{dx}(4\sin(x)) = 4\cos(x)$. ‖ ‖ step 3 ⋮ Take the limit as $x$ approaches $0$ of the derivatives: $\lim_{x o 0} \frac{e^x + \sin(x)}{4\cos(x)}$. ‖ ‖ step 4 ⋮ Substitute $x = 0$ into the limit: $\frac{e^0 + \sin(0)}{4\cos(0)} = \frac{1 + 0}{4 \cdot 1} = \frac{1}{4}$. ‖ ∻Answer∻ ⚹ (A) $\frac{1}{4}$ ⚹ ∻Key Concept∻ ⚹L'Hôpital's Rule for Indeterminate Forms⚹ ∻Explanation∻ ⚹When faced with a limit that results in an indeterminate form such as $\frac{0}{0}$, L'Hôpital's Rule can be applied by taking the derivative of the numerator and the derivative of the denominator separately and then evaluating the limit.⚹◊What is the rule used to find the limit of a function as it approaches a specific value?⍭ Generate me a similar question◊

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