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Posted 3 months ago
Find limx0excosx4sinx\lim _{x \rightarrow 0} \frac{e^{x}-\cos x}{4 \sin x}.
Choose 1 answer:
(A) 14\frac{1}{4}
(B) e14\frac{e-1}{4}
(C) 12-\frac{1}{2}
(D) The limit doesn't exist.
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Answer from Sia
Posted 3 months ago
Solution by Steps
step 1
Apply L'Hôpital's Rule since the limit is of the indeterminate form 00\frac{0}{0}
step 2
Differentiate the numerator and denominator separately: ddx(excos(x))=ex+sin(x)\frac{d}{dx}(e^x - \cos(x)) = e^x + \sin(x) and ddx(4sin(x))=4cos(x)\frac{d}{dx}(4\sin(x)) = 4\cos(x)
step 3
Take the limit as xx approaches 00 of the derivatives: limx0ex+sin(x)4cos(x)\lim_{x \to 0} \frac{e^x + \sin(x)}{4\cos(x)}
step 4
Substitute x=0x = 0 into the limit: e0+sin(0)4cos(0)=1+041=14\frac{e^0 + \sin(0)}{4\cos(0)} = \frac{1 + 0}{4 \cdot 1} = \frac{1}{4}
(A) 14\frac{1}{4}
Key Concept
L'Hôpital's Rule for Indeterminate Forms
When faced with a limit that results in an indeterminate form such as 00\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.

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