Question

Math

Posted 3 months ago

```
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.
```

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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 $\frac{0}{0}$

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: $\lim_{x \to 0} \frac{e^x + \sin(x)}{4\cos(x)}$

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.

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