Question

Math

Posted 3 months ago

```
Find $\lim _{h \rightarrow 0} \frac{\cos \left(\frac{\pi}{6}+h\right)-\cos \left(\frac{\pi}{6}\right)}{h}$.
Choose 1 answer:
(A) $-\frac{1}{2}$
(B) $\frac{1}{2}$
(C) 2
(D) The limit doesn't exist
```

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Answer from Sia

Posted 3 months ago

Solution by Steps

step 1

Apply the limit definition to the given function

step 2

Use the Taylor series expansion of $\cos(x)$ around $\frac{\pi}{6}$ to find the limit

step 3

The Taylor series expansion of $\cos(x)$ around $\frac{\pi}{6}$ is $\cos\left(\frac{\pi}{6}\right) - \frac{\sqrt{3}}{2}h - \frac{h^2}{4} + \frac{\sqrt{3}h^3}{24} - \frac{h^4}{48} + \mathcal{O}(h^5)$

step 4

Substitute the Taylor series into the limit expression and simplify

step 5

The terms involving $h$ will approach 0 as $h$ approaches 0

step 6

The constant term in the Taylor series expansion is $-\frac{\sqrt{3}}{2}$, which is the coefficient of the first degree term in $h$

step 7

The limit is equal to the coefficient of the linear term in $h$ in the Taylor series expansion, which is $-\frac{\sqrt{3}}{2}$

Answer

$-\frac{\sqrt{3}}{2}$

Key Concept

Limit of a function using Taylor series expansion

Explanation

The limit of the function as $h$ approaches 0 is found by identifying the linear term in the Taylor series expansion of the function around the point of interest.

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