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Posted 3 months ago
Find limh0cos(π6+h)cos(π6)h\lim _{h \rightarrow 0} \frac{\cos \left(\frac{\pi}{6}+h\right)-\cos \left(\frac{\pi}{6}\right)}{h}.
Choose 1 answer:
(A) 12-\frac{1}{2}
(B) 12\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)\cos(x) around π6\frac{\pi}{6} to find the limit
step 3
The Taylor series expansion of cos(x)\cos(x) around π6\frac{\pi}{6} is cos(π6)32hh24+3h324h448+O(h5)\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 hh will approach 0 as hh approaches 0
step 6
The constant term in the Taylor series expansion is 32-\frac{\sqrt{3}}{2}, which is the coefficient of the first degree term in hh
step 7
The limit is equal to the coefficient of the linear term in hh in the Taylor series expansion, which is 32-\frac{\sqrt{3}}{2}
Key Concept
Limit of a function using Taylor series expansion
The limit of the function as hh 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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