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

Question

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

Accepted Answer

∻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.⚹◊What is the formula for the derivative of $\cos(x)$?⍭ Generate me a similar question◊

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