Question

evaluate the following limit or state that it does not exist. (hint: identify the limit as the derivative at a point a.)
lim_{h→0} \frac{cos(\frac{pi}{6}+h)-\frac{sqrt{3}}{2}}{h}
which of the following is one possible f and a such that f(a) is equal to the given limit?
a. f(x)=sin x, a = -\frac{pi}{6}
b. f(x)=sin x, a=\frac{pi}{6}
c. f(x)=cos x, a = -\frac{pi}{6}
d. f(x)=cos x, a=\frac{pi}{6}

Explanation:

Step1: Recall the derivative definition

The derivative of a function $y = f(x)$ at a point $a$ is given by $f^{\prime}(a)=\lim_{h
ightarrow0}\frac{f(a + h)-f(a)}{h}$. We know that $\cos(\frac{\pi}{6})=\frac{\sqrt{3}}{2}$. So, the given limit $\lim_{h
ightarrow0}\frac{\cos(\frac{\pi}{6}+h)-\frac{\sqrt{3}}{2}}{h}$ is in the form of the derivative of the function $y = f(x)=\cos x$ at $x = a=\frac{\pi}{6}$, since $f(a + h)=\cos(\frac{\pi}{6}+h)$ and $f(a)=\cos(\frac{\pi}{6})=\frac{\sqrt{3}}{2}$.

Step2: Find the derivative of $f(x)=\cos x$

The derivative of $f(x)=\cos x$ is $f^{\prime}(x)=-\sin x$.

Step3: Evaluate the derivative at $x = a=\frac{\pi}{6}$

Substitute $x=\frac{\pi}{6}$ into $f^{\prime}(x)=-\sin x$. We get $f^{\prime}(\frac{\pi}{6})=-\sin(\frac{\pi}{6})=-\frac{1}{2}$.

For the multiple - choice part:
We know from the above analysis that $f(x)=\cos x$ and $a = \frac{\pi}{6}$ makes $f^{\prime}(a)$ equal to the given limit.

Answer:

The value of the limit is $-\frac{1}{2}$.
D. $f(x)=\cos x, a=\frac{\pi}{6}$