Question

Evaluate the following limits or state that they do not exist. (Hint: Identify each limit as the derivative of a function at a point.) $$\lim _{h \rightarrow 0} \frac{\cos \left(\frac{\pi}{6}+h\right)-\frac{\sqrt{3}}{2}}{h}$$

Short answer

Question: Evaluate the limit $\lim_{h \rightarrow 0} \frac{\cos \left(\frac{\pi}{6}+h\right)-\frac{\sqrt{3}}{2}}{h}.$ Answer: $-\frac{1}{2}$

Step-by-step solution

Identify the function and point

Observe that the given limit is in the form of a derivative: $$\lim_{h \rightarrow 0} \frac{f(a+h) - f(a)}{h}$$ where \(f(x) = \cos(x)\) and \(a = \frac{\pi}{6}.\) Thus, our goal is to find the derivative of the function \(f(x) = \cos(x)\) at the point \(x = \frac{\pi}{6}.\)

Find the derivative of the given function

We'll differentiate the function \(f(x) = \cos(x)\) with respect to \(x\). Using the chain rule, we have: $$f'(x) = \frac{d}{dx} \cos(x) = -\sin(x).$$

Evaluate the derivative at the given point

Now, we'll find the value of the derivative at the point \(x = \frac{\pi}{6}\). Plug this value into our expression for the derivative: $$f'\left(\frac{\pi}{6}\right) = -\sin\left(\frac{\pi}{6}\right) = -\frac{1}{2}$$

Evaluate the given limit

Lastly, we'll use the found value of the derivative to evaluate the given limit: $$\lim _{h \rightarrow 0} \frac{\cos \left(\frac{\pi}{6}+h\right)-\frac{\sqrt{3}}{2}}{h} = f'\left(\frac{\pi}{6}\right) = -\frac{1}{2}$$ So, the value of the limit is: $$\lim _{h \rightarrow 0} \frac{\cos \left(\frac{\pi}{6}+h\right)-\frac{\sqrt{3}}{2}}{h} = -\frac{1}{2}$$

Key concepts

Derivative

A derivative represents the rate at which a function changes as its input changes. For a function \(f(x)\), the derivative is often denoted by \(f'(x)\) or \(\frac{df}{dx}\). It tells us how steep a curve is at any given point along the curve.

To find the derivative of a function at a specific point, we often use the limit definition:

• \(f'(a) = \lim_{h \rightarrow 0} \frac{f(a+h) - f(a)}{h}\)

This expression computes the average rate of change as \(h\) approaches zero, effectively giving us the instantaneous rate of change.

In our exercise above, the limit \(\lim_{h \rightarrow 0} \frac{\cos\left(\frac{\pi}{6}+h\right)-\frac{\sqrt{3}}{2}}{h}\) is identified as the derivative of \(f(x) = \cos(x)\) at the point \(a = \frac{\pi}{6}\). The result of this derivative is \(-\frac{1}{2}\), meaning the steeper the slope of \(\cos(x)\) at this point, the more it directs downward.

Trigonometric Functions

Trigonometric functions are a family of functions important in mathematics, marked by sine, cosine, and tangent—and their inverses. One of them, the cosine function \(\cos(x)\), oscillates between -1 and 1, making it useful for modeling periodic phenomena.

Some key identities for cosine include:

• \(\cos^2(x) + \sin^2(x) = 1\)
• \(\cos(0) = 1\)
• \(\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}\)

These identities help us calculate the function for different angles.

In calculus, trigonometric functions are fundamental when computing derivatives and integrals. For instance, the derivative of \(\cos(x)\) is \(-\sin(x)\). Understanding these basic trigonometric derivatives allows us to solve many types of calculus problems, as shown in our exercise.

Chain Rule

The chain rule is a fundamental concept in calculus used to find the derivative of composite functions. A composite function is where one function is applied to the result of another, like \((f \circ g)(x) = f(g(x))\).

The chain rule states:

• If \(y = f(u)\) and \(u = g(x)\), then \(\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\).

This rule effectively "chains" the derivatives together, allowing us to differentiate complex functions neatly.

In our step-by-step solution, the chain rule helps derive \(-\sin(x)\) from \(\cos(x)\). The rule itself wasn't directly necessary for simple trigonometric functions here, but it's crucial for understanding how more complicated expressions like nested functions are handled. Mastering this allows math students to tackle advanced derivative problems efficiently.