Question

The following limits equal the derivative of a function \(f\) at a point \(a\) a. Find one possible \(f\) and \(a\) b. Evaluate the limit. $$\lim _{h \rightarrow 0} \frac{\cos \left(\frac{\pi}{6}+h\right)-\frac{\sqrt{3}}{2}}{h}$$

Short answer

Answer: The limit is related to the derivative of the function \(f(x) = \cos(x)\) at the point \(a = \frac{\pi}{6}\). Its value is \(-\frac{1}{2}\).

Step-by-step solution

Identify the function f(x) and the point a

Since the limit looks like the derivative of the cosine function, we will first differentiate the cosine function and check if it matches our limit. In this case, it seems like the function f(x) could be the cosine function, \(f(x)=\cos(x)\), and the point a could be \(\frac{\pi}{6}\). Let's find the derivative of the cosine function and see if it matches the given limit.

Find the derivative of the function

To find the derivative of the cosine function, use the definition of a derivative: $$f'(x) = \lim_{h \rightarrow 0} \frac{f(x + h) - f(x)}{h}$$ For the cosine function, we have $$f'(x) = \lim_{h \rightarrow 0} \frac{\cos(x+h) - \cos(x)}{h}$$ Now, we want to know if the derivative of the cosine function matches the given limit at the point \(a=\frac{\pi}{6}\).

Evaluate the limit of the derivative

We need to find $$\lim_{h \rightarrow 0} \frac{\cos \left(\left(\frac{\pi}{6}\right) + h\right) - \cos \left(\frac{\pi}{6}\right)}{h}$$ We know that \(\cos \left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}\), so we have $$\lim_{h \rightarrow 0} \frac{\cos \left(\left(\frac{\pi}{6}\right) + h\right) - \frac{\sqrt{3}}{2}}{h}$$ This limit matches the given limit! So, we can see that \(f(x)=\cos(x)\) and \(a=\frac{\pi}{6}\) satisfy the requirement.

Evaluate the limit

To evaluate the limit, we will find \(f'(x)\) for the cosine function: $$f'(x) = -\sin(x)$$ Now, we will evaluate the derivative at \(\frac{\pi}{6}\): $$f'\left(\frac{\pi}{6}\right) = -\sin \left(\frac{\pi}{6}\right)$$ $$f'\left(\frac{\pi}{6}\right) = -\frac{1}{2}$$ The limit is: $$\lim_{h \rightarrow 0} \frac{\cos \left(\frac{\pi}{6} + h\right) - \frac{\sqrt{3}}{2}}{h} = -\frac{1}{2}$$ #Summary# So, one possible function \(f(x)\) is \(f(x) = \cos(x)\), the point \(a\) is \(\frac{\pi}{6}\), and the limit is equal to \(-\frac{1}{2}\).

Key concepts

Understanding the Derivative of a Function

The derivative of a function represents the rate at which the function's value changes as its input changes. In mathematical terms, the derivative is a fundamental tool used to understand the behavior of functions and can be denoted as \( f'(x) \) or \( \frac{df}{dx} \). It measures how the output value of a function changes as the input changes incrementally.
The process to find a derivative begins with what is known as the "limit definition of a derivative." It is expressed as:

• \( f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \)

This formula calculates the slope of the tangent line to the function's graph at point \( x \), representing how steep the graph is at that specific point.
By applying this definition systematically, you can determine how functions behave locally and predict their behavior with greater accuracy.

Exploring the Cosine Function

The cosine function is a fundamental trigonometric function denoted by \( \cos(x) \). It represents the x-coordinate of a point on the unit circle as the angle \( x \) changes. The cosine function is periodic, repeating its values in regular intervals \( 2\pi \).
Some key properties of the cosine function include:

• It ranges from -1 to 1.
• It is an even function, meaning \( \cos(-x) = \cos(x) \).
• The derivative of the cosine function is \( -\sin(x) \), describing how the slope of the cosine graph changes at a point.

Understanding these properties helps us grasp how the cosine function behaves under rotation and transformations, which is vital for applications in trigonometry and calculus.
For instance, at \( x = \frac{\pi}{6} \), we know \( \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \). This specific value becomes significant when solving problems involving derivatives and limits related to the cosine function.

Evaluating Limits in Calculus

Evaluating limits is a vital concept in calculus, providing the foundation for understanding change and continuity in functions. A limit helps determine what value a function approaches as the input gets closer to a specific point.
The standard form to express a limit in mathematics is:

• \( \lim_{x \to a} f(x) \)

This indicates what \( f(x) \) tends towards as \( x \) approaches \( a \). If the result exists and is finite, the limit is said to be evaluable.
Limits are integral for defining derivatives, as they measure how a function grows or shrinks as the input changes. By computing limits of difference quotients, we find derivatives, allowing us to model and analyze real-world phenomena precisely.
In the given problem, evaluating \( \lim_{h \to 0} \frac{\cos\left(\frac{\pi}{6}+h\right)-\frac{\sqrt{3}}{2}}{h} \) illustrates how the derivative and limits intertwine. It transforms the description of instantaneous rates of change, underpinning the power of limits in understanding the behavior of trigonometric functions at specific points.