Question

Find the function The following limits represent the slope of a curve \(y=f(x)\) at the point \((a, f(a)) .\) Determine a possible function \(f\) and number a; then calculate the limit. $$\lim _{h \rightarrow 0} \frac{\sqrt{2+h}-\sqrt{2}}{h}$$

Short answer

The limit of the given expression as h approaches 0 is (1/2√2).

Step-by-step solution

Derivative definition recognition

The limit we are given is in the form: $$\lim_{h \to 0} \frac{\sqrt{2+h}-\sqrt{2}}{h}$$ This limit is very similar to the definition of a derivative, which is given by: $$f'(a) =\lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$$ In our case, we have \(\sqrt{2}\) in place of \(f(a)\) and \(\sqrt{2+h}\) in place of \(f(a+h)\). Thus, we can write the limit as: $$f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}=\lim_{h \to 0} \frac{\sqrt{2+h}-\sqrt{2}}{h}$$

Finding the function and the point 'a'

Now, we know that our function is in the form of \(f(x) = \sqrt{x+C}\), where \(C\) is some constant. Since \(f(a) = \sqrt{2}\), we can write the equation: $$f(a) = \sqrt{a+C} = \sqrt{2}$$ Square both sides to get: $$a+C=2$$ We don't have enough information to directly find \(C\), but one possible solution for \(a\) and \(C\) can be \(a=2\) and \(C=0\). Therefore, the function is: $$f(x) = \sqrt{x}$$

Finding the derivative of the function

Now that we have determined our function to be \(f(x) = \sqrt{x}\), we need to find its derivative. The derivative of the square root function is given by: $$f'(x) = \frac{1}{2\sqrt{x}}$$

Calculate the limit

We know that \(f'(a)=\lim_{h \to 0} \frac{\sqrt{2+h}-\sqrt{2}}{h}\), and we found \(a=2\) and \(f'(x) = \frac{1}{2\sqrt{x}}\). Thus, after substituting \(a=2\) into our derivative formula, we can find the limit: $$f'(2) = \frac{1}{2\sqrt{2}}$$ The limit of the given expression as \(h\) approaches \(0\) is \(\frac{1}{2\sqrt{2}}\). In conclusion, a possible function is \(f(x)=\sqrt{x}\) and the point at which we are determining the slope is \(a=2\). The limit of the given expression is \(\frac{1}{2\sqrt{2}}\).

Key concepts

Derivative

A derivative is essentially the rate at which a function is changing at any point. Mathematically, it represents the slope of the tangent line to the curve of the function at a given point. This concept is foundational in calculus and helps us understand how functions behave.
The formal definition involves limits. If you have a function \( f(x) \), the derivative at a point \( a \), denoted \( f'(a) \), can be found using:

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

This formula calculates how the function \( f(x) \) changes as we move infinitesimally close to \( a \) by a tiny amount \( h \). It's the mathematical way to find the slope of the original function at that specific point.
When you see a limit like \( \lim_{h \to 0} \frac{\sqrt{2+h}-\sqrt{2}}{h} \), it resembles this derivative form, indicating a calculation of change or slope around a value.

Limit

Limits are fundamental to the concept of derivatives in calculus. They consider what happens to a function as we get infinitely close to a certain point. Limits help capture the behavior of functions in situations where straightforward calculations are not possible due to undefined operations, like division by zero.
In our example problem,

• We see the limit \( \lim_{h \to 0} \frac{\sqrt{2+h}-\sqrt{2}}{h} \).
• This limit examines the behavior of the expression as \( h \) approaches zero, which resembles the derivative formula.

By this process, limits allow us to precisely calculate derivatives, making it possible to find the slope of the curve \( y=f(x) \) at any specific point \( a \). It's a way of peering into what happens right at that moment when you might otherwise hit a mathematical wall.
Understanding limits is key to grasping how and why derivatives work.

Function

A function in calculus is a relation that uniquely assigns one output for each input. In simpler terms, it’s like a machine where you put in a value, and it spits out a result according to a precise rule or formula.
Functions can be simple, like \( f(x) = x^2 \), or more complex, and they are used to model relationships between variables. In our exercise:

• The function we examined turned out to be \( f(x) = \sqrt{x} \).
• This function takes any input \( x \) and gives back the square root of \( x \).

Identifying the correct function is crucial because it determines how we can analyze the problem using calculus tools, like derivatives and limits. In this scenario, resolving \( f(a) = \sqrt{2} \) helped us deduce that the function was indeed \( \sqrt{x} \), which in turn allowed for the computation of the derivative at a given point.

Slope of a Curve

The slope of a curve at a specific point can show how steep or flat the curve is at that particular location. In calculus, this is captured by the derivative of the function at that point.
For a curve represented by \( y=f(x) \), the slope at any point \( (a, f(a)) \) is determined by the derivative \( f'(a) \).

• In our exercise, when calculating the slope at \( a=2 \), we used \( f'(2) = \frac{1}{2\sqrt{2}} \), indicating the exact steepness of the curve \( y=\sqrt{x} \) at the point where \( x=2 \).

The slope is particularly valuable in understanding the behavior of graphs, optimizing functions, or even solving real-world physics problems where change over time is measured.
Each slope we calculate offers an insight into how dynamic the function is at a precise moment, helping in both theoretical and practical applications of calculus.