Problem 111 - Calculus 3

Question

Given \(f(x, y)=x^{2}-4 y\) find $$ \lim _{h \rightarrow 0} \frac{f(1+h, y)-f(1, y)}{h} $$

Step-by-Step Solution

Answer

The limit is 2.

Step 1: Understand the Limit Expression

We're asked to find the limit \( \lim_{h \rightarrow 0} \frac{f(1+h, y) - f(1, y)}{h} \). This is essentially the definition of the partial derivative of \( f \) with respect to \( x \) at the point where \( x = 1 \). We will focus on simplifying this expression.

Step 2: Find \( f(1+h, y) \)

Substitute \( x = 1 + h \) into the function: \[ f(1+h, y) = (1+h)^2 - 4y. \] Simplify \( (1+h)^2 \) to get \( 1 + 2h + h^2 \). So, \[ f(1+h, y) = 1 + 2h + h^2 - 4y. \]

Step 3: Find \( f(1, y) \)

Substitute \( x = 1 \) into the function: \[ f(1, y) = 1^2 - 4y = 1 - 4y. \]

Step 4: Substitute the Function Values into the Limit Expression

Place the values from Steps 2 and 3 into the limit: \[ \lim_{h \rightarrow 0} \frac{(1 + 2h + h^2 - 4y) - (1 - 4y)}{h}. \] Simplify the numerator: \[ \lim_{h \rightarrow 0} \frac{1 + 2h + h^2 - 4y - 1 + 4y}{h} = \lim_{h \rightarrow 0} \frac{2h + h^2}{h}. \]

Step 5: Simplify the Limit Expression

Factor \( h \) out from the numerator: \[ \lim_{h \rightarrow 0} \frac{h(2 + h)}{h}. \] Cancel \( h \) in the numerator and the denominator: \[ \lim_{h \rightarrow 0} (2 + h). \]

Step 6: Calculate the Limit

Evaluate the limit as \( h \) approaches 0: \[ 2 + 0 = 2. \] Thus, the limit is 2.

Key Concepts

Limit Expression

The concept of a limit is fundamental in calculus, serving as a bridge to understanding derivatives and integrals. When we talk about a limit expression, especially in the context of partial derivatives, we are typically looking at the approach of a function to a particular value as one of its variables approaches a certain point. In this exercise, we are dealing with the limit expression:

• \( \lim_{h \rightarrow 0} \frac{f(1+h, y) - f(1, y)}{h} \)

This limit expression is crucial because it represents how the function \(f(x, y) = x^2 - 4y\) changes with respect to the variable \(x\) at a specific point. Here, the limit expression is specifically designed to find the partial derivative of the function with respect to \(x\) at \(x = 1\).
Think of it as examining the slope of the tangent line to the curve in the \(x\)-direction, while keeping \(y\) constant. By simplifying and evaluating this limit, we essentially grasp how changes in \(x\) influence the function's behavior at the micro level at the given point.

Function Simplification

Function simplification is a necessary step in evaluating partial derivatives, as it makes complex expressions manageable and easier to understand. Here, we simplify the function \(f(x, y) = x^2 - 4y \) by substituting \(x = 1 + h\) to examine how the function behaves as \(h\) approaches 0.
Let's observe how the simplification occurs:

• Substitute \(x = 1 + h\) into the function to get \(f(1+h, y) = (1+h)^2 - 4y\).
• Simplify \((1+h)^2\) to obtain \(1 + 2h + h^2\).
• Thus, the function becomes \(f(1+h, y) = 1 + 2h + h^2 - 4y\).

The next step involves substituting \(x = 1\) into the original function to get \(f(1, y) = 1 - 4y\). This simplification is crucial, as it prepares us to place these function values into the limit expression.
Doing so highlights the changes in the function value concerning \(h\), thereby enabling us to isolate terms that contribute to the partial derivative.

Differentiation with Respect to x

Differentiation is a core operation in calculus that gives us the rate of change of a function. When we perform differentiation with respect to \(x\) in a function of two variables like \(f(x, y)\), we are interested in understanding how the function changes as \(x\), the independent variable, varies, with \(y\) held constant.
In finding the partial derivative of the function \(f(x, y) = x^2 - 4y\) with respect to \(x\) at \(x = 1\), we start with the expression:

• \(\lim_{h \rightarrow 0} \frac{(1 + 2h + h^2 - 4y) - (1 - 4y)}{h}\)

This is simplified to:

• \(\lim_{h \rightarrow 0} \frac{2h + h^2}{h}\)
• Which further simplifies to \(\lim_{h \rightarrow 0} (2 + h)\)

Finally, as \(h\) approaches zero, the expression evaluates to 2. This result represents the partial derivative \(\frac{\partial f}{\partial x}\) at the point \(x = 1\), meaning that the function increases by 2 units along the \(x\)-axis for a tiny interval. In solving such exercises, recognizing every step's role helps students fully grasp the impactful nature of calculus.