Composite Functions And Limit Evaluation A Comprehensive Guide

In the realm of mathematics, particularly in calculus and precalculus, composite functions play a crucial role in understanding the interplay between different functions. Composite functions, formed by combining two or more functions, allow us to analyze complex relationships and model real-world phenomena effectively. Understanding how to evaluate composite functions is a fundamental skill for any aspiring mathematician or scientist. Let's embark on a journey to unravel the intricacies of composite functions and learn how to find the value of $(h ext{ ∘ } g ext{ ∘ } f)(3)$. Composite functions are essentially functions within functions, where the output of one function becomes the input of another. This sequential application of functions creates a chain reaction, allowing us to model intricate relationships between variables. The notation $(h ext{ ∘ } g ext{ ∘ } f)(x)$ represents the composition of three functions, where $f$ is applied first, then $g$, and finally $h$. To evaluate $(h ext{ ∘ } g ext{ ∘ } f)(3)$, we need to follow a step-by-step approach. First, we evaluate the innermost function, $f(3)$. This will give us a numerical value, which we then use as the input for the next function, $g$. We evaluate $g$ at this new input, and the resulting value becomes the input for the outermost function, $h$. Finally, we evaluate $h$ at this input to obtain the final result. In essence, we are working our way from the inside out, applying each function sequentially. This process highlights the importance of understanding the order of operations in mathematics. Failing to follow the correct order can lead to incorrect results. Let's delve deeper into the concept with a concrete example. Suppose we have three functions: $f(x) = x + 1$, $g(x) = x^2$, and $h(x) = 2x - 3$. To find $(h ext{ ∘ } g ext{ ∘ } f)(3)$, we start by evaluating $f(3)$. Substituting 3 into $f(x)$, we get $f(3) = 3 + 1 = 4$. Now, we use this result as the input for $g$. So, we evaluate $g(4)$, which gives us $g(4) = 4^2 = 16$. Finally, we use 16 as the input for $h$. Evaluating $h(16)$, we get $h(16) = 2(16) - 3 = 32 - 3 = 29$. Therefore, $(h ext{ ∘ } g ext{ ∘ } f)(3) = 29$. This example demonstrates the step-by-step process of evaluating composite functions. It's crucial to pay attention to the order of functions and the inputs at each stage. With practice, evaluating composite functions becomes a straightforward task. The beauty of composite functions lies in their ability to model complex relationships. Imagine a scenario where a company's profit depends on its revenue, which in turn depends on the number of units sold. This can be modeled using composite functions, where one function represents the relationship between units sold and revenue, and another function represents the relationship between revenue and profit. By composing these functions, we can directly relate the number of units sold to the company's profit. This is just one example of how composite functions can be applied in real-world situations. Their versatility makes them a powerful tool in mathematics, science, and engineering.

(a) $\lim_{x \rightarrow 1} \frac{x^2-\sqrt{x}}{\sqrt{x}-1}$

Limits are a cornerstone of calculus, providing a foundation for understanding continuity, derivatives, and integrals. Evaluating limits allows us to analyze the behavior of functions as their input approaches a specific value. In essence, limits help us to understand what happens to a function as we get arbitrarily close to a particular point, without necessarily reaching that point. This concept is crucial for dealing with situations where direct substitution leads to undefined expressions, such as division by zero. The limit $\lim_{x \rightarrow 1} \frac{x^2-\sqrt{x}}{\sqrt{x}-1}$ presents an interesting challenge. If we try to directly substitute $x = 1$ into the expression, we get $\frac{1^2-\sqrt{1}}{\sqrt{1}-1} = \frac{1-1}{1-1} = \frac{0}{0}$, which is an indeterminate form. This means that direct substitution doesn't work, and we need to employ algebraic manipulation techniques to simplify the expression and evaluate the limit. One common technique for dealing with expressions involving square roots is to multiply by the conjugate. The conjugate of $\sqrt{x} - 1$ is $\sqrt{x} + 1$. Multiplying both the numerator and denominator by the conjugate, we get:

$\frac{x^2-\sqrt{x}}{\sqrt{x}-1} \cdot \frac{\sqrt{x}+1}{\sqrt{x}+1} = \frac{(x^2-\sqrt{x})(\sqrt{x}+1)}{x-1}$

Now, we need to further simplify the numerator. We can factor out $\sqrt{x}$ from the first term in the numerator:

$\frac{\sqrt{x}(x^{3/2}-1)(\sqrt{x}+1)}{x-1}$

Notice that $x^{3/2}$ can be written as $(\sqrt{x})^3$. Let's make a substitution to simplify the expression further. Let $y = \sqrt{x}$. Then, $x = y^2$, and the expression becomes:

$\frac{y(y^3-1)(y+1)}{y^2-1}$

We can factor the difference of cubes in the numerator and the difference of squares in the denominator:

$\frac{y(y-1)(y^2+y+1)(y+1)}{(y-1)(y+1)}$

Now, we can cancel out the common factors $(y-1)$ and $(y+1)$:

$y(y^2+y+1)$

Substituting back $y = \sqrt{x}$, we get:

$\sqrt{x}(x+\sqrt{x}+1)$

Now, we can evaluate the limit as $x$ approaches 1:

$\lim_{x \rightarrow 1} \sqrt{x}(x+\sqrt{x}+1) = \sqrt{1}(1+\sqrt{1}+1) = 1(1+1+1) = 3$

Therefore, $\lim_{x \rightarrow 1} \frac{x^2-\sqrt{x}}{\sqrt{x}-1} = 3$. This example demonstrates the importance of algebraic manipulation in evaluating limits. By multiplying by the conjugate and factoring, we were able to simplify the expression and eliminate the indeterminate form. This allowed us to find the limit successfully. The concept of limits is fundamental to understanding the behavior of functions and is a cornerstone of calculus.

(b) $\lim_{x \rightarrow 0} \frac{x^2+2|x|}{x^2-2|x|}$

This limit involves the absolute value function, which introduces a piecewise nature to the expression. To evaluate $\lim_{x \rightarrow 0} \frac{x^2+2|x|}{x^2-2|x|}$, we need to consider the behavior of the absolute value function near $x = 0$. Recall that the absolute value function is defined as:

$|x| = \begin{cases} x, & \text{if } x \geq 0 \\ -x, & \text{if } x < 0 \end{cases}$

This means we need to consider the limit from the right (as $x$ approaches 0 from positive values) and the limit from the left (as $x$ approaches 0 from negative values) separately. If the limits from the right and left are equal, then the overall limit exists and is equal to that value. If the limits are different, then the overall limit does not exist. Let's first consider the limit from the right, denoted as $\lim_{x \rightarrow 0^+} \frac{x^2+2|x|}{x^2-2|x|}$. As $x$ approaches 0 from the right, $x$ is positive, so $|x| = x$. Therefore, the expression becomes:

$\lim_{x \rightarrow 0^+} \frac{x^2+2x}{x^2-2x}$

We can factor out $x$ from both the numerator and denominator:

$\lim_{x \rightarrow 0^+} \frac{x(x+2)}{x(x-2)}$

Now, we can cancel out the common factor $x$:

$\lim_{x \rightarrow 0^+} \frac{x+2}{x-2}$

Now, we can directly substitute $x = 0$:

$\frac{0+2}{0-2} = \frac{2}{-2} = -1$

So, the limit from the right is -1. Next, let's consider the limit from the left, denoted as $\lim_{x \rightarrow 0^-} \frac{x^2+2|x|}{x^2-2|x|}$. As $x$ approaches 0 from the left, $x$ is negative, so $|x| = -x$. Therefore, the expression becomes:

$\lim_{x \rightarrow 0^-} \frac{x^2-2x}{x^2+2x}$

We can factor out $x$ from both the numerator and denominator:

$\lim_{x \rightarrow 0^-} \frac{x(x-2)}{x(x+2)}$

Now, we can cancel out the common factor $x$:

$\lim_{x \rightarrow 0^-} \frac{x-2}{x+2}$

Now, we can directly substitute $x = 0$:

$\frac{0-2}{0+2} = \frac{-2}{2} = -1$

So, the limit from the left is also -1. Since the limits from the right and left are both equal to -1, the overall limit exists and is equal to -1. Therefore, $\lim_{x \rightarrow 0} \frac{x^2+2|x|}{x^2-2|x|} = -1$. This example highlights the importance of considering one-sided limits when dealing with functions involving absolute values or other piecewise definitions. By analyzing the behavior of the function from both sides, we can determine whether the overall limit exists and find its value. Understanding one-sided limits is crucial for a comprehensive understanding of limits and continuity.

This exploration has delved into the fundamental concepts of composite functions and limit evaluation. We've seen how composite functions allow us to model complex relationships by combining multiple functions, and we've learned the step-by-step process of evaluating them. We've also tackled the intricacies of limits, including dealing with indeterminate forms and the importance of one-sided limits when dealing with absolute value functions. These concepts are essential building blocks for further studies in calculus and related fields. By mastering these techniques, students can confidently tackle more complex mathematical problems and gain a deeper appreciation for the power and elegance of calculus.