Question

Challenge your understanding of limits but can be evaluated using the knowledge gained in this section. Let \(f(x)=0\) and \(g(x)=\frac{x}{x}\) (a) Show why \(\lim _{x \rightarrow 2} f(x)=0\). (b) Show why \(\lim _{x \rightarrow 0} g(x)=1\). (c) Show why \(\lim _{x \rightarrow 2} g(f(x))\) does not exist. (d) Show why the answer to part (c) does not violate the Composition Rule of Theorem 1.3.1.

Short answer

(a) 0, (b) 1, (c) Does not exist, (d) Not a violation.

Step-by-step solution

Evaluate Limit of f(x) as x approaches 2

Since the function \(f(x) = 0\) is constant for all \(x\), the limit can be directly evaluated. As \(x\) approaches 2, \(f(x) = 0\) remains unchanged. Thus, \(\lim_{x \to 2} f(x) = 0\).

Simplify and Evaluate Limit of g(x) as x approaches 0

For the function \(g(x) = \frac{x}{x}\), it simplifies to \(1\) for any \(x eq 0\) since \(\frac{x}{x} = 1\). As \(x\) approaches 0, \(g(x)\) becomes undefined at exactly \(x = 0\), but the limit from both sides is \(1\). Hence, \(\lim_{x \to 0} g(x) = 1\) since the function approaches this value from either side of zero without including zero itself.

Analyze g(f(x)) as x approaches 2 and its Limit

Here, \(g(f(x)) = g(0)\) because \(f(x) = 0\). Inside the function \(g(x) = \frac{x}{x}\), substituting \(x = 0\) makes it \(\frac{0}{0}\), which is undefined. Therefore, \(\lim_{x \to 2} g(f(x))\) does not exist because \(g(0)\) is not defined.

Examine the Composition Rule of Theorem 1.3.1

Theorem 1.3.1 for composition of limits requires that \(g(L)\) be defined if \(\lim_{x \to c} f(x) = L\). In part (c), \(g(0)\) is not defined, which is why the limit does not exist. Hence, this does not violate the composition rule because the conditions for the rule (namely, that \(g\) must be defined at \(L\)) are not met.

Key concepts

Composition Rule

In calculus, the composition rule is fundamental when dealing with the limits of composed functions. It essentially states that if you have a function that is a composition of two other functions, the limit of the composite function can be found by taking the limits of the inner and outer functions individually. However, there’s a catch. The outer function must be defined at the point you are evaluating.

In the context of the problem, when evaluating \(lim_{x \to 2} g(f(x))\), it's important to recognize that this expression results in attempting to evaluate \(g(0)\). Since \(g(x) = \frac{x}{x}\), g is not defined at zero because it results in the indeterminate form \(\frac{0}{0}\). Therefore, the composition rule cannot be successfully applied here because \(g(0)\) does not exist.

This situation demonstrates precisely why the composition rule has the specific requirements it does, protecting against undefined expressions.

Limit evaluation

Limits are a core concept in calculus and act as the foundation for derivatives and integrals. To evaluate limits, you often use techniques like substitution, simplification, or applying specific limit laws. These methods help in calculating the value that a function approaches as the input approaches a certain point.

For example, when finding \(\lim_{x \to 2} f(x)\) where \(f(x) = 0\), the function is constant. This makes the limit evaluation straightforward as \(f(x)\) is always zero regardless of \(x\), resulting in the limit being zero. On the other hand, with \(g(x) = \frac{x}{x}\), you first simplify it to 1 for all \(x eq 0\). When \(x \to 0\), the limit becomes 1 as it approaches from both sides, even though the function itself is undefined at zero.

Limit evaluation often requires careful handling of such undefined points or discontinuities to derive meaningful results.

Continuous functions

A function is continuous if you can draw it without lifting your pencil off the paper. That means, around every point in the domain, the function doesn’t have abrupt jumps or holes. Specifically, the limit of the function as it approaches that point must be equal to the function's value at that point.

For instance, \(f(x) = 0\) is continuous everywhere because it’s constant; there's no variation in its value. However, \(g(x) = \frac{x}{x}\) is only continuous where it's defined, which is not at \(x = 0\) because it forms an indeterminate expression there. Being undefined at a specific point breaks the continuity at that point.

This distinction between continuous and discontinuous points in functions is critical in calculus as it often determines the ability to evaluate limits smoothly.

Indeterminate forms

Indeterminate forms crop up frequently in calculus, especially in limits. They typically arise in expressions like \(\frac{0}{0}\), \(\frac{\infty}{\infty}\), \(0 \cdot \infty\), and other ambiguous expressions resulting from direct substitution in limits.

These forms are labeled indeterminate because simply knowing an individual part (like having 0 in the numerator and denominator) doesn't provide enough information to determine the overall limit. In our case, \(g(f(x)) = g(0) = \frac{0}{0}\), which is indeterminate.

To handle such situations, calculus provides methods such as L'Hôpital's Rule, algebraic manipulation, or considering limits from both sides. Handling indeterminate forms is crucial for accurately understanding the behavior of a function as it approaches specific points.