Question

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. Assume \(a\) and \(L\) are finite numbers. a. If \(\lim _{x \rightarrow a} f(x)=L,\) then \(f(a)=L\) b. If \(\lim _{x \rightarrow a^{-}} f(x)=L,\) then \(\lim _{x \rightarrow a^{+}} f(x)=L\) c. If \(\lim _{x \rightarrow a} f(x)=L\) and \(\lim _{x \rightarrow a} g(x)=L,\) then \(f(a)=g(a)\) d. The limit \(\lim _{x \rightarrow a} \frac{f(x)}{g(x)}\) does not exist if \(g(a)=0\) e. If \(\lim _{x \rightarrow 1^{+}} \sqrt{f(x)}=\sqrt{\lim _{x \rightarrow 1^{+}} f(x)},\) it follows that \(\lim _{x \rightarrow 1} \sqrt{f(x)}=\sqrt{\lim _{x \rightarrow 1} f(x)}\)

Short answer

a. If \(\lim _{x \rightarrow a} f(x)=L,\) then \(f(a)=L\) b. If \(\lim _{x \rightarrow a^{-}} f(x)=L,\) then \(\lim _{x \rightarrow a^{+}} f(x)=L\) c. If \(\lim _{x \rightarrow a} f(x)=L\) and \(\lim _{x \rightarrow a} g(x)=L,\) then \(f(a)=g(a)\) d. The limit \(\lim _{x \rightarrow a} \frac{f(x)}{g(x)}\) does not exist if \(g(a)=0\) e. If \(\lim _{x \rightarrow 1^{+}} \sqrt{f(x)}=\sqrt{\lim _{x \rightarrow 1^{+}} f(x)},\) it follows that \(\lim _{x \rightarrow 1} \sqrt{f(x)}=\sqrt{\lim _{x \rightarrow 1} f(x)}\) Answer: All of the given statements are not necessarily true. Here are the counterexamples or explanations for each statement: a. Counterexample: \(f(x) = \frac{x^2 - 1}{x - 1}\): \(\lim_{x \to 1} f(x) = 2\), but \(f(1)\) is undefined. b. Counterexample: \(f(x) = \begin{cases} x, & x < 0 \\ -x, & x \geq 0 \end{cases}\): \(\lim_{x\to0^-} f(x) = 0\), but \(\lim_{x\to0^+} f(x) = 0\). c. Counterexample: \(f(x) = x\) and \(g(x) = |x|\): \(\lim_{x \to 0} f(x) = 0\) and \(\lim_{x \to 0} g(x) = 0\), but \(f(0) = 0\) and \(g(0) = 0\). d. Counterexample: \(\lim_{x \to 0} \frac{x^2}{x}\) exists and equals 0, even though \(g(0) = 0\). e. Counterexample: \(f(x) = \begin{cases} x, & x \leq 1 \\ x^2, & x > 1 \end{cases}\): \(\lim_{x \to 1^{+}}\sqrt{f(x)} = 1\), but \(\lim_{x \to 1}\sqrt{f(x)}\) does not exist.

Step-by-step solution

Statement a: If \(\lim _{x \rightarrow a} f(x)=L,\) then \(f(a)=L\)

Statement a is not always true. A function can have a limit at a point without being defined at that point. For example, consider the function \(f(x)= \frac{x^2 - 1}{x - 1}\). Here, when we remove the discontinuity at \(x = 1\), we get \(f(x) = x + 1\). Now, \(\lim_{x \rightarrow 1} f(x) = 2\), but \(f(1)\) is undefined.

Statement b: If \(\lim _{x \rightarrow a^{-}} f(x)=L,\) then \(\lim _{x \rightarrow a^{+}} f(x)=L\)

Statement b is not true in general. For example, consider a piecewise function $f(x) = \begin{cases} x, & x < 0 \\ -x, & x \geq 0 \end{cases}\(. Here, we have \)\lim_{x\to0^-} f(x) = 0\( but \)\lim_{x\to0^+} f(x) = 0$ as well. The existence of a left-hand limit doesn't imply a same right-hand limit.

Statement c: If \(\lim _{x \rightarrow a} f(x)=L\) and \(\lim _{x \rightarrow a} g(x)=L,\) then \(f(a)=g(a)\)

Statement c is not always true either. We need to find two functions with the same limit at a certain point, but different values at the same point. Consider functions \(f(x) = x\) and \(g(x) = |x|\). Both have limits of 0 at \(x = 0\), however, \(f(0) = 0\) and \(g(0) = 0\). So \(f(a) = g(a)\).

Statement d: The limit \(\lim _{x \rightarrow a} \frac{f(x)}{g(x)}\) does not exist if \(g(a)=0\)

Statement d is false. \(\lim_{x \to a} \frac{f(x)}{g(x)}\) can exist even if \(g(a) = 0\). As an example, consider the function \(\frac{x^2}{x}\) with \(a = 0\). In this case, we can simplify the function to just \(x\) when \(x\neq0\). Then observe that \(\lim_{x \to 0} \frac{x^2}{x}\) exists and equals 0, despite the fact that \(g(a) = 0\). However, it is also possible for the limit not to exist when \(g(a) = 0\).

Statement e: If \(\lim _{x \rightarrow 1^{+}} \sqrt{f(x)}=\sqrt{\lim _{x \rightarrow 1^{+}} f(x)},\) it follows that \(\lim _{x \rightarrow 1} \sqrt{f(x)}=\sqrt{\lim _{x \rightarrow 1} f(x)}\)

Statement e is false. Just because the one-sided limits agree (the limit from the right in this case), it doesn't guarantee that the two-sided limit exists. For example, let $f(x) = \begin{cases} x, & x \leq 1 \\ x^2, & x > 1 \end{cases}\(. In this case, \)\lim_{x \to 1^{+}}\sqrt{f(x)} = \sqrt{\lim_{x \to 1^{+}}f(x)} = 1\(, but \)\lim_{x \to 1}\sqrt{f(x)}\( does not exist, as \)\lim_{x \to 1^{-}}\sqrt{f(x)} \neq \lim_{x \to 1^{+}}\sqrt{f(x)}$.

Key concepts

Limit Definition

Understanding limits is a fundamental aspect of calculus. A limit describes the behavior of a function as it approaches a particular point. More formally, the limit of a function \( f(x) \) as \( x \) approaches a value \( a \), written as \( \lim_{x \to a} f(x) = L \), means that as \( x \) gets closer to \( a \) (but is not necessarily equal to \( a \)), \( f(x) \) gets arbitrarily close to \( L \).
It is crucial to note that the existence of a limit \( \lim_{x \to a} f(x) = L \) does not imply that \( f(a) = L \). The function \( f(x) \) may not even be defined at \( a \), yet the limit can still exist as demonstrated in the example \( f(x) = \frac{x^2 - 1}{x - 1} \). While \( \lim_{x \to 1} f(x) = 2 \), \( f(1) \) is actually undefined due to a discontinuity, showcasing that the behavior of a function near a point and the function's value at that point can be independent of each other.

Discontinuity in Functions

Discontinuities occur in a function when there is a break, gap, or jump at a specific point or interval. These are types of points where the function does not behave in a smooth or continuous manner. Discontinuities can arise when:

• The function is not defined at that point.
• The left-hand limit and the right-hand limit do not match.
• The function has a gap or jump, causing a break in continuity.

The original problem highlights scenarios like limits involving functions with discontinuities, such as in the equation \( \lim_{x \to 1} f(x) \), where discontinuity at \( x = 1 \) prevents \( f(1) \) from being well-defined unless the function is redefined or factored to remove the discontinuity. Understanding and identifying discontinuities is essential, especially when determining limits, as they often require more careful consideration to evaluate correctly.

Piecewise Functions

Piecewise functions are defined by different expressions based on the input value. Their definition changes according to different intervals of the domain. These functions are incredibly useful in representing systems with sudden changes or distinct modes of operation, such as tax brackets or conditional rates.
For example, consider a function \( f(x) \):
\[ f(x) = \begin{cases} x, & x < 0 \ -x, & x \geq 0 \end{cases} \]
Learning to work with piecewise functions is important as it involves calculating limits independently for each piece or interval. This often means dealing with potential discontinuities or different behaviors at the transitioning points between pieces, requiring a separate analysis for the left-hand and right-hand limits at those transition points.

One-Sided Limits

One-sided limits examine the behavior of a function as the input approaches a specific point from one direction only. There are two types of one-sided limits:

• The left-hand limit \( \lim_{x \to a^-} f(x) \), where \( x \) approaches \( a \) from the left (values less than \( a \)).
• The right-hand limit \( \lim_{x \to a^+} f(x) \), where \( x \) approaches \( a \) from the right (values greater than \( a \)).

In the step-by-step solution regarding statement b, it was shown that one-sided limits can differ, especially in piecewise functions or functions with jumps or discontinuities. The existence of a one-sided limit does not ensure the opposite side's limit matches, nor does it ensure the overall limit exists. For example, with the piecewise function \( f(x) = \begin{cases} x, & x \leq 1 \ x^2, & x > 1 \end{cases} \), \( \lim_{x \to 1^-} \) and \( \lim_{x \to 1^+} \) may differ, highlighting how one-sided analyses offer crucial insights into the function’s overall behavior at specific points.