Question

Determine whether the following statements are true and give an explanation or counterexample. a. The value of \(\lim _{x \rightarrow 3} \frac{x^{2}-9}{x-3}\) does not exist. b. The value of \(\lim _{x \rightarrow a} f(x)\) is always found by computing \(f(a)\) c. The value of \(\lim _{x \rightarrow a} f(x)\) does not exist if \(f(a)\) is undefined. d. \(\lim _{x \rightarrow 0} \sqrt{x}=0\) \(\lim _{x \rightarrow \pi / 2} \cot x=0\)

Short answer

In summary, we analyzed the four given statements involving limits and determined that statement a is false, statement b is false, statement c is false, and statement d is true. For statements a, b, and c, we provided counterexamples to show that they are not true in general, while for statement d, we showed both parts of the statement are true.

Step-by-step solution

Statement a: Checking the Limit Existence

The given limit is: $$ \lim_{x \rightarrow 3} \frac{x^2 - 9}{x - 3} $$ We have an indeterminate form (0/0) when x approaches 3, so we can use factoring to simplify the expression: $$ \lim_{x \rightarrow 3} \frac{(x - 3)(x + 3)}{x - 3} $$ Now, we can cancel out the (x - 3) term: $$ \lim_{x \rightarrow 3} (x + 3) $$ Now, since the expression is no more indeterminate, we can just substitute x = 3 to find the limit: $$ \lim_{x \rightarrow 3} (x + 3) = 3 + 3 = 6 $$ Since the limit exists and equals 6, statement a is false.

Statement b: Checking the General Rule

This statement claims that for any function f(x) and any value a, the limit is found by simply computing f(a). This is not true in general, as we have different types of discontinuities (for example, removable, jump, or infinite) that may prevent the limit from being equal to the function's value at the point. For example, consider the following function: $$ g(x) = \begin{cases} x^2 & x \neq 2 \\ 0 & x = 2 \end{cases} $$ We have: $$ \lim_{x \rightarrow 2} g(x) = \lim_{x \rightarrow 2} x^2 = 4 $$ But, \(g(2) = 0\). In this case, the limit does not equal the function's value at a. So, statement b is false.

Statement c: Checking the Limit's Existence

The statement claims that if \(f(a)\) is undefined, then the limit \(\lim_{x \rightarrow a} f(x)\) does not exist. This is not true because, in some cases, the limit can exist even if the value of the function is undefined at a specific point. Consider the following function as an example: $$ h(x) = \begin{cases} \frac{x^2 - 1}{x - 1} & x \neq 1 \\ \text{undefined} & x = 1 \end{cases} $$ We can simplify the function by factoring: $$ h(x) = \begin{cases} \frac{(x - 1)(x + 1)}{x - 1} & x \neq 1 \\ \text{undefined} & x = 1 \end{cases} $$ After canceling out (x - 1), we have: $$ h(x) = \begin{cases} x + 1 & x \neq 1 \\ \text{undefined} & x = 1 \end{cases} $$ Now, the function is continuous at x = 1, and we can find the limit: $$ \lim_{x \rightarrow 1} h(x) = 1 + 1 = 2 $$ Thus, statement c is false.

Statement d: Checking the Square Root Limit and Trigonometric Limit

We need to check if the following limits are true: 1. \(\lim_{x \rightarrow 0} \sqrt{x} = 0\): Since the function \(\sqrt{x}\) is continuous on its domain, and the domain includes x = 0, we can substitute x = 0: $$ \lim_{x \rightarrow 0} \sqrt{x} = \sqrt{0} = 0 $$ This part of the statement is true. 2. \(\lim_{x \rightarrow \frac{\pi}{2}} \cot x = 0\): We can rewrite the cotangent function as \(\cot x = \frac{1}{\tan x}\). As x approaches \(\frac{\pi}{2}\), \(\tan x\) approaches infinity, and therefore, \(\frac{1}{\tan x} \rightarrow 0\): $$ \lim_{x \rightarrow \frac{\pi}{2}} \cot x = \lim_{x \rightarrow \frac{\pi}{2}} \frac{1}{\tan x} = 0 $$ This part of the statement is also true. Therefore, statement d is true.

Key concepts

Limit Existence

Understanding when a limit exists is crucial in calculus. A limit \(\lim_{x \rightarrow c} f(x)\) exists if as x approaches c from either direction, the function f(x) approaches a specific value L. Here's how you can determine if a limit exists:

• Check if the function approaches the same value from the right \(x \rightarrow c^+\) and from the left \(x \rightarrow c^-\).
• If the one-sided limits agree, the two-sided limit exists.
• If the function has a jump, vertical asymptote, or oscillates without approaching a specific value near c, the limit does not exist.

In our exercise showing step 1, \(\frac{x^2 - 9}{x - 3}\) simplifies to \(x + 3\) by factoring and canceling, demonstrating that a limit can exist even if the initial form suggests an indeterminate form like 0/0.

Indeterminate Forms

Indeterminate forms often occur in calculus when evaluating limits and can be misleading. Common indeterminate forms include \(\frac{0}{0}\), \(\frac{\infty}{\infty}\), \(0\times\infty\), \(\infty - \infty\), \(0^0\), \(\infty^0\), and \(1^\infty\).

Despite their initial ambiguity, many indeterminate forms can be resolved using algebraic simplification, L'Hôpital's rule, or other methods to find a determinate value. For example, in the solution to step 1, we handle the \(\frac{0}{0}\) form by factoring and simplifying to identify the limit's existence and value.

Continuous Functions

A function is continuous at a point \(x = c\) if the following three conditions are met:

• The function \(f(c)\) is defined. That's to say, there is a point on the graph at x = c.
• The limit \(\lim_{x \rightarrow c} f(x)\) exists, and both the left and right-hand limits match.
• The value of the function at c and the limit as x approaches c are equal, \(f(c) = \lim_{x \rightarrow c} f(x)\).

Even if the function is undefined at a particular point, as shown in step 3, the limit may still exist. Continuous functions provide us with a smooth graph, without any breaks, jumps, or holes.

Trigonometric Limits

Calculating limits involving trigonometric functions is a common task in calculus, and it can feature some unique challenges.

Trigonometric limits, like those in step 4 of our exercise, often require special techniques or trigonometric identities to simplify and evaluate. For instance, recognizing that \(\cot x = \frac{1}{\tan x}\) and knowing the behavior of the tangent function as x approaches \(\frac{\pi}{2}\) allow us to determine that \(\lim_{x \rightarrow \frac{\pi}{2}} \cot x = 0\). Understanding how trigonometric functions behave as they approach certain critical angles is key to solving these limits.