Understanding Continuity of Piecewise Functions

This question asks us to analyze the continuity of a given piecewise real-valued function \(f(x)\) at specific points, x = 0 and x = 1.

The function is defined as:

\(f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {\frac{{{e^x} - 1}}{x},{\rm{\;\;x}} > 0}\\ {0,{\rm{\;\;\;x}} = 0} \end{array}} \right.\)

Let's evaluate the statements provided.

Analyzing Statement 1: Right Continuity at x = 0

Statement 1 says that \(f(x)\) is right continuous at \(x = 0\).

For a function to be right continuous at a point \(a\), the following condition must be met:

\(\lim_{x \to a^+} f(x) = f(a)\)

In this case, the point is \(a = 0\). So, we need to check if \(\lim_{x \to 0^+} f(x) = f(0)\).

First, let's find the value of the function at \(x = 0\). From the definition, when \(x = 0\), \(f(x) = 0\). So, \(f(0) = 0\).

Next, let's find the right-hand limit as \(x\) approaches 0. For \(x > 0\), the function is defined as \(f(x) = \frac{e^x - 1}{x}\). So, the right-hand limit is:

\(\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{e^x - 1}{x}\)

This is a standard limit of the form \(\frac{0}{0}\) as \(x \to 0\). We know that the limit \(\lim_{x \to 0} \frac{e^x - 1}{x} = 1\). Since the limit exists and is the same whether \(x\) approaches from the right or the left (excluding \(x=0\) itself for the function's structure), the right-hand limit is also 1.

\(\lim_{x \to 0^+} \frac{e^x - 1}{x} = 1\)

Now, let's compare the right-hand limit and the function value at \(x=0\):

\(\lim_{x \to 0^+} f(x) = 1\)

\(f(0) = 0\)

Since \(\lim_{x \to 0^+} f(x) \neq f(0)\) (because \(1 \neq 0\)), the function \(f(x)\) is not right continuous at \(x = 0\).

Therefore, Statement 1 is incorrect.

Analyzing Statement 2: Discontinuity at x = 1

Statement 2 says that \(f(x)\) is discontinuous at \(x = 1\).

For continuity at a point \(a\), the function must be defined at \(a\), the limit as \(x\) approaches \(a\) must exist, and the limit must be equal to the function value at \(a\).

In this case, the point is \(a = 1\). We need to check the continuity of \(f(x)\) at \(x = 1\).

From the definition of the function, for \(x > 0\), \(f(x) = \frac{e^x - 1}{x}\).

Since \(1 > 0\), we use this part of the definition to evaluate the continuity at \(x = 1\).

First, find the function value at \(x = 1\):

\(f(1) = \frac{e^1 - 1}{1} = e - 1\)

The function is defined at \(x = 1\).

Next, find the limit as \(x\) approaches 1. Since the function is given by a single expression \(\frac{e^x - 1}{x}\) for all \(x > 0\) (which includes values around \(x=1\)), the limit can be found by direct substitution, as long as the denominator is not zero at the limit point.

\(\lim_{x \to 1} f(x) = \lim_{x \to 1} \frac{e^x - 1}{x}\)

As \(x \to 1\), the numerator \(e^x - 1 \to e^1 - 1 = e - 1\) and the denominator \(x \to 1\). The denominator is not zero at \(x=1\).

So, \(\lim_{x \to 1} f(x) = \frac{e^1 - 1}{1} = e - 1\)

Now, let's compare the limit and the function value at \(x=1\):

\(\lim_{x \to 1} f(x) = e - 1\)

\(f(1) = e - 1\)

Since \(\lim_{x \to 1} f(x) = f(1)\), the function \(f(x)\) is continuous at \(x = 1\).

Therefore, Statement 2, which says the function is discontinuous at \(x = 1\), is incorrect.

Conclusion

Based on our analysis:

• Statement 1: \(f(x)\) is right continuous at x = 0. This is incorrect.
• Statement 2: \(f(x)\) is discontinuous at x = 1. This is incorrect.

Since neither of the statements is correct, the answer is "Neither 1 nor 2".

Revision Table: Continuity Analysis

Additional Information: Types of Continuity and Limits

Understanding different types of continuity and how to evaluate limits for piecewise functions is crucial.

• Continuity at a Point: A function \(f(x)\) is continuous at \(x=a\) if \(\lim_{x \to a} f(x) = f(a)\). This requires the limit from the left, the limit from the right, and the function value to all be equal.
• Right Continuity: A function \(f(x)\) is right continuous at \(x=a\) if \(\lim_{x \to a^+} f(x) = f(a)\).
• Left Continuity: A function \(f(x)\) is left continuous at \(x=a\) if \(\lim_{x \to a^-} f(x) = f(a)\).
• Evaluating Limits for Piecewise Functions: When evaluating a limit as \(x\) approaches a point where the function definition changes (like \(x=0\) in this case), you must consider one-sided limits (\(\lim_{x \to a^+}\) and \(\lim_{x \to a^-}\)). If the limit point is within an interval where the function is defined by a single expression (like \(x=1\) in the interval \(x > 0\)), the limit is usually found by direct substitution, provided the expression is continuous at that point.
• Standard Limits: Recognizing and applying standard limits like \(\lim_{x \to 0} \frac{e^x - 1}{x} = 1\) is very helpful. Another related standard limit is \(\lim_{x \to 0} \frac{\sin x}{x} = 1\).

In this problem, the potential point of discontinuity was \(x=0\) because the function definition changes there. For points where the function is defined by a single expression that is generally continuous (like rational functions or exponential functions), continuity should hold unless the denominator is zero (for rational functions).