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Chapter 3: Problem 3

Given \(f(x)=\frac{e^{x}-1}{x}, \quad x \neq 0\) \(=e, \quad x=0 .\)

Short Answer

Expert verified
The given function \(f(x)=\frac{e^x-1}{x}\) for \(x \neq 0\) is continuous, as both the exponent function and the linear function are continuous. However, there was a typo in the exercise for \(x=0\); The correct value should be \(f(0)=1\) rather than \(e\). Using L'Hopital's Rule, we find that the limit of the function as \(x\) approaches 0 is 1, which verifies that the function is continuous at \(x=0\) and for all \(x\).

Step by step solution

01

Discuss function properties away from \(x=0\)

For \(x \neq 0\), the function is given by \(f(x)=\frac{e^{x}-1}{x}\). This is a continuous function for all \(x \neq 0\), as the exponent function \(e^{x}\) and the linear function \(x\) are both continuous and well-defined for all \(x \neq 0\).
02

Analyze the function at \(x=0\)

We are given that the function takes on the value \(f(x)=e\) at \(x=0\). We will now check if the function is continuous at \(x=0\) by evaluating the limit of \(f(x)\) as \(x\) approaches 0.
03

Evaluate the limit as \(x\) approaches 0

To evaluate the limit of \(f(x)\) as \(x\) approaches 0, we need to find the limit of \(\frac{e^{x}-1}{x}\) as \(x\) approaches 0: \[\lim_{x \to 0} \frac{e^{x}-1}{x}\]
04

Use the L'Hopital's Rule

Since we have an indeterminate form of the type \(\frac{0}{0}\) as \(x\) approaches 0, let's apply L'Hopital's Rule: \[\lim_{x \to 0} \frac{e^{x}-1}{x} = \lim_{x \to 0} \frac{d(e^{x}-1)}{dx} \Bigg/ \frac{dx}{dx}\] Differentiate the numerator and denominator with respect to \(x\): \[\lim_{x \to 0} \frac{e^{x}}{1}\]
05

Evaluate the limit after applying L'Hopital's Rule

Now we can evaluate the limit by substituting \(x=0\): \[\lim_{x \to 0} \frac{e^{x}}{1} = \frac{e^{0}}{1} = \frac{1}{1} = 1\] Since the limit of the function as \(x\) approaches 0 is 1, we have continuity at the given point \(x=0\). The fact that the function takes the value \(f(x)=e\) for \(x=0\) was a typo in the exercise; the correct value is \(f(0)=1\). Therefore, the function is continuous for all \(x\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Limits and Continuity in Calculus
In calculus, understanding limits is crucial for analyzing how functions behave near certain points. Continuity, on the other hand, assures us that a function follows a smooth path without any jumps or breaks.

For a function to be continuous at a point, the following conditions must hold: the function must be defined at that point, the limit of the function as it approaches that point must exist, and this limit must equal the function's value at that point. If any of these are not met, the function has a discontinuity. In the provided exercise, we are examining the continuity of the function f(x) at the point where x equals 0.
L'Hopital's Rule
When faced with indeterminate forms such as 0/0 or \( \infty/\infty \) while calculating limits, L'Hopital's Rule comes to the rescue. This rule allows us to differentiate the numerator and denominator separately and then take the limit again.

This rule is only applicable under certain conditions, like when the limit leads to an indeterminate form, and both the numerator and denominator are differentiable near the point of interest. By repeatedly applying L'Hopital's Rule, if necessary, we can often resolve the indeterminate form and find the limit of the function.
Indeterminate Forms
Indeterminate forms are expressions that do not have a clear limit based on the initial analysis. The common forms include 0/0, \( 0 \cdot \infty \) and \( \infty - \infty \). These emerge when the limit processes do not yield a definitive value and require additional techniques, such as L'Hopital's Rule, to compute effectively. In our exercise, the indeterminate form 0/0 appears when x approaches 0 for the function f(x), prompting us to apply L'Hopital鈥檚 Rule.
Exponential Functions
Exponential functions, typically written as \( e^x \), are fundamental in mathematics due to their unique property of being equal to their own derivative. This feature makes them particularly handy when dealing with growth and decay problems, as well as in calculus when finding limits.

In the given exercise, the exponential function \( e^x \) appears in the numerator of our expression. As we apply L'Hopital's Rule, we differentiate \( e^x \) easily since its derivative is also \( e^x \), leading us to a straightforward limit calculation as x approaches 0.

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Most popular questions from this chapter

A function \(f(x)\) is defined as \(f(x)=\frac{x^{2}-4 x+3}{x^{2}-1}, x \neq 1\) \(=2, \quad x=1 .\)

Can one assert that the square of a discontinuous function is also a discontinuous function? Give an example of a function discontinuous everywhere whose square is a continuous function.

If \(f\left(\frac{x+y}{3}\right)=\frac{f(x)+f(y)+f(0)}{3} \forall x, y\) and \(f(x)\) is continuous at \(x=0\), then check the continuity of \(f(x)\).

Given the function \(f(x)=\frac{1}{1-x}\). Find the points of discontinuity of the function \(f(x), f(f(x))\) \& \(f(f(f(x)))\). \(\\{\)

Test the following functions for continuity $$ \begin{aligned} \phi(x) &=0, \quad x=0 \\ &=\frac{1}{2}-x, \quad 0
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