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Chapter 7: Problem 14

\lim _{x \rightarrow 2} \frac{e^{x}-e^{2}}{x-2}

Short Answer

Expert verified
The value of the given limit is \(e^{2}\) or approximately 7.389.

Step by step solution

01

Check if the given limit is in the form of 0/0 or 鈭�/鈭�

We substitute \(x = 2\) into \(\frac{e^{x}-e^{2}}{x-2}\) to check this. The result is \(\frac{e^{2}-e^{2}}{2-2}\), which simplifies to 0/0. Therefore, the given limit is an indeterminate form and we can apply L'H么pital's rule.
02

Apply L'H么pital's rule

L'H么pital's rule states that \(\lim_{x \rightarrow a} \frac{f(x)}{g(x)} = \lim_{x \rightarrow a} \frac{f'(x)}{g'(x)}\), as long as the limit on the right side exists. The derivatives with respect to \(x\) of \(e^{x}\) and \(x-2\) are respectively \(e^{x}\) and 1.
03

Substitute the value of x back to the new limit

We substitute \(x = 2\) into the new limit \(\frac{e^{x}}{1}\). After substituting, we get \(e^{2}\) or approximately 7.389.

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Key Concepts

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

Indeterminate Forms
An indeterminate form arises when evaluating a limit results in a value that is undefined or without a clear meaning. The most common cases include the forms 0/0 and 鈭�/鈭�.

In our exercise, upon substituting 2 into the expression \(\frac{e^{x} - e^{2}}{x-2}\), we get the form \(\frac{0}{0}\). This is deemed an indeterminate form because the numerator and denominator both approach zero, making it impossible to straightforwardly determine the limit.

Indeterminate forms often require techniques like L'H么pital's Rule to resolve and find the actual limit value, avoiding division by zero while still analyzing the behavior of the function near the point of interest.
Differential Calculus
Differential calculus is a fundamental tool in mathematics that deals with the concept of derivatives. A derivative represents the rate of change of a function with respect to a variable.

In the exercise, we used derivatives to apply L'H么pital's Rule. This involved differentiating the numerator and denominator separately.
  • The derivative of \(e^x\) is \(e^x\).
  • The derivative of \(x-2\) is 1.
This simplification allowed us to ascertain the behavior of the original expression as \(x\) approaches 2. By substituting back into the simplified expression, the resulting value provided the limit of the original function.
Exponential Function
The exponential function, denoted by \(e^x\), is a mathematical function based on the constant \(e\), which is approximately 2.718.

The exponential function is ubiquitous in mathematics because of its unique property: its derivative with respect to \(x\) is itself, \(e^x\). This feature makes it particularly useful in differential calculus.

In the context of the exercise, we utilized the exponential function \(e^x\) in both calculating the original and the differentiated forms. Understanding that the rate of growth of \(e^x\) is proportional to its current value is crucial when working with limits and indeterminate forms that involve exponential functions.

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Most popular questions from this chapter

Look back at Example \(7.6 .\) When approximating the slope of \(x^{2}\) at \(x=2\), we end up with the expression \(\left(4 h+h^{2}\right) / h .\) If we assume \(h \neq 0\), then we can cancel the \(h \mathrm{~s}\), arriving at \(4+h\). In this problem, we will investigate a function like \(\left(4 h+h^{2}\right) / h\). Consider the following pay scale for employees at the nepotistic Nelson Nattle Company. Let \(D\) be the date of hire (we use the start-up date of the Nattle Company as our benchmark time, \(D=0, D\) measured in years) and let \(S\) be the starting annual salary. (a) Sketch the graph of \(S(D) .\) The graph looks a little weird until you nd out that Nelson Nattle is the only person hired at \(D=0\) and that his son, Nelson Nattle, Jr., is due back from college exactly one year after the Nattle Company s startup date. The expression \(\frac{(15000+200 D) D(D-1)}{D(D-1)}\) is equal to \(15000+200 D\) as long as \(D \neq 0\) and \(D \neq 1 .\) For \(D=0\) and \(D=1, \frac{(15000+200 D) D(D-1)}{D(D-1)}\) is unde ned. (b) Nelson has just received a letter from his brother Nathaniel asking for a position in the company. Nathaniel s projected date of hire is \(D=1.5 .\) Nelson is thinking of offering his brother a starting salary of \(\$ 40,000\). Adjust \(S(D)\) to de ne \(S(1.5)\) appropriately.

Each of the functions in Problems 5 through 10 is either continuous on \((-\infty, \infty)\) or has a point of discontinuity at some point \((s) x=a .\) Determine any point \((s)\) of discontinuity. Is the point of discontinuity removable? In other words, can the function be made continuous by defining or redefining the function at the point of discontinuity? $$ f(x)=\frac{1}{x^{2}+2} $$

(a) Suppose you are interested in nding \(\lim _{x \rightarrow 2} f(x)\), where the function \(f(x)\) is explicitly given by a formula. What approaches might you take to investigate this limit? (b) Suppose you re now interested in nding \(\lim _{x \rightarrow \infty} f(x)\), where \(f(x)\) again is explicitly given by a formula. What approaches might you take to investigate this limit?

Each of the functions in Problems 5 through 10 is either continuous on \((-\infty, \infty)\) or has a point of discontinuity at some point \((s) x=a .\) Determine any point \((s)\) of discontinuity. Is the point of discontinuity removable? In other words, can the function be made continuous by defining or redefining the function at the point of discontinuity? $$ f(x)=\frac{x^{2}-4}{x+2} $$

\(f(x)=\left\\{\begin{array}{ll}|x|, & x \neq 3 \\ 0, & x=3\end{array}\right.\) (a) \(\lim _{x \rightarrow 0} f(x)\) (b) \(\lim _{x \rightarrow 3} f(x)\) (c) \(\lim _{x \rightarrow-\infty} f(x)\)
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