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Use continuity to evaluate the limit. $$\lim _{x \rightarrow 1} e^{x^{2}-x}$$

Short Answer

Expert verified
The limit is 1.

Step by step solution

01

Identify the Continuity of the Function

The first step is to determine whether the function \( e^{x^2 - x} \) is continuous at \( x = 1 \). The exponential function \( e^u \) is continuous for all real numbers \( u \), and since polynomials like \( x^2 - x \) are also continuous everywhere, their composition is also continuous.
02

Substitute the Limit Value into the Function

Since \( e^{x^2 - x} \) is continuous at \( x = 1 \), we can evaluate the limit by direct substitution. Substitute \( x = 1 \) into the expression \( x^2 - x \).
03

Evaluate the Expression

Calculate \( 1^2 - 1 = 0 \). This means we can substitute back into the exponential function: \( e^{0} = 1 \).
04

Final Evaluation

Since we have replaced the expression with its value at the continuous point, the limit simplifies to \( e^{0} \), which equals 1.

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Key Concepts

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

Continuous Functions
Continuous functions are a core concept in calculus and mathematics, representing functions that do not have breaks, jumps, or holes in their graphs. If you can draw the graph of a function without lifting your pencil from the paper, the function is likely continuous. Essentially, a function \( f(x) \) is continuous at a point \( x = a \) if the limit of \( f(x) \) as \( x \) approaches \( a \) is equal to the function's value at \( a \), or formally:
  • \( \lim_{x \to a} f(x) = f(a) \)
This property allows us to evaluate limits simply by substitution, as seen in the context of the original exercise. Many functions, like polynomials or exponential functions, are continuous over their entire domain. Appreciating this concept is vital, as it enables straightforward computation in calculus without additional algebraic manipulation.
Exponential Function
The exponential function, denoted as \( e^x \), is one of the most important functions in mathematics because of its remarkable properties and applications in various fields. This function shows rapid growth or decay, depending on whether the exponent is positive or negative.
An exponential function is always continuous and differentiable, which means you can evaluate limits involving exponential functions efficiently by substitution. For any real number \( u \), \( e^u \) is well-defined, real, and continuous at every point.
  • For example, \( e^{x^2-x} \) in our exercise includes a polynomial as the exponent, ensuring continuity across all \( x \)-values.
  • Exponential growth is critical in modeling real-world phenomena like population growth, investment growth, and radioactive decay.
Recognizing the continuity of \( e^x \) greatly simplifies the process of finding limits.
Polynomials
Polynomials are algebraic expressions consisting of variables raised to non-negative integer powers and coefficients. They form a fundamental class of functions, which are particularly easy to work with due to their simplicity and predictable behavior. A polynomial like \( x^2 - x \) has the general form:
  • \( a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 \)
These functions are continuous everywhere for all real numbers, meaning they have plotted graphs without any breaks. This universal continuity is because they can be expressed as sums of continuous functions themselves.
The polynomial \( x^2 - x \) in the original problem serves as the exponent in the exponential function. This characteristic makes it easy to evaluate at any point, such as substituting \( x = 1 \) directly without worrying about undefined expressions.
Understanding polynomials' continuity and simplicity allows for effective problem-solving in calculus, as it helps streamline evaluating limits, derivatives, and integrals.

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Most popular questions from this chapter

Let \(f(x)=x^{2}.\) (a) Estimate the values of \(f^{\prime}(0), f^{\prime}\left(\frac{1}{2}\right), f^{\prime}(1),\) and \(f^{\prime} 2)\) by using a graphing device to zoom in on the graph of f. (b) Use symmetry to deduce the values of \(f^{\prime}\left(-\frac{1}{2}\right), f^{\prime}(-1)\) and \(f^{\prime}(-2).\) (c) Use the results from parts (a) and (b) to guess a formula for \(f^{\prime}(x)\). (d) Use the definition of a derivative to prove that your guess in part (c) is correct.

Find the derivative of the function using the definition of derivative. State the domain of the function and the domain of its derivative. \(f(x)=1.5 x^{2}-x+3.7\)

Let \(P(t)\) be the percentage of Americans under the age of 18 at time \(t .\) The table gives values of this function in census years from 1950 to \(2000.\) $$\begin{array}{|c|c|c|c|}\hline t & {P(t)} & {t} & {P(t)} \\ \hline 1950 & {31.1} & {1980} & {28.0} \\ {1960} & {35.7} & {1990} & {25.7} \\ {1970} & {34.0} & {2000} & {25.7} \\ \hline\end{array}$$ (a) What is the meaning of \(\mathrm{P}^{\prime}(\mathrm{t}) ?\) What are its units? (b) Construct a table of estimated values for \(\mathrm{P}^{\prime}(\mathrm{t}) .\) (c) Graph \(\mathrm{P}\) and \(\mathrm{P}^{\prime}\) (d) How would it be possible to get more accurate values for \(\mathrm{P}^{\prime}(\mathrm{t}) ?\)

(a) Use numerical and graphical evidence to guess the value of the limit $$\lim _{x \rightarrow 1} \frac{x^{3}-1}{\sqrt{x}-1}$$ (b) How close to 1 does \(x\) have to be to ensure that the function in part (a) is within a distance 0.5 of its limit?

$$ \begin{array}{l}{\text { For the limit }} \\ {\quad \lim _{x \rightarrow-\infty} \frac{\sqrt{4 x^{2}+1}}{x+1}=-2} \\ {\text { illustrate Dcfinition } 8 \text { by finding values of } N \text { that correspond }} \\\ {\text { to } \varepsilon=0.5 \text { and } \varepsilon=0.1}\end{array} $$

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