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The Precise Definition of a Limit
Describe the epsilon-delta definition of a limit.
Quantifying Closeness
Before stating the formal definition of a limit, we must introduce a few preliminary ideas. Recall that the distance between two points a and b on a number line is given by \(|a-b|.\)
- The statement \(|f(x)-L|<\epsilon\) may be interpreted as: The distance between \(f(x)\) and L is less than ε.
- The statement \(0<|x-a|<\delta\) may be interpreted as: \(x\ne a\) and the distance between x and a is less than δ.
It is also important to look at the following equivalences for absolute value:
- The statement \(|f(x)-L|<\epsilon\) is equivalent to the statement L-\epsilon
- The statement \(0<|x-a|<\delta\) is equivalent to the statement a-\delta
- The statement \(0<|x-a|<\delta\) is equivalent to the statement a-\delta
With these clarifications, we can state the formal epsilon-delta definition of the limit.
This definition may seem rather complex from a mathematical point of view, but it becomes easier to understand if we break it down phrase by phrase. The statement itself involves something called a universal quantifier (for every \(\epsilon >0\text{),}\) an existential quantifier (there exists a \(\delta >0\text{),}\) and, last, a conditional statement (if \(0<|x-a|<\delta ,\) then \(|f(x)-L|<\epsilon \text{).}\) Let’s take a look at , which breaks down the definition and translates each part.
| Definition | Translation |
| 1. For every \(\epsilon >0,\) | 1. For every positive distance ε from L, |
| 2. there exists a \(\delta >0,\) | 2. There is a positive distance \(\delta\) from a, |
| 3. such that | 3. such that |
| 4. if \(0<|x-a|<\delta ,\) then \(|f(x)-L|<\epsilon .\) | 4. if x is closer than \(\delta\) to a and \(x\ne a,\) then \(f(x)\) is closer than ε to L. |
We can get a better handle on this definition by looking at the definition geometrically. shows possible values of \(\delta\) for various choices of \(\epsilon >0\) for a given function \(f(x),\) a number a, and a limit L at a. Notice that as we choose smaller values of ε (the distance between the function and the limit), we can always find a \(\delta\) small enough so that if we have chosen an x value within \(\delta\) of a, then the value of \(f(x)\) is within ε of the limit L.
shows how you can use this definition to prove a statement about the limit of a specific function at a specified value.
The following Problem-Solving Strategy summarizes the type of proof we worked out in .
Condensed: the full section is in OpenStax Calculus Volume 1.
Proving Limit Laws
We now demonstrate how to use the epsilon-delta definition of a limit to construct a rigorous proof of one of the limit laws. The triangle inequality is used at a key point of the proof, so we first review this key property of absolute value.
We prove the following limit law: If \(\underset{x\to a}{\text{lim}}f(x)=L\) and \(\underset{x\to a}{\text{lim}}g(x)=M,\) then \(\underset{x\to a}{\text{lim}}(f(x)+g(x))=L+M.\)
Let \(\epsilon >0.\)
Choose \({\delta }_{1}>0\) so that if \(0<|x-a|<{\delta }_{1},\) then \(|f(x)-L|<\epsilon \text{/}2.\)
Choose \({\delta }_{2}>0\) so that if \(0<|x-a|<{\delta }_{2},\) then \(|g(x)-M|<\epsilon \text{/}2.\)
Choose \(\delta =\text{min}\{{\delta }_{1},{\delta }_{2}\}.\)
Assume \(0<|x-a|<\delta .\)
Thus,
\[0<|x-a|<{\delta }_{1}\ \text{and}\ 0<|x-a|<{\delta }_{2}.\]Hence,
\[\begin{array}{ll}|(f(x)+g(x))-(L+M)| & =|(f(x)-L)+(g(x)-M)| \\ & \le |f(x)-L|+|g(x)-M| \\ & <\frac{\epsilon }{2}+\frac{\epsilon }{2}=\epsilon .\end{array}\]□
We now explore what it means for a limit not to exist. The limit \(\underset{x\to a}{\text{lim}}f(x)\) does not exist if there is no real number L for which \(\underset{x\to a}{\text{lim}}f(x)=L.\) Thus, for all real numbers L, \(\underset{x\to a}{\text{lim}}f(x)\ne L.\) To understand what this means, we look at each part of the definition of \(\underset{x\to a}{\text{lim}}f(x)=L\) together with its opposite. A translation of the definition is
| Definition | Opposite |
| 1. For every \(\epsilon >0,\) | 1. There exists \(\epsilon >0\) so that |
| 2. there exists a \(\delta >0,\) so that | 2. for every \(\delta >0,\) |
| 3. if \(0<|x-a|<\delta ,\) then \(|f(x)-L|<\epsilon .\) | 3. There is an x satisfying \(0<|x-a|<\delta\) so that \(|f(x)-L|\ge \epsilon .\) |
Finally, we may state what it means for a limit not to exist. The limit \(\underset{x\to a}{\text{lim}}f(x)\) does not exist if for every real number L, there exists a real number \(\epsilon >0\) so that for all \(\delta >0,\) there is an x satisfying \(0<|x-a|<\delta ,\) so that \(|f(x)-L|\ge \epsilon .\) Let’s apply this in to show that a limit does not exist.
Example
Try it.
Show that \(\underset{x\to 0}{\text{lim}}\frac{|x|}{x}\) does not exist. The graph of \(f(x)=|x|\text{/}x\) is shown here:
Solution
Suppose that L is a candidate for a limit. Choose \(\epsilon =1\text{/}2.\)
Let \(\delta >0.\) Either \(L\ge 0\) or \(L<0.\) If \(L\ge 0,\) then let \(x=-\delta \text{/}2.\) Thus,
\[|x-0|=|-\frac{\delta }{2}-0|=\frac{\delta }{2}<\delta\]and
\[|\frac{|-\frac{\delta }{2}|}{-\frac{\delta }{2}}-L|=|-1-L|=L+1\ge 1>\frac{1}{2}=\epsilon .\]On the other hand, if \(L<0,\) then let \(x=\delta \text{/}2.\) Thus,
\[|x-0|=|\frac{\delta }{2}-0|=\frac{\delta }{2}<\delta\]and
\[|\frac{|\frac{\delta }{2}|}{\frac{\delta }{2}}-L|=|1-L|=|L|+1\ge 1>\frac{1}{2}=\epsilon .\]Thus, for any value of L, \(\underset{x\to 0}{\text{lim}}\frac{|x|}{x}\ne L.\)
One-Sided and Infinite Limits
Just as we first gained an intuitive understanding of limits and then moved on to a more rigorous definition of a limit, we now revisit one-sided limits. To do this, we modify the epsilon-delta definition of a limit to give formal epsilon-delta definitions for limits from the right and left at a point. These definitions only require slight modifications from the definition of the limit. In the definition of the limit from the right, the inequality 0
Example
Try it.
Prove that \(\underset{x\to {4}^{+}}{\text{lim}}\sqrt{x-4}=0.\)
Solution
Let \(\epsilon >0.\)
Choose \(\delta ={\epsilon }^{2}.\) Since we ultimately want \(|\sqrt{x-4}-0|<\epsilon ,\) we manipulate this inequality to get \(\sqrt{x-4}<\epsilon\) or, equivalently, 0 Assume 0 Therefore, \(\underset{x\to {4}^{+}}{\text{lim}}\sqrt{x-4}=0.\)
We conclude the process of converting our intuitive ideas of various types of limits to rigorous formal definitions by pursuing a formal definition of infinite limits. To have \(\underset{x\to a}{\text{lim}}f(x)=+\infty ,\) we want the values of the function \(f(x)\) to get larger and larger as x approaches a. Instead of the requirement that \(|f(x)-L|<\epsilon\) for arbitrarily small ε when \(0<|x-a|<\delta\) for small enough \(\delta ,\) we want \(f(x)>M\) for arbitrarily large positive M when \(0<|x-a|<\delta\) for small enough \(\delta .\) illustrates this idea by showing the value of \(\delta\) for successively larger values of M.
Key Concepts
- The intuitive notion of a limit may be converted into a rigorous mathematical definition known as the epsilon-delta definition of the limit.
- The epsilon-delta definition may be used to prove statements about limits.
- The epsilon-delta definition of a limit may be modified to define one-sided limits.
The Precise Definition of a Limit
In the following exercises, write the appropriate \(\epsilon\)-\(\delta\) definition for each of the given statements.
The following graph of the function f satisfies \(\underset{x\to 2}{\text{lim}}f(x)=2.\) In the following exercises, determine a value of \(\delta >0\) that satisfies each statement.
The following graph of the function f satisfies \(\underset{x\to 3}{\text{lim}}f(x)=-1.\) In the following exercises, determine a value of \(\delta >0\) that satisfies each statement.
The following graph of the function f satisfies \(\underset{x\to 3}{\text{lim}}f(x)=2.\) In the following exercises, for each value of ε, find a value of \(\delta >0\) such that the precise definition of limit holds true.
[T] In the following exercises, use a graphing calculator to find a number \(\delta\) such that the statements hold true.
In the following exercises, use the precise definition of limit to prove the given limits.
In the following exercises, use the precise definition of limit to prove the given one-sided limits.
Condensed: the full section is in OpenStax Calculus Volume 1.
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实践 (40)
先在纸上试一个。 将答案显示为检查; 校验的答案可以在求解器中打开, 每一步都可用 。
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Prove that \(\underset{x\to 1}{\text{lim}}(2x+1)=3.\)
发送答案
Let \(\epsilon >0.\)
The first part of the definition begins “For every \(\epsilon >0\text{.”}\) This means we must prove that whatever follows is true no matter what positive value of ε is chosen. By stating “Let \(\epsilon >0\text{,”}\) we signal our intent to do so.
Choose \(\delta =\frac{\epsilon }{2}.\)
The definition continues with “there exists a \(\delta >0.\)” The phrase “there exists” in a mathematical statement is always a signal for a scavenger hunt. In other words, we must go and find \(\delta .\) So, where exactly did \(\delta =\epsilon \text{/}2\) come from? There are two basic approaches to tracking down \(\delta .\) One method is purely algebraic and the other is geometric.
We begin by tackling the problem from an algebraic point of view. Since ultimately we want \(|(2x+1)-3|<\epsilon ,\) we begin by manipulating this expression: \(|(2x+1)-3|<\epsilon\) is equivalent to \(|2x-2|<\epsilon ,\) which in turn is equivalent to \(|2||x-1|<\epsilon .\) Last, this is equivalent to \(|x-1|<\epsilon \text{/}2.\) Thus, it would seem that \(\delta =\epsilon \text{/}2\) is appropriate.
We may also find \(\delta\) through geometric methods. demonstrates how this is done.
Assume \(0<|x-1|<\delta .\) When \(\delta\) has been chosen, our goal is to show that if \(0<|x-1|<\delta ,\) then \(|(2x+1)-3|<\epsilon .\) To prove any statement of the form “If this, then that,” we begin by assuming “this” and trying to get “that.”
Thus,
\(\begin{array}{lllll}|(2x+1)-3| & =|2x-2| & & & \text{property of absolute value} \\ & =|2(x-1)| \\ & =|2||x-1| & & & |2|=2 \\ & =2|x-1| & & & \\ & <2\cdot \delta & & & \text{here’s where we use the assumption that}\ 0<|x-1|<\delta \\ & =2\cdot \frac{\epsilon }{2}=\epsilon & & & \text{here’s where we use our choice of}\ \delta =\epsilon \text{/}2\end{array}\)
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Complete the proof that \(\underset{x\to -1}{\text{lim}}(4x+1)=-3\) by filling in the blanks.
Let _____.
Choose \(\delta =\text{\underline{\hspace{1.5em}}.}\)
Assume \(0<|x-\underline{\hspace{1.5em}}|<\delta .\)
Thus, \(|\text{\underline{\hspace{1.5em}}}-\text{\underline{\hspace{1.5em}}}|=\underline{\hspace{1.5em}}\epsilon .\)
发送答案
We begin by filling in the blanks where the choices are specified by the definition. Thus, we have
Let \(\epsilon >0.\)
Choose \(\delta =\text{\underline{\hspace{1.5em}}.}\)
Assume \(0<|x-(-1)|<\delta .\) (or equivalently, \(0<|x+1|<\delta \text{.)}\)
Thus, \(|(4x+1)-(-3)|=|4x+4|=|4||x+1|<4\delta \text{\underline{\hspace{1.5em}}}\epsilon .\)
Focusing on the final line of the proof, we see that we should choose \(\delta =\frac{\epsilon }{4}.\)
We now complete the final write-up of the proof:
Let \(\epsilon >0.\)
Choose \(\delta =\frac{\epsilon }{4}.\)
Assume \(0<|x-(-1)|<\delta\) (or equivalently, \(0<|x+1|<\delta \text{.)}\)
Thus, \(|(4x+1)-(-3)|=|4x+4|=|4||x+1|<4\delta =4(\epsilon \text{/}4)=\epsilon .\)
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Complete the proof that \(\underset{x\to 2}{\text{lim}}(3x-2)=4\) by filling in the blanks.
Let _______.
Choose \(\delta =\text{\underline{\hspace{1.5em}}}.\)
Assume \(0<|x-\text{\underline{\hspace{1.5em}}}|<\text{\underline{\hspace{1.5em}}}.\)
Thus,
\(|\text{\underline{\hspace{1.5em}}}-\text{\underline{\hspace{1.5em}}}|=\text{\underline{\hspace{1.5em}}}\epsilon .\)
Therefore, \(\underset{x\to 2}{\text{lim}}(3x-2)=4.\)
发送答案
Let \(\epsilon >0;\) choose \(\delta =\frac{\epsilon }{3};\) assume \(0<|x-2|<\delta .\)
Thus, \(|(3x-2)-4|=|3x-6|=|3|\cdot |x-2|<3\cdot \delta =3\cdot (\epsilon \text{/}3)=\epsilon .\)
Therefore, \(\underset{x\to 2}{\text{lim}}3x-2=4.\)
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Prove that \(\underset{x\to 2}{\text{lim}}{x}^{2}=4.\)
发送答案
- Let \(\epsilon >0.\) The first part of the definition begins “For every \(\epsilon >0\text{,”}\) so we must prove that whatever follows is true no matter what positive value of ε is chosen. By stating “Let \(\epsilon >0\text{,”}\) we signal our intent to do so.
- Without loss of generality, assume \(\epsilon \le 4.\) Two questions present themselves: Why do we want \(\epsilon \le 4\) and why is it okay to make this assumption? In answer to the first question: Later on, in the process of solving for \(\delta ,\) we will discover that \(\delta\) involves the quantity \(\sqrt{4-\epsilon }.\) Consequently, we need \(\epsilon \le 4.\) In answer to the second question: If we can find \(\delta >0\) that “works” for \(\epsilon \le 4,\) then it will “work” for any \(\epsilon >4\) as well. Keep in mind that, although it is always okay to put an upper bound on ε, it is never okay to put a lower bound (other than zero) on ε.
- Choose \(\delta =\text{min}\{2-\sqrt{4-\epsilon },\sqrt{4+\epsilon }-2\}.\) shows how we made this choice of \(\delta .\)
- We must show: If \(0<|x-2|<\delta ,\) then \(|{x}^{2}-4|<\epsilon ,\) so we must begin by assuming
\[0<|x-2|<\delta .\]
We don’t really need \(0<|x-2|\) (in other words, \(x\ne 2)\) for this proof. Since \(0<|x-2|<\delta ⇒|x-2|<\delta ,\) it is okay to drop \(0<|x-2|.\)
\[|x-2|<\delta .\]
Hence,
\[-\delta
\[-(2-\sqrt{4-\epsilon })\le -\delta
\[-2+\sqrt{4-\epsilon }
\[\sqrt{4-\epsilon }
\[4-\epsilon <{x}^{2}<4+\epsilon .\]
We subtract 4 from all parts of the inequality:
\[-\epsilon <{x}^{2}-4<\epsilon .\]
Last,
\[|{x}^{2}-4|<\epsilon .\] - Therefore,
\[\underset{x\to 2}{\text{lim}}{x}^{2}=4.\]
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Find δ corresponding to \(\epsilon >0\) for a proof that \(\underset{x\to 9}{\text{lim}}\sqrt{x}=3.\)
发送答案
Choose \(\delta =\text{min}\{9-{(3-\epsilon )}^{2},{(3+\epsilon )}^{2}-9\}.\)
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Prove that \(\underset{x\to -1}{\text{lim}}({x}^{2}-2x+3)=6.\)
发送答案
Let’s use our outline from the Problem-Solving Strategy:
- Let \(\epsilon >0.\)
- Choose \(\delta =\text{min}\{1,\epsilon \text{/}5\}.\) This choice of \(\delta\) may appear odd at first glance, but it was obtained by taking a look at our ultimate desired inequality: \(|({x}^{2}-2x+3)-6|<\epsilon .\) This inequality is equivalent to \(|x+1|\cdot |x-3|<\epsilon .\) At this point, the temptation simply to choose \(\delta =\frac{\epsilon }{x-3}\) is very strong. Unfortunately, our choice of \(\delta\) must depend on ε only and no other variable. If we can replace \(|x-3|\) by a numerical value, our problem can be resolved. This is the place where assuming \(\delta \le 1\) comes into play. The choice of \(\delta \le 1\) here is arbitrary. We could have just as easily used any other positive number. In some proofs, greater care in this choice may be necessary. Now, since \(\delta \le 1\) and \(|x+1|<\delta \le 1,\) we are able to show that \(|x-3|<5.\) Consequently, \(|x+1|\cdot |x-3|<|x+1|\cdot 5.\) At this point we realize that we also need \(\delta \le \epsilon \text{/}5.\) Thus, we choose \(\delta =\text{min}\{1,\epsilon \text{/}5\}.\)
- Assume \(0<|x+1|<\delta .\) Thus,
\[|x+1|<1\ \text{and}\ |x+1|<\frac{\epsilon }{5}.\]
Since \(|x+1|<1,\) we may conclude that -1
Therefore,
\[\underset{x\to -1}{\text{lim}}({x}^{2}-2x+3)=6.\]
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Complete the proof that \(\underset{x\to 1}{\text{lim}}{x}^{2}=1.\)
Let \(\epsilon >0;\) choose \(\delta =\text{min}\{1,\epsilon \text{/}3\};\) assume \(0<|x-1|<\delta .\)
Since \(|x-1|<1,\) we may conclude that -1
发送答案
\(|{x}^{2}-1|=|x-1|\cdot |x+1|<\epsilon \text{/}3\cdot 3=\epsilon\)
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Show that \(\underset{x\to 0}{\text{lim}}\frac{|x|}{x}\) does not exist. The graph of \(f(x)=|x|\text{/}x\) is shown here:
发送答案
Suppose that L is a candidate for a limit. Choose \(\epsilon =1\text{/}2.\)
Let \(\delta >0.\) Either \(L\ge 0\) or \(L<0.\) If \(L\ge 0,\) then let \(x=-\delta \text{/}2.\) Thus,
\[|x-0|=|-\frac{\delta }{2}-0|=\frac{\delta }{2}<\delta\]and
\[|\frac{|-\frac{\delta }{2}|}{-\frac{\delta }{2}}-L|=|-1-L|=L+1\ge 1>\frac{1}{2}=\epsilon .\]On the other hand, if \(L<0,\) then let \(x=\delta \text{/}2.\) Thus,
\[|x-0|=|\frac{\delta }{2}-0|=\frac{\delta }{2}<\delta\]and
\[|\frac{|\frac{\delta }{2}|}{\frac{\delta }{2}}-L|=|1-L|=|L|+1\ge 1>\frac{1}{2}=\epsilon .\]Thus, for any value of L, \(\underset{x\to 0}{\text{lim}}\frac{|x|}{x}\ne L.\)
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Prove that \(\underset{x\to {4}^{+}}{\text{lim}}\sqrt{x-4}=0.\)
发送答案
Let \(\epsilon >0.\)
Choose \(\delta ={\epsilon }^{2}.\) Since we ultimately want \(|\sqrt{x-4}-0|<\epsilon ,\) we manipulate this inequality to get \(\sqrt{x-4}<\epsilon\) or, equivalently, 0
Assume 0
Therefore, \(\underset{x\to {4}^{+}}{\text{lim}}\sqrt{x-4}=0.\)
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Find \(\delta\) corresponding to ε for a proof that \(\underset{x\to {1}^{-}}{\text{lim}}\sqrt{1-x}=0.\)
发送答案
\(\delta ={\epsilon }^{2}\)
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\(\underset{x\to a}{\text{lim}}f(x)=N\)
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\(\underset{t\to b}{\text{lim}}g(t)=M\)
发送答案
For every \(\epsilon >0,\) there exists a \(\delta >0,\) so that if \(0<|t-b|<\delta ,\) then \(|g(t)-M|<\epsilon\)
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\(\underset{x\to c}{\text{lim}}h(x)=L\)
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\(\underset{x\to a}{\text{lim}}\phi (x)=A\)
发送答案
For every \(\epsilon >0,\) there exists a \(\delta >0,\) so that if \(0<|x-a|<\delta ,\) then \(|\phi (x)-A|<\epsilon\)
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\(\epsilon =1.5\)
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\(\epsilon =3\)
发送答案
\(\delta \le 1\)
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\(|\text{sin}\ (2x)-\frac{1}{2}|<0.1,\) whenever \(|x-\frac{\pi }{12}|<\delta\)
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\(|\sqrt{x-4}-2|<0.1,\ \text{whenever}\ |x-8|<\delta\)
发送答案
\(\delta <0.3900\)
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\(\underset{x\to 2}{\text{lim}}(5x+8)=18\)
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\(\underset{x\to 3}{\text{lim}}\frac{{x}^{2}-9}{x-3}=6\)
发送答案
Let \(\delta =\epsilon .\) If \(0<|x-3|<\epsilon ,\) then \(|x+3-6|=|x-3|<\epsilon .\)
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\(\underset{x\to 2}{\text{lim}}\frac{2{x}^{2}-3x-2}{x-2}=5\)
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\(\underset{x\to 0}{\text{lim}}{x}^{4}=0\)
发送答案
Let \(\delta =\sqrt[4]{\epsilon }.\) If \(0<|x|<\sqrt[4]{\epsilon },\) then \(|{x}^{4}|={x}^{4}<\epsilon .\)
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\(\underset{x\to 2}{\text{lim}}({x}^{2}+2x)=8\)
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\(\underset{x\to {5}^{-}}{\text{lim}}\sqrt{5-x}=0\)
发送答案
Let \(\delta ={\epsilon }^{2}.\) If 5-{\epsilon }^{2}
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\(\underset{x\to {0}^{+}}{\text{lim}}f(x)=-2,\ \text{where}\ f(x)=\{\begin{array}{l}8x-3,\ \text{if}\ x<0 \\ 4x-2,\ \text{if}\ x\ge 0\end{array}.\)
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\(\underset{x\to {1}^{-}}{\text{lim}}f(x)=3,\ \text{where}\ f(x)=\{\begin{array}{l}5x-2,\ \text{if}\ x<1 \\ 7x-1,\ \text{if}\ x\ge 1\end{array}.\)
发送答案
Let \(\delta =\epsilon \text{/}5.\) If 1-\epsilon \text{/}5
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\(\underset{x\to 0}{\text{lim}}\frac{1}{{x}^{2}}=\infty\)
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\(\underset{x\to -1}{\text{lim}}\frac{3}{{(x+1)}^{2}}=\infty\)
发送答案
Let \(\delta =\sqrt{\frac{3}{M}}.\) If \(0<|x+1|<\sqrt{\frac{3}{M}},\) then \(f(x)=\frac{3}{{(x+1)}^{2}}>M.\)
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\(\underset{x\to 2}{\text{lim}}-\frac{1}{{(x-2)}^{2}}=\text{-}\infty\)
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An engineer is using a machine to cut a flat square of Aerogel of area 144 cm2. If there is a maximum error tolerance in the area of 8 cm2, how accurately must the engineer cut on the side, assuming all sides have the same length? How do these numbers relate to \(\delta ,\) ε, a, and L?
发送答案
The engineer must cut within 0.328 cm of 12 cm on each side; \(\epsilon =8,\delta =0.328,a=12,L=144\)
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Use the precise definition of limit to prove that the following limit does not exist: \(\underset{x\to 1}{\text{lim}}\frac{|x-1|}{x-1}.\)
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Using precise definitions of limits, prove that \(\underset{x\to 0}{\text{lim}}f(x)\) does not exist, given that \(f(x)\) is the ceiling function. (Hint: Try any \(\delta <1\text{.)}\)
发送答案
Answers may vary.
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Using precise definitions of limits, prove that \(\underset{x\to 0}{\text{lim}}f(x)\) does not exist: \(f(x)=\{\begin{array}{l}1\ \text{if}\ x\ \text{is rational} \\ 0\ \text{if}\ x\ \text{is irrational}\end{array}.\) (Hint: Think about how you can always choose a rational number 0
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Using precise definitions of limits, determine \(\underset{x\to 0}{\text{lim}}f(x)\) for \(f(x)=\{\begin{array}{l}x\ \text{if}\ x\ \text{is rational} \\ 0\ \text{if}\ x\ \text{is irrational}\end{array}.\) (Hint: Break into two cases, x rational and x irrational.)
发送答案
0
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Using the function from the previous exercise, use the precise definition of limits to show that \(\underset{x\to a}{\text{lim}}f(x)\) does not exist for \(a\ne 0.\)
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\(\underset{x\to a}{\text{lim}}(f(x)+g(x))=L+M\)
发送答案
\(\begin{array}{l}\underset{x\to a}{\lim }\left(f\left(x\right)\right)+\underset{x\to a}{\lim }\left(g\left(x\right)\right) \\ =L+M\end{array}\)
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\(\underset{x\to a}{\text{lim}}[cf(x)]=cL\) for any real constant c (Hint: Consider two cases: \(c=0\) and \(c\ne 0\text{.)}\)
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\(\underset{x\to a}{\text{lim}}[f(x)g(x)]=LM.\) (Hint: \(|f(x)g(x)-LM|=\) \(|f(x)g(x)-f(x)M+f(x)M-LM|\le |f(x)||g(x)-M|+|M||f(x)-L|\text{.)}\)
发送答案
Answers may vary.
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A function has to be continuous at \(x=a\) if the \(\underset{x\to a}{\text{lim}}f(x)\) exists.
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You can use the quotient rule to evaluate \(\underset{x\to 0}{\text{lim}}\frac{\text{sin}\ x}{x}.\)
发送答案
False
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输入完整定义、图片和其中每个字母的含义的符号。
如何: The Precise Definition of a Limit
- Describe the epsilon-delta definition of a limit.
- Apply the epsilon-delta definition to find the limit of a function.
- Describe the epsilon-delta definitions of one-sided limits and infinite limits.
- Use the epsilon-delta definition to prove the limit laws.
- The statement
- The statement
- The statement
- The statement
人们对人们提出的问题
What is a derivative in one sentence?
The slope of the graph at a point, the rate at which the output is changing there. Speed is the derivative of position.
What is an integral in one sentence?
The accumulated total of a rate: the area under the curve. Distance travelled is the integral of speed.
Why are derivatives and integrals opposites?
That is the fundamental theorem of calculus: accumulating a rate and then measuring how fast the accumulation grows gets you back the rate. Integration undoes differentiation up to a constant.
When do I use substitution and when integration by parts?
Substitution when part of the integrand is the derivative of another part (u and du both present). Parts when the integrand is a product of two unrelated kinds of function, a polynomial times an exponential, log or trig function.
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LimitsDerivativesIntegralsDefinite integralsTaylor seriesSeries and sumsMaxima and minimaThe chain ruleImplicit differentiationRelated rates and optimisationIntegration techniques: substitution, parts, partial fractionsApplications of integration: area, volume, arc lengthInfinite series and convergence tests