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2.5E: Exercises for Section 2.5

  • Page ID
    102701
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    In exercises 1 - 4, write the appropriate \( ε − δ\) definition for each of the given statements.

    1) \(\displaystyle \lim_{x →a}f(x)=N\)

    2) \(\displaystyle \lim_{t →b}g(t)=M\)

    Answer
    For every \( ε >0\), there exists a \( δ >0\), so that if \(0 <|t −b| < δ\), then \(|g(t) −M| < ε\)

    3) \(\displaystyle \lim_{x →c}h(x)=L\)

    4) \(\displaystyle \lim_{x →a} φ(x)=A\)

    Answer
    For every \( ε >0\), there exists a \( δ >0\), so that if \(0 <|x −a| < δ\), then \(| φ(x) −A| < ε\)

    The following graph of the function \(f\) satisfies \(\displaystyle \lim_{x →2}f(x)=2\). In the following exercises, determine a value of \( δ >0\) that satisfies each statement.

    A function drawn in quadrant one for x  > 0. It is an increasing concave up function, with points approximately (0,0), (1, .5), (2,2), and (3,4).

    5) If \(0 <|x −2| < δ\), then \(|f(x) −2| <1\).

    6) If \(0 <|x −2| < δ\), then \(|f(x) −2| <0.5\).

    Answer
    \( δ ≤0.25\)

    The following graph of the function \(f\) satisfies \(\displaystyle \lim_{x →3}f(x)= −1\). In the following exercises, determine a value of \( δ >0\) that satisfies each statement.

    A graph of a decreasing linear function, with points (0,2), (1,1), (2,0), (3,-1), (4,-2), and so on for x  >= 0.

    7) If \(0 <|x −3| < δ\), then \(|f(x)+1| <1\).

    8) If \(0 <|x −3| < δ\), then \(|f(x)+1| <2\).

    Answer
    \( δ ≤2\)

    The following graph of the function \(f\) satisfies \(\displaystyle \lim_{x →3}f(x)=2\). In the following exercises, for each value of \( ε\), find a value of \( δ >0\) such that the precise definition of limit holds true.

    A graph of an increasing linear function intersecting the x axis at about (2.25, 0) and going through the points (3,2) and, approximately, (1,-5) and (4,5).

    9) \( ε=1.5\)

    10) \( ε=3\)

    Answer
    \( δ ≤1\)

    [T] In exercises 11 - 12, use a graphing calculator to find a number \( δ\) such that the statements hold true.

    11) \(\left|\sin(2x) −\frac{1}{2}\right| <0.1\), whenever \(\left|x −\frac{ π}{12}\right| < δ\)

    12) \(\left|\sqrt{x −4} −2\right| <0.1\), whenever \(|x −8| < δ\)

    Answer
    \( δ <0.3900\)

    In exercises 13 - 17, use the precise definition of limit to prove the given limits.

    13) \(\displaystyle \lim_{x →2}\,(5x+8)=18\)

    14) \(\displaystyle \lim_{x →3}\frac{x^2 −9}{x −3}=6\)

    Answer
    Let \( δ= ε\). If \(0 <|x −3| < ε\), then \(\left|\dfrac{x^2 −9}{x −3} - 6\right| = \left|\dfrac{(x+3)(x −3)}{x −3} - 6\right| = |x+3 −6|=|x −3| < ε\).

    15) \(\displaystyle \lim_{x →2}\frac{2x^2 −3x −2}{x −2}=5\)

    16) \(\displaystyle \lim_{x →0}x^4=0\)

    Answer
    Let \( δ=\sqrt[4]{ ε}\). If \(0 <|x| <\sqrt[4]{ ε}\), then \(\left|x^4-0\right|=x^4 < ε\).

    17) \(\displaystyle \lim_{x →2}\,(x^2+2x)=8\)

    In exercises 18 - 20, use the precise definition of limit to prove the given one-sided limits.

    18) \(\displaystyle \lim_{x →5^ −}\sqrt{5 −x}=0\)

    Answer
    Let \( δ= ε^2\). If \(- ε^2 < x - 5 < 0,\) we can multiply through by \(-1\) to get \(0 <5-x < ε^2.\)
    Then \(\left|\sqrt{5 −x} - 0\right|=\sqrt{5 −x} < \sqrt{ ε^2} = ε\).

    19) \(\displaystyle \lim_{x →0^+}f(x)= −2\), where \(f(x)=\begin{cases}8x −3, & \text{if }x <0\\4x −2, & \text{if }x ≥0\end{cases}\).

    20) \(\displaystyle \lim_{x →1^ −}f(x)=3\), where \(f(x)=\begin{cases}5x −2, & \text{if }x <1\\7x −1, & \text{if }x ≥1\end{cases}\).

    Answer
    Let \( δ= ε/5\). If \( − ε/5 < x - 1 <0,\) we can multiply through by \(-1\) to get \(0 <1-x < ε/5.\)
    Then \(|f(x) −3|=|5x-2-3| = |5x −5| = 5(1-x),\) since \(x <1\) here.
    And \(5(1-x) < 5( ε/5) = ε\).

    In exercises 21 - 23, use the precise definition of limit to prove the given infinite limits.

    21) \(\displaystyle \lim_{x →0}\frac{1}{x^2}= ∞\)

    22) \(\displaystyle \lim_{x → −1}\frac{3}{(x+1)^2}= ∞\)

    Answer
    Let \( δ=\sqrt{\frac{3}{N}}\). If \(0 <|x+1| <\sqrt{\frac{3}{N}}\), then \(f(x)=\frac{3}{(x+1)^2} >N\).

    23) \(\displaystyle \lim_{x →2} −\frac{1}{(x −2)^2}= − ∞\)

    24) An engineer is using a machine to cut a flat square of Aerogel of area \(144 \,\text{cm}^2\). If there is a maximum error tolerance in the area of \(8 \,\text{cm}^2\), how accurately must the engineer cut on the side, assuming all sides have the same length? How do these numbers relate to \( δ\), \( ε\), \(a\), and \(L\)?

    Answer
    \(0.033 \text{ cm}, \, ε=8,\, δ=0.33,\,a=12,\,L=144\)

    25) Use the precise definition of limit to prove that the following limit does not exist: \(\displaystyle \lim_{x →1}\frac{|x −1|}{x −1}.\)

    26) Using precise definitions of limits, prove that \(\displaystyle \lim_{x →0}f(x)\) does not exist, given that \(f(x)\) is the ceiling function. (Hint: Try any \( δ <1\).)

    Answer
    Answers may very.

    27) Using precise definitions of limits, prove that \(\displaystyle \lim_{x →0}f(x)\) does not exist: \(f(x)=\begin{cases}1, & \text{if }x\text{ is rational}\\0, & \text{if }x\text{ is irrational}\end{cases}\). (Hint: Think about how you can always choose a rational number \(0 <d\), >

    28) Using precise definitions of limits, determine \(\displaystyle \lim_{x →0}f(x)\) for \(f(x)=\begin{cases}x, & \text{if }x\text{ is rational}\\0, & \text{if }x\text{ is irrational}\end{cases}\). (Hint: Break into two cases, \(x\) rational and \(x\) irrational.)

    Answer
    \(0\)

    29) Using the function from the previous exercise, use the precise definition of limits to show that \(\displaystyle \lim_{x →a}f(x)\) does not exist for \(a ≠0\)

    For exercises 30 - 32, suppose that \(\displaystyle \lim_{x →a}f(x)=L\) and \(\displaystyle \lim_{x →a}g(x)=M\) both exist. Use the precise definition of limits to prove the following limit laws:

    30) \(\displaystyle \lim_{x →a}(f(x) −g(x))=L −M\)

    Answer
    \(f(x) −g(x)=f(x)+( −1)g(x)\)

    31) \(\displaystyle \lim_{x →a}[cf(x)]=cL\) for any real constant \(c\) (Hint: Consider two cases: \(c=0\) and \(c ≠0\).)

    32) \(\displaystyle \lim_{x →a}[f(x)g(x)]=LM\). (Hint: \(|f(x)g(x) −LM|= |f(x)g(x) −f(x)M +f(x)M −LM| ≤|f(x)||g(x) −M| +|M||f(x) −L|.)\)

    Answer
    Answers may vary.

    2.5E: Exercises for Section 2.5 is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.

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