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Mathematics LibreTexts

2.7E: Precise Definition of Limit EXERCISES

  • Page ID
    10231
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    2.7: The Precise Definition of a Limit

    In the following exercises, write the appropriate \(ε−δ\) definition for each of the given statements.

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

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

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

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

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

    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.

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

    181) 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.

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

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

    Answer:
    \(δ≤2\)

    In the following exercises, use the precise definition of limit to prove the given limits.

    J3.7.1) \(\displaystyle \lim_{x→5}\,(2x - 1)=9\)

    Answer:

    Let ε\(>0\); choose \(δ=\frac{ε}{2}\); assume \(0<|x−5|<δ\).

    In other words:

    \(0<|x - 5|<\frac{ε}{2}\),

    so \(-\frac{ε}{2}<x - 5<\frac{ε}{2}\),

    then \(-ε<2x - 10<ε\)

    then |\(2x - 10|<ε\),

    then |(\(2x - 1)−9\)|<ε

    Thus, if \(0<|x−5|<δ\), then |(\(2x - 1)−9\)|<ε.

    Therefore, by the definition of limit, \(\displaystyle \lim_{x→5}\,(2x - 1)=9\).

    J3.7.2) \(\displaystyle \lim_{x→-3}\,(5x+2)=-13\)

    J3.7.3) \(\displaystyle \lim_{x→-7^-}\,\frac{1}{x+7}= −∞ \)

    Answer:

    Let \(M>0\); since this is a limit from the left, we need \(-δ<x+7<0\) to lead \(\frac{1}{x+7}<-M\) (since the limit is \(−∞ \))
    Note: since \(x<-7, x+7<0\)

    Proof:

    Let \(M>0\). Choose \(δ=\frac{1}{M}\).

    If \(-δ<x+7<0\), in other words \(-\frac{1}{M}<x+7<0\)

    then \(\frac{1}{M}>-(x+7)>0\) (both \(\frac{1}{M}\) and \(-(x+7)\) are positive)

    then \(M<-\frac{1}{x+7}\)

    then \(-M>\frac{1}{x+7}\)

    so \(\frac{1}{x+7}<-M\)

    Thus, if \(-\frac{1}{M}<x+7<0\) then\(\frac{1}{x+7}<-M\)

    Therefore, by the definition of (infinite, from the left) limit, \(\displaystyle \lim_{x→-7^-}\,\frac{1}{x+7}= −∞ \).

    J3.7.4) \(\displaystyle \lim_{x→2^+}\,\frac{1}{x-2}= ∞ \)

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

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

    Answer:
    \(\frac{x^2−9}{x−3}\) is equivalent to \(x + 3\) by factoring, as long as \(x\) is not \(3\). Since we are looking at the limit as \(x→3\), we do not consider \(x=3\).
    Let \(ε>0\), choose \(δ=ε\). If \(0<|x−3|<ε\), then \(|x+3−6|=|x−3|<ε\). Thus, by the definition of limit, \(\displaystyle \lim_{x→3}\,\frac{x^2−9}{x−3}=6\).

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

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

    Answer:
    Let \(ε>0\), choose\(δ=\sqrt[4]{ε}\) If \(0<|x|<\sqrt[4]{ε}\), then \(∣x^4∣=x^4<ε\). Thus, by the definition of limit, \(\displaystyle\lim_{x→0},x^4=0\).

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

    Chapter Review Exercises

    True or False. In the following exercises, justify your answer with a proof or a counterexample.

    208) A function has to be continuous at \(x=a\) if the \(\displaystyle \lim_{x→a}\,f(x)\) exists.

    209) Evaluate \(\displaystyle \lim_{x→0}\,\frac{sinx}{x}\) = ?

    Answer:
    1

    210) If there is a vertical asymptote at \(x=a\) for the function \(f(x)\), then f is undefined at the point \(x=a\).

    211) If \(\displaystyle \lim_{x→a}\,f(x)\) does not exist, then f is undefined at the point \(x=a\).

    Answer:
    False. A removable discontinuity is possible.

    212) Using the graph, find each limit or explain why the limit does not exist.

    a. \(\displaystyle \lim_{x→−1}\,f(x)\)

    b. \(\displaystyle \lim_{x→1}\,f(x)\)

    c. \(\displaystyle \lim_{x→0^+}\,f(x)\)

    d. \(\displaystyle \lim_{x→2}\,f(x)\)

    In the following exercises, evaluate the limit algebraically or explain why the limit does not exist.

    213) \(\displaystyle \lim_{x→2}\,\frac{2x^2−3x−2}{x−2}\)

    Answer:
    5

    214) \(\displaystyle \lim_{x→0}\,3x^2−2x+4\)

    215) \(\displaystyle \lim_{x→3}\,\frac{x^3−2x^2−1}{3x−2}\)

    Answer:
    8/7

    216) \(\displaystyle \lim_{x→π/2}\,\frac{cotx}{cosx}\)

    217) \(\displaystyle \lim_{x→−5}\,\frac{x^2+25}{x+5}\)

    Answer:
    DNE

    218) \(\displaystyle \lim_{x→2}\,\frac{3x^2−2x−8}{x^2−4}\)

    219) \(\displaystyle \lim_{x→1}\,\frac{x^2−1}{x^3−1}\)

    Answer:
    2/3

    220) \(\displaystyle \lim_{x→1}\,\frac{x^2−1}{\sqrt{x}−1}\)

    221) \(\displaystyle \lim_{x→4}\,\frac{4−x}{\sqrt{x}−2}\)

    Answer:
    −4

    222) \(\displaystyle \lim_{x→4}\,\frac{1}{\sqrt{x}−2}\)

    In the following exercises, use the squeeze theorem to prove the limit.

    223) \(\displaystyle \lim_{x→0}\,x^2cos(2πx)=0\)

    Answer:
    Since \(−1≤cos(2πx)≤1\), then \(−x^2≤x^2cos(2πx)≤x^2\). Since \(lim_{x→0}\,x^2=0=lim_{x→0}\,−x^2\), it follows that \(lim_{x→0}\,x^2cos(2πx)=0\).

    224) \(\displaystyle \lim_{x→0}\,x^3sin(\frac{π}{x})=0\)

    225) Determine the domain such that the function \(f(x)=\sqrt{x−2}+xe^x\) is continuous over its domain.

    Answer:
    \([2,∞]\)

    In the following exercises, determine the value of c such that the function remains continuous. Draw your resulting function to ensure it is continuous.

    226) \(f(x)=\begin{cases}x^2+1 & x>c\\2^x & x≤c\end{cases}\)

    227) \(f(x)=\begin{cases}\sqrt{x+1} & x>−1\\x^2+c & x≤−1\end{cases}\)

    Answer:
    \(c=-1\)

    In the following exercises, use the precise definition of limit to prove the limit.

    228) \(\displaystyle \lim_{x→1}\,(8x+16)=24\)

    229) \(\displaystyle \lim_{x→0}\,x^3=0\)

    Answer:
    \(δ=\sqrt[3]{ε}\) [This is just a piece for constructing the proof.]

    230) A ball is thrown into the air and the vertical position is given by \(x(t)=−4.9t^2+25t+5\). Use the Intermediate Value Theorem to show that the ball must land on the ground sometime between 5 sec and 6 sec after the throw.

    231) A particle moving along a line has a displacement according to the function \(x(t)=t^2−2t+4\), where x is measured in meters and t is measured in seconds. Find the average velocity over the time period \(t=[0,2]\).

    Answer:
    \(0\) m/sec

    232) From the previous exercises, estimate the instantaneous velocity at \(t=2\) by checking the average velocity within \(t=0.01\) sec.