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2.7: Chapter 2 Review Exercises

  • Page ID
    25932
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    True or False. In exercises 1 - 4, justify your answer with a proof or a counterexample.

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

    2) You can use the quotient rule to evaluate \(\displaystyle \lim_{x→0}\frac{\sin x}{x}\).

    Answer
    False, since we cannot have \(\displaystyle \lim_{x→0}x=0\) in the denominator.

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

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

    Answer
    False. A jump discontinuity is possible.

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

    A graph of a piecewise function with several segments. The first is a decreasing concave up curve existing for x < -1. It ends at an open circle at (-1, 1). The second is an increasing linear function starting at (-1, -2) and ending at (0,-1). The third is an increasing concave down curve existing from an open circle at (0,0) to an open circle at (1,1). The fourth is a closed circle at (1,-1). The fifth is a line with no slope existing for x > 1, starting at the open circle at (1,1).

    In exercises 6 - 15, evaluate the limit algebraically or explain why the limit does not exist.

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

    Answer
    \(5\)

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

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

    Answer
    \(8/7\)

    9) \(\displaystyle \lim_{x→π/2}\frac{\cot x}{\cos x}\)

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

    Answer
    DNE

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

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

    Answer
    \(2/3\)

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

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

    Answer
    \(−4\)

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

    In exercises 16 - 17, use the squeeze theorem to prove the limit.

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

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

    17) \(\displaystyle \lim_{x→0}x^3\sin\left(\frac{π}{x}\right)=0\)

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

    Answer
    \([2,∞]\)

    In exercises 19 - 20, determine the value of \(c\) such that the function remains continuous. Draw your resulting function to ensure it is continuous.

    19) \(f(x)=\begin{cases}x^2+1, & \text{if } x>c\\2^x, & \text{if } x≤c\end{cases}\)

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

    In exercises 21 - 22, use the precise definition of limit to prove the limit.

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

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

    Answer
    \(δ=\sqrt[3]{ε}\)

    23) 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.

    24) 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

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


    2.7: Chapter 2 Review Exercises is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.

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