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Calculus 3 : Limits

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Example Question #81 : Partial Derivatives

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow-7+}\frac{(12x - 12x^{2} + 672)}{(82x + 2x^{2} - 2x^{3} - 210)}\end{align*}$

Possible Answers:

$-\infty$

$\infty$

$-\frac{3}{4}$

Correct answer:

$-\frac{3}{4}$

Explanation:

$\begin{align*}&\text{We can't just plug in }x=-7\text{, as that'd}\\&\text{create a zero in the denominator, not telling us much.}\\&\text{Yet, note that}\\&\frac{(12x - 12x^{2} + 672)}{(82x + 2x^{2} - 2x^{3} - 210)}\\&\text{can be factored, allowing us to find the limit}\\&\text{taken from the right:}\\&\frac{(12\cdot (x + 7)\cdot (x - 8))}{(2\cdot (x + 7)\cdot (x - 3)\cdot (x - 5))}\\&\frac{(6\cdot (x - 8))}{((x - 3)\cdot (x - 5))}\\&lim_{x\rightarrow-7+}\frac{(12x - 12x^{2} + 672)}{(82x + 2x^{2} - 2x^{3} - 210)}=-\frac{3}{4}\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow-7+}\frac{(24x - 12)}{(6x^{2} - 4x - 82)}=-\frac{3}{4}\end{align*}$

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Example Question #1041 : Calculus 3

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow7+}\frac{(225x + 12x^{2} - 3x^{3} - 1134)}{(235x - 20x^{2} - 5x^{3} + 1050)}\end{align*}$

Possible Answers:

$-\frac{4}{65}$

$\infty$

$\frac{4}{65}$

$-\infty$

Correct answer:

$\frac{4}{65}$

Explanation:

$\begin{align*}&\text{We can't just plug in }x=7\text{, as that'd}\\&\text{create a zero in the denominator, not telling us much.}\\&\text{Yet, note that}\\&\frac{(225x + 12x^{2} - 3x^{3} - 1134)}{(235x - 20x^{2} - 5x^{3} + 1050)}\\&\text{can be factored, allowing us to find the limit}\\&\text{taken from the right:}\\&\frac{(3\cdot (x - 7)\cdot (x - 6)\cdot (x + 9))}{(5\cdot (x - 7)\cdot (x + 5)\cdot (x + 6))}\\&\frac{(3\cdot (x - 6)\cdot (x + 9))}{(5\cdot (x + 5)\cdot (x + 6))}\\&lim_{x\rightarrow7+}\frac{(225x + 12x^{2} - 3x^{3} - 1134)}{(235x - 20x^{2} - 5x^{3} + 1050)}=\frac{4}{65}\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow7+}\frac{(9x^{2} - 24x - 225)}{(40x + 15x^{2} - 235)}=\frac{4}{65}\end{align*}$

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Example Question #1042 : Calculus 3

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow8+}-\frac{(864x + 1044x^{2} + 360x^{3} + 9x^{4} - 9x^{5})}{(7x - 56)}\end{align*}$

Possible Answers:

$-\infty$

$\frac{79200}{7}$

$\infty$

Correct answer:

$\frac{79200}{7}$

Explanation:

$\begin{align*}&\text{We can't just plug in }x=8\text{, as that'd}\\&\text{create a zero in the denominator, not telling us much.}\\&\text{Yet, note that}\\&-\frac{(864x + 1044x^{2} + 360x^{3} + 9x^{4} - 9x^{5})}{(7x - 56)}\\&\text{can be factored, allowing us to find the limit}\\&\text{taken from the right:}\\&\frac{(9x\cdot (x - 8)\cdot (x + 2)^{2}\cdot (x + 3))}{(7\cdot (x - 8))}\\&\frac{(9x\cdot (x + 2)^{2}\cdot (x + 3))}{7}\\&lim_{x\rightarrow8+}-\frac{(864x + 1044x^{2} + 360x^{3} + 9x^{4} - 9x^{5})}{(7x - 56)}=\frac{79200}{7}\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow8+}\frac{(45x^{4} - 1080x^{2} - 36x^{3} - 2088x - 864)}{(7)}=\frac{79200}{7}\end{align*}$

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Example Question #1043 : Calculus 3

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow-3-}\frac{(19x + 57)}{(13x + 39)}\end{align*}$

Possible Answers:

$\infty$

$-\infty$

$\frac{19}{13}$

Correct answer:

$\frac{19}{13}$

Explanation:

$\begin{align*}&\text{We can't just plug in }x=-3\text{, as that'd}\\&\text{create a zero in the denominator, not telling us much.}\\&\text{Yet, note that}\\&\frac{(19x + 57)}{(13x + 39)}\\&\text{can be factored, allowing us to find the limit}\\&\text{taken from the left:}\\&\frac{(19\cdot (x + 3))}{(13\cdot (x + 3))}\\&\frac{19}{13}\\&lim_{x\rightarrow-3-}\frac{(19x + 57)}{(13x + 39)}=\frac{19}{13}\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow-3-}\frac{(19)}{(13)}=\frac{19}{13}\end{align*}$

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Example Question #1044 : Calculus 3

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow-6-}\frac{(144x + 12x^{2} + 432)}{(38016x + 5544x^{2} - 1428x^{3} - 48x^{4} + 12x^{5} - 124416)}\end{align*}$

Possible Answers:

$-\frac{57}{31}$

$-\infty$

$\infty$

Correct answer:

Explanation:

$\begin{align*}&\text{We can't just plug in }x=-6\text{, as that'd}\\&\text{create a zero in the denominator, not telling us much.}\\&\text{Yet, note that}\\&\frac{(144x + 12x^{2} + 432)}{(38016x + 5544x^{2} - 1428x^{3} - 48x^{4} + 12x^{5} - 124416)}\\&\text{can be factored, allowing us to find the limit}\\&\text{taken from the left:}\\&\frac{(12\cdot (x + 6)^{2})}{(12\cdot (x + 6)\cdot (x - 3)\cdot (x - 8)^{2}\cdot (x + 9))}\\&\frac{(x + 6)}{((x - 3)\cdot (x - 8)^{2}\cdot (x + 9))}\\&lim_{x\rightarrow-6-}\frac{(144x + 12x^{2} + 432)}{(38016x + 5544x^{2} - 1428x^{3} - 48x^{4} + 12x^{5} - 124416)}=0\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow-6-}\frac{(24x + 144)}{(11088x - 4284x^{2} - 192x^{3} + 60x^{4} + 38016)}=0\end{align*}$

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Example Question #81 : Limits

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow8-}-\frac{(13200x + 7403x^{2} - 803x^{3} - 99x^{4} + 11x^{5} - 123200)}{(1456x^{2} - 13088x + 128x^{3} - 16x^{4} + 11520)}\end{align*}$

Possible Answers:

$\infty$

$\frac{ 50}{29}$

$-\infty$

Correct answer:

Explanation:

$\begin{align*}&\text{We can't just plug in }x=8\text{, as that'd}\\&\text{create a zero in the denominator, not telling us much.}\\&\text{Yet, note that}\\&-\frac{(13200x + 7403x^{2} - 803x^{3} - 99x^{4} + 11x^{5} - 123200)}{(1456x^{2} - 13088x + 128x^{3} - 16x^{4} + 11520)}\\&\text{can be factored, allowing us to find the limit}\end{align*}$

$\begin{align*}&\text{taken from the left:}\\&\frac{(11\cdot (x - 8)^{2}\cdot (x - 5)\cdot (x + 5)\cdot (x + 7))}{(16\cdot (x - 8)\cdot (x - 1)\cdot (x - 9)\cdot (x + 10))}\\&\frac{(11\cdot (x - 5)\cdot (x + 5)\cdot (x + 7)\cdot (x - 8))}{(16\cdot (x - 1)\cdot (x - 9)\cdot (x + 10))}\\&lim_{x\rightarrow8-}-\frac{(13200x + 7403x^{2} - 803x^{3} - 99x^{4} + 11x^{5} - 123200)}{(1456x^{2} - 13088x + 128x^{3} - 16x^{4} + 11520)}=0\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow8-}\frac{(14806x - 2409x^{2} - 396x^{3} + 55x^{4} + 13200)}{(64x^{3} - 384x^{2} - 2912x + 13088)}=0\end{align*}$

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Example Question #1046 : Calculus 3

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow3-}-\frac{(616x + 966x^{2} - 14x^{4} - 9408)}{(540x + 300x^{2} - 220x^{3} + 20x^{4})}\end{align*}$

Possible Answers:

$-\infty$

$-\frac{245}{72}$

$\infty$

$\frac{245}{72}$

Correct answer:

$\frac{245}{72}$

Explanation:

$\begin{align*}&\text{We can't just plug in }x=3\text{, as that'd}\\&\text{create a zero in the denominator, not telling us much.}\\&\text{Yet, note that}\\&-\frac{(616x + 966x^{2} - 14x^{4} - 9408)}{(540x + 300x^{2} - 220x^{3} + 20x^{4})}\\&\text{can be factored, allowing us to find the limit}\\&\text{taken from the left:}\\&\frac{(14\cdot (x - 3)\cdot (x + 4)\cdot (x + 7)\cdot (x - 8))}{(20x\cdot (x - 3)\cdot (x + 1)\cdot (x - 9))}\\&\frac{(7\cdot (x + 4)\cdot (x + 7)\cdot (x - 8))}{(10x\cdot (x + 1)\cdot (x - 9))}\\&lim_{x\rightarrow3-}-\frac{(616x + 966x^{2} - 14x^{4} - 9408)}{(540x + 300x^{2} - 220x^{3} + 20x^{4})}=\frac{245}{72}\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow3-}\frac{(56x^{3} - 1932x - 616)}{(600x - 660x^{2} + 80x^{3} + 540)}=\frac{245}{72}\end{align*}$

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Example Question #1047 : Calculus 3

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow-8+}\frac{(15x + 120)}{(704x + 176x^{2} + 11x^{3})}\end{align*}$

Possible Answers:

$\infty$

$-\infty$

$\frac{ 37}{43}$

Correct answer:

$-\infty$

Explanation:

$\begin{align*}&\text{We can't just plug in }x=-8\text{, as that'd}\\&\text{create a zero in the denominator, not telling us much.}\\&\text{Yet, note that}\\&\frac{(15x + 120)}{(704x + 176x^{2} + 11x^{3})}\\&\text{can be factored, allowing us to find the limit}\\&\text{taken from the right:}\\&\frac{(15\cdot (x + 8))}{(11x\cdot (x + 8)^{2})}\\&\frac{15}{(11x\cdot (x + 8))}\\&lim_{x\rightarrow-8+}\frac{(15x + 120)}{(704x + 176x^{2} + 11x^{3})}=-\infty\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow-8+}\frac{(15)}{(352x + 33x^{2} + 704)}=-\infty\end{align*}$

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Example Question #1048 : Calculus 3

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow-5+}\frac{(3x + 15)}{(6x + 3x^{2} - 45)}\end{align*}$

Possible Answers:

$-\infty$

$-\frac{1}{8}$

$\infty$

Correct answer:

$-\frac{1}{8}$

Explanation:

$\begin{align*}&\text{We can't just plug in }x=-5\text{, as that'd}\\&\text{create a zero in the denominator, not telling us much.}\\&\text{Yet, note that}\\&\frac{(3x + 15)}{(6x + 3x^{2} - 45)}\\&\text{can be factored, allowing us to find the limit}\\&\text{taken from the right:}\\&\frac{(3\cdot (x + 5))}{(3\cdot (x + 5)\cdot (x - 3))}\\&\frac{1}{(x - 3)}\\&lim_{x\rightarrow-5+}\frac{(3x + 15)}{(6x + 3x^{2} - 45)}=-\frac{1}{8}\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow-5+}\frac{(3)}{(6x + 6)}=-\frac{1}{8}\end{align*}$

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Example Question #1049 : Calculus 3

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow5+}-\frac{(2160x - 1212x^{2} + 216x^{3} - 12x^{4})}{(75x - 30x^{2} + 3x^{3})}\end{align*}$

Possible Answers:

$\infty$

$-\infty$

$-\frac{10}{19}$

Correct answer:

$-\infty$

Explanation:

$\begin{align*}&\text{We can't just plug in }x=5\text{, as that'd}\\&\text{create a zero in the denominator, not telling us much.}\\&\text{Yet, note that}\\&-\frac{(2160x - 1212x^{2} + 216x^{3} - 12x^{4})}{(75x - 30x^{2} + 3x^{3})}\\&\text{can be factored, allowing us to find the limit}\\&\text{taken from the right:}\\&\frac{(12x\cdot (x - 5)\cdot (x - 4)\cdot (x - 9))}{(3x\cdot (x - 5)^{2})}\\&\frac{(4\cdot (x - 4)\cdot (x - 9))}{(x - 5)}\\&lim_{x\rightarrow5+}-\frac{(2160x - 1212x^{2} + 216x^{3} - 12x^{4})}{(75x - 30x^{2} + 3x^{3})}=-\infty\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow5+}\frac{(2424x - 648x^{2} + 48x^{3} - 2160)}{(9x^{2} - 60x + 75)}=-\infty\end{align*}$

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Calculus 3 : Limits

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #81 : Partial Derivatives

Example Question #1041 : Calculus 3

Example Question #1042 : Calculus 3

Example Question #1043 : Calculus 3

Example Question #1044 : Calculus 3

Example Question #81 : Limits

Example Question #1046 : Calculus 3

Example Question #1047 : Calculus 3

Example Question #1048 : Calculus 3

Example Question #1049 : Calculus 3

All Calculus 3 Resources

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