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Calculus 3 : Limits

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Example Questions

Example Question #71 : Limits

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow10-}-\frac{(21x - 7x^{2} + 490)}{(7x^{2} - 126x + 560)}\end{align*}$

Possible Answers:

$-\infty$

$\infty$

$\frac{17}{2}$

Correct answer:

$\frac{17}{2}$

Explanation:

$\begin{align*}&\text{We cannot simply plug in the value of }x=10\text{, as that would}\\&\text{create a zero value in the denominator, and not tell us much as is.}\\&\text{However, note that the expression}\\&-\frac{(21x - 7x^{2} + 490)}{(7x^{2} - 126x + 560)}\\&\text{can be factored. This will allow us to find the limit}\\&\text{taken from the left:}\\&\frac{(7\cdot (x - 10)\cdot (x + 7))}{(7\cdot (x - 10)\cdot (x - 8))}\\&\frac{(x + 7)}{(x - 8)}\\&lim_{x\rightarrow10-}-\frac{(21x - 7x^{2} + 490)}{(7x^{2} - 126x + 560)}=\frac{17}{2}\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow10-}\frac{(14x - 21)}{(14x - 126)}=\frac{17}{2}\end{align*}$

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Example Question #72 : Limits

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow8+}\frac{(172x - 34x^{2} + 2x^{3} - 224)}{(7x - 56)}\end{align*}$

Possible Answers:

$-\infty$

$\frac{12}{7}$

$\infty$

$-\frac{12}{7}$

Correct answer:

$\frac{12}{7}$

Explanation:

$\begin{align*}&\text{We cannot simply plug in the value of }x=8\text{, as that would}\\&\text{create a zero value in the denominator, and not tell us much as is.}\\&\text{However, note that the expression}\\&\frac{(172x - 34x^{2} + 2x^{3} - 224)}{(7x - 56)}\\&\text{can be factored. This will allow us to find the limit}\\&\text{taken from the right:}\\&\frac{(2\cdot (x - 8)\cdot (x - 2)\cdot (x - 7))}{(7\cdot (x - 8))}\\&\frac{(2\cdot (x - 2)\cdot (x - 7))}{7}\\&lim_{x\rightarrow8+}\frac{(172x - 34x^{2} + 2x^{3} - 224)}{(7x - 56)}=\frac{12}{7}\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow8+}\frac{(6x^{2} - 68x + 172)}{(7)}=\frac{12}{7}\end{align*}$

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Example Question #73 : Limits

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow-3-}-\frac{(16x - 96x^{2} - 16x^{3} + 480)}{(9x^{3} - 356x^{2} - 292x + 14x^{4} + x^{5} + 1680)}\end{align*}$

Possible Answers:

$\frac{8}{49}$

$\infty$

$-\frac{8}{49}$

$-\infty$

Correct answer:

$-\frac{8}{49}$

Explanation:

$\begin{align*}&\text{We cannot simply plug in the value of }x=-3\text{, as that would}\\&\text{create a zero value in the denominator, and not tell us much as is.}\\&\text{However, note that the expression}\\&-\frac{(16x - 96x^{2} - 16x^{3} + 480)}{(9x^{3} - 356x^{2} - 292x + 14x^{4} + x^{5} + 1680)}\\&\text{can be factored. This will allow us to find the limit}\\&\text{taken from the left:}\\&\frac{(16\cdot (x + 3)\cdot (x - 2)\cdot (x + 5))}{((x - 2)\cdot (x + 3)\cdot (x - 4)\cdot (x + 7)\cdot (x + 10))}\\&\frac{(16\cdot (x + 5))}{((x - 4)\cdot (x + 7)\cdot (x + 10))}\\&lim_{x\rightarrow-3-}-\frac{(16x - 96x^{2} - 16x^{3} + 480)}{(9x^{3} - 356x^{2} - 292x + 14x^{4} + x^{5} + 1680)}=-\frac{8}{49}\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow-3-}\frac{(192x + 48x^{2} - 16)}{(27x^{2} - 712x + 56x^{3} + 5x^{4} - 292)}=-\frac{8}{49}\end{align*}$

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Example Question #74 : Limits

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow-8-}-\frac{(2970x - 693x^{2} - 54x^{3} + 9x^{4} + 3600)}{(312x + 6x^{2} - 6x^{3} - 960)}\end{align*}$

Possible Answers:

$\frac{63}{4}$

$-\frac{63}{4}$

$\infty$

$-\infty$

Correct answer:

$-\frac{63}{4}$

Explanation:

$\begin{align*}&\text{We cannot simply plug in the value of }x=-8\text{, as that would}\\&\text{create a zero value in the denominator, and not tell us much as is.}\\&\text{However, note that the expression}\\&-\frac{(2970x - 693x^{2} - 54x^{3} + 9x^{4} + 3600)}{(312x + 6x^{2} - 6x^{3} - 960)}\\&\text{can be factored. This will allow us to find the limit}\\&\text{taken from the left:}\\&\frac{(9\cdot (x + 8)\cdot (x + 1)\cdot (x - 5)\cdot (x - 10))}{(6\cdot (x + 8)\cdot (x - 4)\cdot (x - 5))}\\&\frac{(3\cdot (x + 1)\cdot (x - 10))}{(2\cdot (x - 4))}\\&lim_{x\rightarrow-8-}-\frac{(2970x - 693x^{2} - 54x^{3} + 9x^{4} + 3600)}{(312x + 6x^{2} - 6x^{3} - 960)}=-\frac{63}{4}\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow-8-}\frac{(36x^{3} - 162x^{2} - 1386x + 2970)}{(18x^{2} - 12x - 312)}=-\frac{63}{4}\end{align*}$

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Example Question #75 : Limits

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow8+}-\frac{(8x - 64)}{(7x - x^{2} + 8)}\end{align*}$

Possible Answers:

$\frac{8}{9}$

$-\frac{8}{9}$

$\infty$

$-\infty$

Correct answer:

$\frac{8}{9}$

Explanation:

$\begin{align*}&\text{We cannot simply plug in the value of }x=8\text{, as that would}\\&\text{create a zero value in the denominator, and not tell us much as is.}\\&\text{However, note that the expression}\\&-\frac{(8x - 64)}{(7x - x^{2} + 8)}\\&\text{can be factored. This will allow us to find the limit}\\&\text{taken from the right:}\\&\frac{(8\cdot (x - 8))}{((x + 1)\cdot (x - 8))}\\&\frac{8}{(x + 1)}\\&lim_{x\rightarrow8+}-\frac{(8x - 64)}{(7x - x^{2} + 8)}=\frac{8}{9}\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow8+}\frac{(8)}{(2x - 7)}=\frac{8}{9}\end{align*}$

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Example Question #76 : Limits

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow9+}\frac{(4x^{2} - 72x + 324)}{(38556x + 4964x^{2} - 1853x^{3} - 68x^{4} + 17x^{5} + 44064)}\end{align*}$

Possible Answers:

$\frac{ 73}{82}$

$\infty$

$-\infty$

Correct answer:

Explanation:

$\begin{align*}&\text{We cannot simply plug in the value of }x=9\text{, as that would}\\&\text{create a zero value in the denominator, and not tell us much as is.}\\&\text{However, note that the expression}\\&\frac{(4x^{2} - 72x + 324)}{(38556x + 4964x^{2} - 1853x^{3} - 68x^{4} + 17x^{5} + 44064)}\\&\text{can be factored. This will allow us to find the limit}\\&\text{taken from the right:}\\&\frac{(4\cdot (x - 9)^{2})}{(17\cdot (x - 9)\cdot (x + 2)^{2}\cdot (x - 8)\cdot (x + 9))}\\&\frac{(4\cdot (x - 9))}{(17\cdot (x + 2)^{2}\cdot (x - 8)\cdot (x + 9))}\\&lim_{x\rightarrow9+}\frac{(4x^{2} - 72x + 324)}{(38556x + 4964x^{2} - 1853x^{3} - 68x^{4} + 17x^{5} + 44064)}=0\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow9+}\frac{(8x - 72)}{(9928x - 5559x^{2} - 272x^{3} + 85x^{4} + 38556)}=0\end{align*}$

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Example Question #77 : Limits

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow-4-}\frac{(51x^{2} - 442x + 102x^{3} + 17x^{4} - 408)}{(238x^{3} - 1232x^{2} - 3360x + 140x^{4} + 14x^{5})}\end{align*}$

Possible Answers:

$-\infty$

$-\frac{86}{69}$

$\infty$

Correct answer:

$\infty$

Explanation:

$\begin{align*}&\text{We can't just plug in }x=-4\text{, as that'd}\\&\text{create a zero value in the denominator, not telling us much.}\\&\text{Yet, note that}\\&\frac{(51x^{2} - 442x + 102x^{3} + 17x^{4} - 408)}{(238x^{3} - 1232x^{2} - 3360x + 140x^{4} + 14x^{5})}\\&\text{can be factored. This'll allow us to find the limit}\\&\text{taken from the left:}\\&\frac{(17\cdot (x + 4)\cdot (x + 1)\cdot (x - 2)\cdot (x + 3))}{(14x\cdot (x + 4)^{2}\cdot (x - 3)\cdot (x + 5))}\\&\frac{(17\cdot (x + 1)\cdot (x - 2)\cdot (x + 3))}{(14x\cdot (x - 3)\cdot (x + 4)\cdot (x + 5))}\\&lim_{x\rightarrow-4-}\frac{(51x^{2} - 442x + 102x^{3} + 17x^{4} - 408)}{(238x^{3} - 1232x^{2} - 3360x + 140x^{4} + 14x^{5})}=\infty\\&\text{Another method, since numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow-4-}\frac{(102x + 306x^{2} + 68x^{3} - 442)}{(714x^{2} - 2464x + 560x^{3} + 70x^{4} - 3360)}=\infty\end{align*}$

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Example Question #78 : Limits

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow2+}\frac{(874x^{2} - 3268x + 133x^{3} - 19x^{4} + 2280)}{(52x - 91x^{2} - 13x^{3} + 364)}\end{align*}$

Possible Answers:

$-\infty$

$-\frac{304}{117}$

$\infty$

Correct answer:

$-\frac{304}{117}$

Explanation:

$\begin{align*}&\text{We can't just plug in }x=2\text{, as that'd}\\&\text{create a zero in the denominator, not telling us much.}\\&\text{Yet, note that}\\&\frac{(874x^{2} - 3268x + 133x^{3} - 19x^{4} + 2280)}{(52x - 91x^{2} - 13x^{3} + 364)}\\&\text{can be factored, allowing us to find the limit}\\&\text{taken from the right:}\\&\frac{(19\cdot (x - 2)\cdot (x - 1)\cdot (x + 6)\cdot (x - 10))}{(13\cdot (x - 2)\cdot (x + 2)\cdot (x + 7))}\\&\frac{(19\cdot (x - 1)\cdot (x + 6)\cdot (x - 10))}{(13\cdot (x + 2)\cdot (x + 7))}\\&lim_{x\rightarrow2+}\frac{(874x^{2} - 3268x + 133x^{3} - 19x^{4} + 2280)}{(52x - 91x^{2} - 13x^{3} + 364)}=-\frac{304}{117}\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow2+}\frac{(76x^{3} - 399x^{2} - 1748x + 3268)}{(182x + 39x^{2} - 52)}=-\frac{304}{117}\end{align*}$

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Example Question #79 : Limits

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow-2+}\frac{(160x^{2} + 260x^{3} + 110x^{4} + 10x^{5})}{(18x + 36)}\end{align*}$

Possible Answers:

$-\frac{40}{3}$

$-\infty$

$\frac{40}{3}$

$\infty$

Correct answer:

$-\frac{40}{3}$

Explanation:

$\begin{align*}&\text{We can't just plug in }x=-2\text{, as that'd}\\&\text{create a zero in the denominator, not telling us much.}\\&\text{Yet, note that}\\&\frac{(160x^{2} + 260x^{3} + 110x^{4} + 10x^{5})}{(18x + 36)}\\&\text{can be factored, allowing us to find the limit}\\&\text{taken from the right:}\\&\frac{(10x^{2}\cdot (x + 2)\cdot (x + 1)\cdot (x + 8))}{(18\cdot (x + 2))}\\&\frac{(5x^{2}\cdot (x + 1)\cdot (x + 8))}{9}\\&lim_{x\rightarrow-2+}\frac{(160x^{2} + 260x^{3} + 110x^{4} + 10x^{5})}{(18x + 36)}=-\frac{40}{3}\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow-2+}\frac{(320x + 780x^{2} + 440x^{3} + 50x^{4})}{(18)}=-\frac{40}{3}\end{align*}$

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Example Question #80 : Limits

$\begin{align*}&\text{Evaluate the following limit:}\\&lim_{x\rightarrow0+}-\frac{(10x - 2x^{2})}{(1050x + 665x^{2} + 126x^{3} + 7x^{4})}\end{align*}$

Possible Answers:

$\frac{1}{105}$

$\infty$

$-\frac{1}{105}$

$-\infty$

Correct answer:

$-\frac{1}{105}$

Explanation:

$\begin{align*}&\text{We can't just plug in }x=0\text{, as that'd}\\&\text{create a zero in the denominator, not telling us much.}\\&\text{Yet, note that}\\&-\frac{(10x - 2x^{2})}{(1050x + 665x^{2} + 126x^{3} + 7x^{4})}\\&\text{can be factored, allowing us to find the limit}\\&\text{taken from the right:}\\&\frac{(2x\cdot (x - 5))}{(7x\cdot (x + 3)\cdot (x + 5)\cdot (x + 10))}\\&\frac{(2\cdot (x - 5))}{(7\cdot (x + 3)\cdot (x + 5)\cdot (x + 10))}\\&lim_{x\rightarrow0+}-\frac{(10x - 2x^{2})}{(1050x + 665x^{2} + 126x^{3} + 7x^{4})}=-\frac{1}{105}\\&\text{Another method, since both numerator and denominator go}\\&\text{to zero initially, is to use L'Hopital's rule:}\\&lim\frac{f(x)}{g(x)}=lim\frac{f'(x)}{g'(x)}\\&lim_{x\rightarrow0+}\frac{(4x - 10)}{(1330x + 378x^{2} + 28x^{3} + 1050)}=-\frac{1}{105}\end{align*}$

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Calculus 3 : Limits

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #71 : Limits

Example Question #72 : Limits

Example Question #73 : Limits

Example Question #74 : Limits

Example Question #75 : Limits

Example Question #76 : Limits

Example Question #77 : Limits

Example Question #78 : Limits

Example Question #79 : Limits

Example Question #80 : Limits

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