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Use the graph to estimate the specified limit. -Find lim x →\rightarrow 0 f(x) Use the graph to estimate the specified limit. -Find lim x \rightarrow 0 f(x) A) - \frac { 2 \sqrt { 3 } } { 3 } B) - 3 C) \frac { 2 } { 3 } D) \frac { 2 \sqrt { 3 } } { 3 }


A) −233- \frac { 2 \sqrt { 3 } } { 3 }
B) −3- 3
C) 23\frac { 2 } { 3 }
D) 233\frac { 2 \sqrt { 3 } } { 3 }

E) None of the above
F) All of the above

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Use the slopes of UQ, UR, US, and UT to estimate the rate of change of y at the specified value of x. -x = 2.5 Use the slopes of UQ, UR, US, and UT to estimate the rate of change of y at the specified value of x. -x = 2.5 A) 1.25 B) 3.75 C) 7.5 D) 0


A) 1.25
B) 3.75
C) 7.5
D) 0

E) A) and C)
F) A) and B)

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Provide an appropriate response. -Write the formal notation for the principle "the limit of a quotient is the quotient of the limits" and include a statement of any restrictions on the principle.


A) lim⁡x→ag(x) f(x) =g(a) f(a) \lim _ { x \rightarrow a } \frac { g ( x ) } { f ( x ) } = \frac { g ( a ) } { f ( a ) } .
B) If lim⁡x→ag(x) =M\lim _ { x \rightarrow a } g ( x ) = M and lim⁡x→af(x) =L\lim _ { x \rightarrow a } f ( x ) = L , then lim⁡x→ag(x) f(x) =lim⁡x→ag(x) lim⁡x→af(x) =ML\lim _ { x \rightarrow a } \frac { g ( x ) } { f ( x ) } = \frac { \lim _ { x \rightarrow a } g ( x ) } { \lim _ { x \rightarrow a } f ( x ) } = \frac { M } { L } , provided that f(a) ≠0f ( a ) \neq 0 .
C) lim⁡x→ag(x) f(x) =g(a) f(a) \lim _ { x \rightarrow a } \frac { g ( x ) } { f ( x ) } = \frac { g ( a ) } { f ( a ) } , provided that f(a) ≠0f ( a ) \neq 0 .
D) If lim⁡x→ag(x) =M\lim _ { x \rightarrow a } g ( x ) = M and lim⁡x→af(x) =L\lim _ { x \rightarrow a } f ( x ) = L , then lim⁡x→ag(x) f(x) =lim⁡x→ag(x) lim⁡x→af(x) =ML\lim _ { x \rightarrow a } \frac { g ( x ) } { f ( x ) } = \frac { \lim _ { x \rightarrow a } g ( x ) } { \lim _ { x \rightarrow a } f ( x ) } = \frac { M } { L } , provided that L≠0L \neq 0 .

E) A) and D)
F) C) and D)

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 Find the limit using lim⁡x=0sin⁡xx=1. \text { Find the limit using } \lim _ { x = 0 } \frac { \sin x } { x } = 1 \text {. } - lim⁡x→0sin⁡5xsin⁡4x\lim _ { x \rightarrow 0 } \frac { \sin 5 x } { \sin 4 x }


A) does not exist
B) 45\frac { 4 } { 5 }
C) 54\frac { 5 } { 4 }
D) 0

E) A) and D)
F) A) and B)

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Prove the limit statement -Select the correct statement for the definition of the limit: lim⁡x→0f(x) =L\lim _ { x \rightarrow 0 } f ( x ) = L means that___________


A) if given any number ε>0\varepsilon > 0 , there exists a number δ>0\delta > 0 , such that for all xx , 0<∣x−x0∣<ε0 < \left| x - x _ { 0 } \right| < \varepsilon implies ∣f(x) −L∣>δ| f ( x ) - L | > \delta .
B) if given any number ε>0\varepsilon > 0 , there exists a number δ>0\delta > 0 , such that for all xx , 0<∣x−x0∣<ε0 < \left| x - x _ { 0 } \right| < \varepsilon implies ∣f(x) −L∣<δ| f ( x ) - L | < \delta .
C) if given a number ε>0\varepsilon > 0 , there exists a number δ>0\delta > 0 , such that for all xx , 0<∣x−x0∣<δ0 < \left| x - x _ { 0 } \right| < \delta implies ∣f(x) −L∣>ε| f ( x ) - L | > \varepsilon .
D) if given any number ε>0\varepsilon > 0 , there exists a number δ>0\delta > 0 , such that for all xx , 0<∣x−x0∣<δ0 < \left| x - x _ { 0 } \right| < \delta implies ∣f(x) −L∣<ε| f ( x ) - L | < \varepsilon .

E) B) and C)
F) C) and D)

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Find the limit, if it exists. - lim⁡x→2x2−4x2−3x+2\lim _ { x \rightarrow 2 } \frac { x ^ { 2 } - 4 } { x ^ { 2 } - 3 x + 2 }


A) 0
B) 2
C) Does not exist
D) 4

E) All of the above
F) A) and D)

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 Find the limit using lim⁡x=0sin⁡xx=1. \text { Find the limit using } \lim _ { x = 0 } \frac { \sin x } { x } = 1 \text {. } - lim⁡x→0sin⁡xcos⁡4xx+xcos⁡5x\lim _ { x \rightarrow 0 } \frac { \sin x \cos 4 x } { x + x \cos 5 x }


A) 0
B) 45\frac { 4 } { 5 }
C) 12\frac { 1 } { 2 }
D) does not exist

E) All of the above
F) B) and C)

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Use the graph to evaluate the limit. - lim⁡x→1f(x) \lim _ { x \rightarrow 1 } f ( x ) Use the graph to evaluate the limit. - \lim _ { x \rightarrow 1 } f ( x ) A) \frac { 3 } { 4 } B) \infty C) - 1 D) - \frac { 3 } { 4 }


A) 34\frac { 3 } { 4 }
B) ∞\infty
C) −1- 1
D) −34- \frac { 3 } { 4 }

E) All of the above
F) None of the above

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Use the graph to estimate the specified limit. -  Find lim⁡x→0f(x) \text { Find } \lim _ { x \rightarrow 0 } f ( x ) Use the graph to estimate the specified limit. - \text { Find } \lim _ { x \rightarrow 0 } f ( x ) A) does not exist B) -1 C) 0 D) 6


A) does not exist
B) -1
C) 0
D) 6

E) C) and D)
F) B) and C)

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Use the table to estimate the rate of change of y at the specified value of x. -x = 1. xy000.20.020.40.080.60.180.80.321.00.51.20.721.40.98\begin{array} { c | c } \mathrm { x } & \mathrm { y } \\\hline 0 & 0 \\0.2 & 0.02 \\0.4 & 0.08 \\0.6 & 0.18 \\0.8 & 0.32 \\1.0 & 0.5 \\1.2 & 0.72 \\1.4 & 0.98\end{array}


A) 0.5
B) 1
C) 2
D) 1.5

E) None of the above
F) A) and D)

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Find the limit if it exists. - lim⁡x→19x(x+4) (x−4) \lim _ { x \rightarrow 1 } 9 x ( x + 4 ) ( x - 4 )


A) -225
B) -135
C) -81
D) 135

E) A) and B)
F) A) and C)

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Use the table of values of f to estimate the limit. -  Let f(x) =x2−4x−5x2−7x+10, find lim⁡x→5f(x) . \text { Let } f(x) =\frac{x^{2}-4 x-5}{x^{2}-7 x+10} \text {, find } \lim _{x \rightarrow 5} f(x) \text {. } x4.94.994.9995.0015.015.1f(x) \begin{array}{c|l|l|l|l|l|l}\mathrm{x} & 4.9 & 4.99 & 4.999 & 5.001 & 5.01 & 5.1 \\\hline \mathrm{f}(\mathrm{x}) & & & & & &\end{array}


A)
x4.94.994.9995.0015.015.1f(x) 2.13452.10332.10032.09972.09672.0677 limit =2.1\begin{array}{c|cccccc}\mathrm{x} & 4.9 & 4.99 & 4.999 & 5.001 & 5.01 & 5.1 \\\hline \mathrm{f}(\mathrm{x}) & 2.1345 & 2.1033 & 2.1003 & 2.0997 & 2.0967 & 2.0677\end{array} \text { limit }=2.1

B)
x4.94.994.9995.0015.015.1f(x) 1.93451.90331.90031.89971.89671.8677 limit = 1.9\begin{array}{c|cccccc}\mathrm{x} & 4.9 & 4.99 & 4.999 & 5.001 & 5.01 & 5.1 \\\hline \mathrm{f}(\mathrm{x}) & 1.9345 & 1.9033 & 1.9003 & 1.8997 & 1.8967 & 1.8677\end{array} \text { limit = } 1.9

C)
x4.94.994.9995.0015.015.1f(x) 2.03452.00332.00031.99971.99671.9677 limit = 2 \begin{array}{c|cccccc}\mathrm{x} & 4.9 & 4.99 & 4.999 & 5.001 & 5.01 & 5.1 \\\hline \mathrm{f}(\mathrm{x}) & 2.0345 & 2.0033 & 2.0003 & 1.9997 & 1.9967 & 1.9677\end{array} \text { limit = 2 }

D)
x4.94.994.9995.0015.015.1f(x) 0.57750.57200.57150.57140.57080.5652 limit =0.5714\begin{array}{c|cccccc}\mathrm{x} & 4.9 & 4.99 & 4.999 & 5.001 & 5.01 & 5.1 \\\hline \mathrm{f}(\mathrm{x}) & 0.5775 & 0.5720 & 0.5715 & 0.5714 & 0.5708 & 0.5652\end{array} \text { limit }=0.5714

E) C) and D)
F) A) and C)

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Use a CAS to plot the function near the point x0 being approached. From your plot guess the value of the limit. - lim⁡x→01−x−1x\lim _ { x \rightarrow 0 } \frac { \sqrt { 1 - x } - 1 } { x }


A) 2
B) 12\frac { 1 } { 2 }
C) 1
D) −12- \frac { 1 } { 2 }

E) A) and B)
F) None of the above

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Find the limit LL for the given function ff , the point x0x _ { 0 } , and the positive number ε\varepsilon . Then find a number δ>0\delta > 0 such that, for all xt0<∣x−x0∣<δ⇒∣f(x) −L∣<εx _ { t } 0 < \left| x - x _ { 0 } \right| < \delta \Rightarrow | f ( x ) - L | < \varepsilon . - f(x) =18−x,L=4,x0=2, and ε=1f ( x ) = \sqrt { 18 - x } , L = 4 , x _ { 0 } = 2 , \text { and } \varepsilon = 1


A) -9
B) 9
C) 7
D) 12

E) C) and D)
F) B) and D)

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Provide an appropriate response. -Given ε>0\varepsilon > 0 , find an interval I=(5−δ,5) ,δ>0I = ( 5 - \delta , 5 ) , \delta > 0 , such that if xx lies in II , then 5−x<ε\sqrt { 5 - x } < \varepsilon . What limit is being verified and what is its value?


A) lim⁡x→5−x=5\lim _ { x \rightarrow 5 ^ { - } } \sqrt { x } = 5
B) lim⁡x−5+5−x=0\lim _ { x - 5 ^ { + } } \sqrt { 5 - x } = 0
C) lim⁡x→0−5−x=0\lim _ { x \rightarrow 0 ^ { - } } \sqrt { 5 - x } = 0
D) lim⁡x→5−5−x=0\lim _ { x \rightarrow 5 ^ { - } } \sqrt { 5 - x } = 0

E) C) and D)
F) All of the above

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Find the limit. - lim⁡x→2(x2+8x−2) \lim _ { x \rightarrow 2 } \left( x ^ { 2 } + 8 x - 2 \right)


A) does not exist
B) 0
C) 18
D) -18

E) A) and B)
F) C) and D)

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The graph below shows the number of tuberculosis deaths in the United States from 1989 to 1998. The graph below shows the number of tuberculosis deaths in the United States from 1989 to 1998. Estimate the average rate of change in tuberculosis deaths from 1993 to 1995. A) About -150 deaths per year B) About -300 deaths per year C) About -1 deaths per year D) About -80 deaths per year Estimate the average rate of change in tuberculosis deaths from 1993 to 1995.


A) About -150 deaths per year
B) About -300 deaths per year
C) About -1 deaths per year
D) About -80 deaths per year

E) A) and D)
F) A) and B)

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Prove the limit statement -You are asked to make some circular cylinders, each with a cross-sectional area of 9 cm29 \mathrm {~cm} ^ { 2 } . To do this, you need to know how much deviation from the ideal cylinder diameter of x0=1.93 cmx _ { 0 } = 1.93 \mathrm {~cm} you can allow and still have the area come within 0.1 cm20.1 \mathrm {~cm} ^ { 2 } of the required 9 cm29 \mathrm {~cm} ^ { 2 } . To find out, let A=π(x2) 2A = \pi \left( \frac { x } { 2 } \right) ^ { 2 } and look for the interval in which you must hold xx to make ∣A−9∣<0.1| \mathrm { A } - 9 | < 0.1 . What interval do you find?


A) (3.3663,3.4039) ( 3.3663,3.4039 )
B) (0.5642,0.5642) ( 0.5642,0.5642 )
C) (2.3803,2.4069) ( 2.3803,2.4069 )
D) (5.9666,6.0332) ( 5.9666,6.0332 )

E) A) and B)
F) All of the above

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Find all points where the function is discontinuous. -Find all points where the function is discontinuous. - A) x = 2 B) None C) x = -2, x = 2 D) x = -2


A) x = 2
B) None
C) x = -2, x = 2
D) x = -2

E) B) and D)
F) None of the above

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Use the table of values of f to estimate the limit. -  Let f(x) =sin⁡(6x) x, find lim⁡x→0f(x) x−0.1−0.01−0.0010.0010.010.1f(x) 5.996400655.99640065\begin{array}{l}\text { Let } f ( x ) = \frac { \sin ( 6 x ) } { x } , \text { find } \lim _ { x \rightarrow 0 } f ( x ) \\\begin{array} { l | c | c | c | c | c | c } x & - 0.1 & - 0.01 & - 0.001 & 0.001 & 0.01 & 0.1 \\\hline f ( x ) & & 5.99640065 & & & 5.99640065 &\end{array}\end{array}


A) limit = 6
B) limit = 5.5
C) limit does not exist
D) limit = 0

E) A) and C)
F) A) and B)

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