Exam 2: Limits and Derivatives

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Find the limit if it exists. - limx3(4x4)\lim _ { x \rightarrow 3 } ( 4 x - 4 )

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Use the table of values of f to estimate the limit. -  Let f(x)=x4x2, find limx4f(x)\text { Let } f(x)=\frac{x-4}{\sqrt{x}-2} \text {, find } \lim _{x \rightarrow 4} f(x) 3.9 3.99 3.999 4.001 4.01 4.1 ()

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Find the limit, if it exists. - limhθ23h+4+2\lim _ { h \rightarrow \theta } \frac { 2 } { \sqrt { 3 h + 4 } + 2 }

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Use the table to estimate the rate of change of y at the specified value of x. -x = 1. 0 0 0.2 0.12 0.4 0.48 0.6 1.08 0.8 1.92 1.0 3 1.2 4.32 1.4 5.88

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Provide an appropriate response. - a=1.317, b=2.713,x0=1.976\mathrm { a } = 1.317 , \mathrm {~b} = 2.713 , \mathrm { x } _ { 0 } = 1.976

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Give an appropriate answer. -Let limx8f(x)=32\lim _ { x \rightarrow 8 } f ( x ) = 32 . Find limx8f(x)5\lim _ { x \rightarrow 8 } \sqrt [ 5 ] { f ( x ) } .

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Provide an appropriate response. -Given limx0f(x)=Ll,limx0+f(x)=Lr\lim _ { x \rightarrow 0^ { - } } \mathrm { f } ( \mathrm { x } ) = \mathrm { L } _ { \mathrm { l } } , \lim _ { \mathrm { x } \rightarrow 0 ^ { + } } \mathrm { f } ( \mathrm { x } ) = \mathrm { L } _ { \mathrm { r } } , and Ll=Lr\mathrm { L } _ { \mathrm { l } } = \mathrm { L } _ { \mathrm { r } } , which of the following statements is false? I. limx0f(x)=L1\lim _ { x \rightarrow 0 } f ( x ) = L _ { 1 } II. limx0f(x)=Lr\lim _ { x \rightarrow0 } f ( x ) = L _ { r } III. limx0f(x)\lim _ { x \rightarrow 0} f ( x ) does not exist.

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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. - limx366x36x\lim _ { x \rightarrow 36 } \frac { 6 - \sqrt { x } } { 36 - x }

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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<xx0<δf(x)L<εx _ { t } 0 < \left| x - x _ { 0 } \right| < \delta \Rightarrow | f ( x ) - L | < \varepsilon . -  Find the limit  L  for the given function  f , the point  x _ { 0 } , and the positive number  \varepsilon . Then find a number  \delta > 0  such that, for all  x _ { t } 0 < \left| x - x _ { 0 } \right| < \delta \Rightarrow | f ( x ) - L | < \varepsilon . -

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Give an appropriate answer. -Let limx9f(x)=49\lim _ { x \rightarrow -9 } f ( x ) = 49 . Find limx9f(x)\lim _ { x \rightarrow -9 } \sqrt { f ( x ) } .

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Find all points where the function is discontinuous. -Does limf(x)\lim f ( x ) exist? f(x)={x2+1,1x<02x,0<x<14,x=12x+41<x<35,3<x<5f ( x ) = \left\{ \begin{array} { l l } - x ^ { 2 } + 1 , & - 1 \leq x < 0 \\2 x , & 0 < x < 1 \\- 4 , & x = 1 \\- 2 x + 4 & 1 < x < 3 \\5 , & 3 < x < 5\end{array} \right.  Find all points where the function is discontinuous. -Does  \lim f ( x )  exist?  f ( x ) = \left\{ \begin{array} { l l }  - x ^ { 2 } + 1 , & - 1 \leq x < 0 \\ 2 x , & 0 < x < 1 \\ - 4 , & x = 1 \\ - 2 x + 4 & 1 < x < 3 \\ 5 , & 3 < x < 5 \end{array} \right.

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Use the table to estimate the rate of change of y at the specified value of x. -x = 1. x y 0 0 0.2 0.01 0.4 0.04 0.6 0.09 0.8 0.16 1.0 0.25 1.2 0.36 1.4 0.49

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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<xx0<δf(x)L<εx _ { t } 0 < \left| x - x _ { 0 } \right| < \delta \Rightarrow | f ( x ) - L | < \varepsilon . -  Find the limit  L  for the given function  f , the point  x _ { 0 } , and the positive number  \varepsilon . Then find a number  \delta > 0  such that, for all  x _ { t } 0 < \left| x - x _ { 0 } \right| < \delta \Rightarrow | f ( x ) - L | < \varepsilon . -

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Find the slope of the curve at the given point P and an equation of the tangent line at P. - y=4x3,(1,5)y = 4 - x ^ { 3 } , ( - 1,5 )

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Find all points where the function is discontinuous. -Find all points where the function is discontinuous. -

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 Find the limit using limx=0sinxx=1\text { Find the limit using } \lim _ { x = 0 } \frac { \sin x } { x } = 1 \text {. } - limx0xsin3x\lim _ { x \rightarrow0 } \frac { x } { \sin 3 x }

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Determine the limit by sketching an appropriate graph. - limx3f(x), where f(x)={4x+1 for x<32x+2 for x3\lim _ { x \rightarrow 3 ^ { - } } f ( x ) \text {, where } f ( x ) = \left\{ \begin{array} { l l } - 4 x + 1 & \text { for } x < 3 \\2 x + 2 & \text { for } x \geq 3\end{array} \right.  Determine the limit by sketching an appropriate graph. - \lim _ { x \rightarrow 3 ^ { - } } f ( x ) \text {, where } f ( x ) = \left\{ \begin{array} { l l }  - 4 x + 1 & \text { for } x < 3 \\ 2 x + 2 & \text { for } x \geq 3 \end{array} \right.

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 Find the limit using limx=0sinxx=1\text { Find the limit using } \lim _ { x = 0 } \frac { \sin x } { x } = 1 \text {. } - limx0sin3xcot4xcot5x\lim _ { x \rightarrow 0 } \frac { \sin 3 x \cot 4 x } { \cot 5 x }

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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<xx0<δf(x)L<εx _ { t } 0 < \left| x - x _ { 0 } \right| < \delta \Rightarrow | f ( x ) - L | < \varepsilon . - f(x)=3x2, L=243,x0=9, and ε=0.5\mathrm { f } ( \mathrm { x } ) = 3 \mathrm { x } ^ { 2 } , \mathrm {~L} = 243 , \mathrm { x } _ { 0 } = 9 \text {, and } \varepsilon = 0.5

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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. - limx06+6x6x\lim _ { x \rightarrow 0 } \frac { \sqrt { 6 + 6 x } - \sqrt { 6 } } { x }

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