Exam 2: Limits and Derivatives

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Find the limit. -If limx0f(x)x2=1\lim _ { x \rightarrow0} \frac { f ( x ) } { x ^ { 2 } } = 1 , find limx0f(x)x\lim _ { x \rightarrow 0 } \frac { f ( x ) } { x } .

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Give an appropriate answer. - f(x)=4x for x0=4f ( x ) = 4 \sqrt { x } \text { for } x _ { 0 } = 4

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Use the graph to evaluate the limit. - limx0f(x)\lim _ { x \rightarrow 0 } f ( x )  Use the graph to evaluate the limit. - \lim _ { x \rightarrow 0 } f ( x )  </sub>

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Give an appropriate answer. -Let limx9f(x)=4\lim _ { x \rightarrow 9 } f ( x ) = 4 . Find limx9(3)f(x)\lim _ { x \rightarrow 9 } ( - 3 ) ^ { f ( x ) } .

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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 {. } - limx1+(xx+9)(3x+8x2+9x)\lim _ { x \rightarrow 1 ^ { + } } \left( \frac { x } { x + 9 } \right) \left( \frac { - 3 x + 8 } { x ^ { 2 } + 9 x } \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 {. } - limx2+(x+5)(x+2x+2)\lim _ { x \rightarrow 2 ^ { + } } ( x + 5 ) \left( \frac { | x + 2 | } { x + 2 } \right)

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Give an appropriate answer. -  Let limx5f(x)=7 and limx5g(x)=10. Find limx5[f(x)g(x)]\text { Let } \lim _ { x \rightarrow 5 } f ( x ) = 7 \text { and } \lim _ { x \rightarrow 5 } g ( x ) = 10 \text {. Find } \lim _ { x \rightarrow 5 } [ f ( x ) \cdot g ( x ) ] \text {. }

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Find the average rate of change of the function over the given interval. - h(t)=sin(4t),[0,π8]h ( t ) = \sin ( 4 t ) , \left[ 0 , \frac { \pi } { 8 } \right]

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Give an appropriate answer. - f(x)=x3+4 for x0=4f ( x ) = \frac { x } { 3 } + 4 \text { for } x _ { 0 } = 4

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Use the table of values of f to estimate the limit. - Let f(\theta)= , find f(\theta) . -0.1 -0.01 -0.001 0.001 0.01 0.1 (\theta) -9.2106099 9.2106099

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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 {. } - limx0sin4xsin5x\lim _ { x \rightarrow 0 } \frac { \sin 4 x } { \sin 5 x }

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Determine the limit by sketching an appropriate graph. - limx4+f(x), where f(x)={x2+3 for x40 for x=4\lim _ { x \rightarrow 4 ^ { + } } f ( x ) , \text { where } f ( x ) = \left\{ \begin{array} { l l } x ^ { 2 } + 3 & \text { for } x \neq 4 \\0 & \text { for } x = 4\end{array} \right.  Determine the limit by sketching an appropriate graph. - \lim _ { x \rightarrow 4 ^ { + } } f ( x ) , \text { where } f ( x ) = \left\{ \begin{array} { l l }  x ^ { 2 } + 3 & \text { for } x \neq 4 \\ 0 & \text { for } x = 4 \end{array} \right.

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Find the limit. - limx2(8x+3)\lim _ { x \rightarrow 2 } ( 8 x + 3 )

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Find the limit. - limx2(3x53x44x3+x2+5)\lim _ { x \rightarrow 2 } \left( 3 x ^ { 5 } - 3 x ^ { 4 } - 4 x ^ { 3 } + x ^ { 2 } + 5 \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 {. } - limx 6(xx) \lim _{x\rightarrow\ 6^{-}}(x-\lfloor x\rfloor)

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Prove the limit statement - limx5(5x3)=22\lim _ { x \rightarrow 5 } ( 5 x - 3 ) = 22

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Provide an appropriate response. - a=39,b=99,x0=49a = \frac { 3 } { 9 } , b = \frac { 9 } { 9 } , x _ { 0 } = \frac { 4 } { 9 }

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Provide an appropriate response. -The inequality 1x22<sinxx<11 - \frac { x ^ { 2 } } { 2 } < \frac { \sin x } { x } < 1 holds when xx is measured in radians and x<1| x | < 1 . Find limxθsinxx\lim _ { x \rightarrow \theta } \frac { \sin x } { x } if it exists.

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limx0f(x)\lim _ { x \rightarrow 0 } f ( x ) \lim _ { x \rightarrow 0 } f ( x )

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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. - limx036+2x6x\lim _ { x \rightarrow 0 } \frac { \sqrt { 36 + 2 x } - 6 } { x }

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