Exam 3: Limits and Continuity

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Find the limit. - limx31x29\lim _ { x \rightarrow 3 ^ { - } } \frac { 1 } { x ^ { 2 } - 9 }

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Find the limit and determine if the function is continuous at the point being approached. - limθ2πtan(πcos(sinθ))\lim _ { \theta \rightarrow 2 \pi } \tan ( \pi \cos ( \sin \theta ) )

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Answer the question. -Is f\mathrm { f } continuous at x=4?x = 4 ? f(x)={x3,2<x02x,0x<22,2<x40,x=2\mathrm { f } ( \mathrm { x } ) = \left\{ \begin{array} { l l } \mathrm { x } ^ { 3 } , & - 2 < \mathrm { x } \leq 0 \\- 2 \mathrm { x } , & 0 \leq \mathrm { x } < 2 \\2 , & 2 < \mathrm { x } \leq 4 \\0 , & \mathrm { x } = 2\end{array} \right.  Answer the question. -Is  \mathrm { f }  continuous at  x = 4 ?   \mathrm { f } ( \mathrm { x } ) = \left\{ \begin{array} { l l }  \mathrm { x } ^ { 3 } , & - 2 < \mathrm { x } \leq 0 \\ - 2 \mathrm { x } , & 0 \leq \mathrm { x } < 2 \\ 2 , & 2 < \mathrm { x } \leq 4 \\ 0 , & \mathrm { x } = 2 \end{array} \right.

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Find the intervals on which the function is continuous. - y=x210y = \sqrt { x ^ { 2 } - 10 }

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Find numbers a and b, or k, so that f is continuous at every point. - f(x)={1,x<2ax+b,2x31,x>3f ( x ) = \left\{ \begin{array} { l l } 1 , & x < 2 \\a x + b , & 2 \leq x \leq 3 \\- 1 , & x > 3\end{array} \right.

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Find the limit and determine if the function is continuous at the point being approached. - limxπ/2cos(5π2cos(tanx))\lim _ { x \rightarrow \pi / 2 } \cos \left( \frac { 5 \pi } { 2 } \cos ( \tan x ) \right)

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Because of their connection with secant lines, tangents, and instantaneous rates, limits of the form limh0f(x+h)f(x)h\lim _ { h \rightarrow 0 } \frac { f ( x + h ) - f ( x ) } { h } occur frequently in calculus. Evaluate this limit for the given value of xx and function ff . - (x)=5x2+4,x=4( x ) = 5 x ^ { 2 } + 4 , x = - 4

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

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

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Find the limit. - limx1x231x\lim _ { x \rightarrow 1 } \frac { x ^ { 2 } } { 3 } - \frac { 1 } { x }

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Divide numerator and denominator by the highest power of x in the denominator to find the limit. - limx3x12x35x2+x5\lim _ { x \rightarrow \infty } \frac { 3 x ^ { - 1 } - 2 x ^ { - 3 } } { 5 x ^ { - 2 } + x ^ { - 5 } }

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Graph the rational function. Include the graphs and equations of the asymptotes. - y=x2+1x3y = \frac { x ^ { 2 } + 1 } { x ^ { 3 } }  Graph the rational function. Include the graphs and equations of the asymptotes. - y = \frac { x ^ { 2 } + 1 } { x ^ { 3 } }

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Provide an appropriate response. -Let limx1f(x)=7\lim _ { x \rightarrow 1 } f ( x ) = 7 and limx1g(x)=8\lim _ { x \rightarrow 1 } g ( x ) = - 8 . Find limx1[2f(x)4g(x)3+g(x)]\lim _ { x \rightarrow 1 } \left[ \frac { - 2 f ( x ) - 4 g ( x ) } { 3 + g ( x ) } \right] .

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Use the graph to evaluate the limit. - limxθf(x)\lim _{x-\theta} f(x)  Use the graph to evaluate the limit. - \lim _{x-\theta} f(x)

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Find the limit if it exists. - limx65x(x+7)(x5)\lim _ { x \rightarrow 6 } 5 x ( x + 7 ) ( x - 5 )

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Because of their connection with secant lines, tangents, and instantaneous rates, limits of the form limh0f(x+h)f(x)h\lim _ { h \rightarrow 0 } \frac { f ( x + h ) - f ( x ) } { h } occur frequently in calculus. Evaluate this limit for the given value of xx and function ff . - (x)=2x,x=5( x ) = \frac { 2 } { x } , x = - 5

(Multiple Choice)
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Determine if the given function can be extended to a continuous function at x = 0. If so, approximate the extended function's value at x = 0 (rounded to four decimal places if necessary). If not, determine whether the function can be continuously extended from the left or from the right and provide the values of the extended functions at x = 0. Otherwise write "no continuous extension." - f(x)=tanxxf ( x ) = \frac { \tan x } { x }

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Find the limit. - limx2(x3+5x27x+1)\lim _ { x \rightarrow 2 } \left( x ^ { 3 } + 5 x ^ { 2 } - 7 x + 1 \right)

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Find the limit, if it exists. - limx+x24x+3x25x+4\lim _ { x \rightarrow + } \frac { x ^ { 2 } - 4 x + 3 } { x ^ { 2 } - 5 x + 4 }

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Divide numerator and denominator by the highest power of x in the denominator to find the limit. - limx3x+x14x2\lim _ { x \rightarrow \infty } \frac { 3 \sqrt { x } + x ^ { - 1 } } { 4 x - 2 }

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