Calculus I - The Limit (Assignment Problems)

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Section 2.2 : The Limit

For the function $\displaystyle g\left( x \right) = \frac{{{x^2} + 6x + 9}}{{{x^2} + 3x}}$ answer each of the following questions.
1. Evaluate the function the following values of $x$ compute (accurate to at least 8 decimal places).
  1. -2.5
  2. -2.9
  3. -2.99
  4. -2.999
  5. -2.9999
  1. -3.5
  2. -3.1
  3. -3.01
  4. -3.001
  5. -3.0001
2. Use the information from (a) to estimate the value of $\displaystyle \mathop {\lim }\limits_{x \to \, - 3} \frac{{{x^2} + 6x + 9}}{{{x^2} + 3x}}$.
For the function $\displaystyle f\left( z \right) = \frac{{10z - 9 - {z^2}}}{{{z^2} - 1}}$ answer each of the following questions.
1. Evaluate the function the following values of $t$ compute (accurate to at least 8 decimal places).
  1. 1.5
  2. 1.1
  3. 1.01
  4. 1.001
  5. 1.0001
  1. 0.5
  2. 0.9
  3. 0.99
  4. 0.999
  5. 0.9999
2. Use the information from (a) to estimate the value of $\displaystyle \mathop {\lim }\limits_{z \to \,1} \frac{{10z - 9 - {z^2}}}{{{z^2} - 1}}$.
For the function $\displaystyle h\left( t \right) = \frac{{2 - \sqrt {4 + 2t} }}{t}$ answer each of the following questions.
1. Evaluate the function the following values of $t$ compute (accurate to at least 8 decimal places). Make sure your calculator is set to radians for the computations.
  1. 0.5
  2. 0.1
  3. 0.01
  4. 0.001
  5. 0.0001
  1. -0.5
  2. -0.1
  3. -0.01
  4. -0.001
  5. -0.0001
2. Use the information from (a) to estimate the value of $\displaystyle \mathop {\lim }\limits_{t \to \,0} \frac{{2 - \sqrt {4 + 2t} }}{t}$.
For the function $\displaystyle g\left( \theta \right) = \frac{{\cos \left( {\theta - 4} \right) - 1}}{{2\theta - 8}}$ answer each of the following questions.
1. Evaluate the function the following values of $\theta $ compute (accurate to at least 8 decimal places). Make sure your calculator is set to radians for the computations.
  1. 4.5
  2. 4.1
  3. 4.01
  4. 4.001
  5. 4.0001
  1. 3.5
  2. 3.9
  3. 3.99
  4. 3.999
  5. 3.9999
2. Use the information from (a) to estimate the value of $\displaystyle \mathop {\lim }\limits_{\theta \to \,4} \frac{{\cos \left( {\theta - 4} \right) - 1}}{{2\theta - 8}}$.
Below is the graph of $f\left( x \right)$. For each of the given points determine the value of $f\left( a \right)$ and $\mathop {\lim }\limits_{x \to a} f\left( x \right)$. If any of the quantities do not exist clearly explain why.
1. $a = - 2$
2. $a = - 1$
3. $a = 2$
4. $a = 3$
$This function has three parts to it. In the domain $-4 \le x <-1$ it is a parabola with a vertex at approximately (-2.2, 4.1) that opens down intersecting the x-axis at approximately (-3.8,0) and ends at (-1,1) in an open dot. Also in this portion there is an open dot at (-2,4) and a closed dot at (-2,2). In the domain $-1 \le x \le 2$ it is a parabola with vertex at approximately (0.25, 5) that opens upwards. The left end is a closed dot at (-1,-3) and the right end is a closed dot at (2,-1). The final portion is in the domain $2 < x \le 4$ and is a parabola with vertex at (3,5) which is indicated by a closed dot and opens downward. The left end of the parabola is an open dot at (2,2).$
Below is the graph of $f\left( x \right)$. For each of the given points determine the value of $f\left( a \right)$ and $\mathop {\lim }\limits_{x \to a} f\left( x \right)$. If any of the quantities do not exist clearly explain why.
1. $a = - 3$
2. $a = - 1$
3. $a = 1$
4. $a = 3$
$This function has three parts to it. In the domain $-5 \le x <-3$ it is a wavy curve that mostly decreases. As the curve approaches x =-3 from the left the curves decreases faster and faster the closer it gets to x=-3. In the domain $-3 \le x < 1$ the graph is a line that starts at (-3,5) in a closed dot and ends at (1,1) in an open dot. There is also an open dot at (-1,3). The final portion is in the domain $2 \le x < 4$ and is a parabola with vertex at approximately (1.8, -3.2) and opens upward The left end of the parabola is at (1,-2) and is a close dot. There is also a closed dot at (3,2).$
Below is the graph of $f\left( x \right)$. For each of the given points determine the value of $f\left( a \right)$ and $\mathop {\lim }\limits_{x \to a} f\left( x \right)$. If any of the quantities do not exist clearly explain why.
1. $a = - 4$
2. $a = - 2$
3. $a = 1$
4. $a = 4$
$This function has three parts to it. In the domain $-5 \le x < 1$ it is a curve that looks pretty much like a parabola with vertex at approximately (-2.5, 3.2) and opens downward. There is a closed dot at (-4,-2), and open dot at (-2,3), a closed dot at (-2,5) and the curve ends at an open dot at (1,-3). The second portion of this graph is in the domain $1 < x < 4$. It starts with an open dot at (1,4), there is a closed dot at (1,2). As the graph moves to the right it starts out fairly flat with a slight decrease but as it gets closer and closer to x=4 from the left the graph decreases faster and faster. The final portion of the graph is in the domain $4 < x < 6$. As the graph approaches x=4 from the right it increases faster and faster and as it moves away from x=4 the graph decreases until it reaches the end of the domain at z=6.$
Explain in your own words what the following equation means.
\[\mathop {\lim }\limits_{x \to 12} f\left( x \right) = 6\]
Suppose we know that $\mathop {\lim }\limits_{x \to \, - 7} f\left( x \right) = 18$. If possible, determine the value of $f\left( { - 7} \right)$. If it is not possible to determine the value explain why not.
Is it possible to have $\mathop {\lim }\limits_{x \to 1} f\left( x \right) = - 23$ and $f\left( 1 \right) = 107$? Explain your answer.