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High School Math : Calculus II — Integrals

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Example Question #1 :Using Limits With Continuity

Function

The above graph depicts a function h(x) . Does $\lim_{x\rightarrow 1} h(x)$ exist, and why or why not?

Possible Answers:

$\lim_{x\rightarrow 1} h(x)$ exists because $f (1 ) = \frac{\lim_{x\rightarrow 1^{-}} h(x) + \lim_{x\rightarrow 1^{+ }} h(x)}{2}$

$\lim_{x\rightarrow 1} h(x)$ does not exist because $f (1 ) \neq \lim_{x\rightarrow 1^{+}} h(x)$

$\lim_{x\rightarrow 1} h(x)$ exists because $\lim_{x\rightarrow 1^{-}} | h(x)| = \lim_{x\rightarrow 1^{+}} |h(x)|$

$\lim_{x\rightarrow 1} h(x)$ does not exist because $f (1 ) \neq \lim_{x\rightarrow 1^{-}} h(x)$

$\lim_{x\rightarrow 1} h(x)$ does not exist because $\lim_{x\rightarrow 1^{-}} h(x) \neq \lim_{x\rightarrow 1^{+}} h(x)$ .

Correct answer:

$\lim_{x\rightarrow 1} h(x)$ does not exist because $\lim_{x\rightarrow 1^{-}} h(x) \neq \lim_{x\rightarrow 1^{+}} h(x)$ .

Explanation:

$\lim_{x\rightarrow 1} h(x)$ exists if and only if $\lim_{x\rightarrow 1^{-}} h(x) = \lim_{x\rightarrow 1^{+}} h(x)$ . As can be seen from the diagram, $\lim_{x\rightarrow 1^{-}} h(x) = 1$ , but $\lim_{x\rightarrow 1^{+}} h(x) = -1$ . Since $\lim_{x\rightarrow 1^{-}} h(x) \neq \lim_{x\rightarrow 1^{+}} h(x)$ , $\lim_{x\rightarrow 1} h(x)$ does not exist.

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Example Question #1 :Calculus Ii — Integrals

Function

The above graph depicts a function f(x) . Does $\lim_{x\rightarrow 1} f(x)$ exist, and why or why not?

Possible Answers:

$\lim_{x\rightarrow 1} f(x)$ does not exist because $f(1) \neq \lim_{x\rightarrow 1^{+}} f(x)$

$\lim_{x\rightarrow 1} f(x)$ exists because $\lim_{x\rightarrow 1^{+}} f(x) = \lim_{x\rightarrow 1^{-}} f(x)$

$\lim_{x\rightarrow 1} f(x)$ does not exist becauseis not continuaous at x = 1 .

$\lim_{x\rightarrow 1} f(x)$ does not exist because $\lim_{x\rightarrow 1^{+}} f(x) \neq \lim_{x\rightarrow 1^{-}} f(x)$

$\lim_{x\rightarrow 1} f(x)$ does not exist because $f(1) \neq \lim_{x\rightarrow 1^{-}} f(x)$

Correct answer:

$\lim_{x\rightarrow 1} f(x)$ exists because $\lim_{x\rightarrow 1^{+}} f(x) = \lim_{x\rightarrow 1^{-}} f(x)$

Explanation:

$\lim_{x\rightarrow 1} f(x)$ exists if and only if $\lim_{x\rightarrow 1^{+}} f(x) = \lim_{x\rightarrow 1^{-}} f(x)$ ;

the actual value of f(1) is irrelevant, as is whetheris continuous there.

As can be seen,

$\lim_{x\rightarrow 1^{+}} f(x) = 1$ and $\lim_{x\rightarrow 1^{-}} f(x) = 1$ ;

therefore, $\lim_{x\rightarrow 1^{+}} f(x) = \lim_{x\rightarrow 1^{-}} f(x)$ ,

and $\lim_{x\rightarrow 1} f(x)$ exists.

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Example Question #2 :Using Limits With Continuity

A function is defined by the following piecewise equation:

$f(x)=\left\{\begin{matrix} x^{2}+4x-3, x<3\\10x-12, x\geq 3 \end{matrix}\right.$

At x=3 , the function is:

Possible Answers:

continuous

discontinuous

Correct answer:

continuous

Explanation:

The first step to determine continuity at a point is to determine if the function is defined at that point. When we substitute in 3 for, we get 18 as our-value. x=3 is thus defined for this function.

The next step is determine if the limit of the function is defined at that point. This means that the left-hand limit must be equal to the right-hand limit at x=3 . Substitution reveals the following:

$\lim_{x\rightarrow 3^{-}}f(x)=(3)^{2}+4(3)-3=9+12-3=18$

$\lim_{x\rightarrow 3^{+}}f(x)=10(3)-12=18$

Both sides of the function, therefore, approach a-value of 18.

Finally, we must ensure that the curve is smooth by checking the limit of the derivative of both sides.

$\lim_{x\rightarrow 3^{-}}f^{'}(x)=2(3)+4=10$

$\lim_{x\rightarrow 3^{+}}f^{'}(x)=10$

Since the function passes all three tests, it is continuous.

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Example Question #1 :Limits

Function

The graph depicts a function g(x) . Does $\lim_{x\rightarrow 1} g(x)$ exist?

Possible Answers:

$\lim_{x\rightarrow 1} g(x)$ does not exist because $\lim_{x\rightarrow 1^{-}} g(x) \neq \lim_{x\rightarrow 1^{+}} g(x)$ .

$\lim_{x\rightarrow 1} g(x)$ does not exist because g(1) is undefined.

$\lim_{x\rightarrow 1} g(x)$ does not exist becauseis not continuous at x = 1 .

$\lim_{x\rightarrow 1} g(x)$ exists because g(x) is constant on $(-\infty , 1)$ .

$\lim_{x\rightarrow 1} g(x)$ exists because $\lim_{x\rightarrow 1^{-}} g(x) = \lim_{x\rightarrow 1^{+}} g(x)$ .

Correct answer:

$\lim_{x\rightarrow 1} g(x)$ exists because $\lim_{x\rightarrow 1^{-}} g(x) = \lim_{x\rightarrow 1^{+}} g(x)$ .

Explanation:

$\lim_{x\rightarrow 1}g(x)$ exists if and only if $\lim_{x\rightarrow 1^{+}} g(x) = \lim_{x\rightarrow 1^{-}} g(x)$ ; the actual value of f(1) is irrelevant.

As can be seen, $\lim_{x\rightarrow 1^{+}} g(x) = 1$ and $\lim_{x\rightarrow 1^{-}}g(x) = 1$ ; therefore, $\lim_{x\rightarrow 1^{+}} g(x) = \lim_{x\rightarrow 1^{-}} g(x)$ , and $\lim_{x\rightarrow 1} g(x)$ exists.

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Example Question #1 :Understanding Polar Coordinates

The polar coordinates of a point are $\left(2.1, \frac{7\pi }{3}\right )$ . Give its-coordinate in the rectangular coordinate system (nearest hundredth).

Possible Answers:

4.20

1.05

-1.05

-1.82

1.82

Correct answer:

1.82

Explanation:

Given the polar coordinates $(r,\theta )$ , the-coordinate is $y= r \sin \theta$ . We can find this coordinate by substituting $r = 2.1, \theta = \frac{7\pi }{3}$ :

$y = r \sin \theta = 2.1 \cdot \sin \frac{7\pi }{3} \approx 2.1 \cdot 0.8660 \approx 1.82$

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Example Question #1 :Calculus Ii — Integrals

The polar coordinates of a point are $\left(2.1, \frac{7\pi }{3}\right )$ . Give its-coordinate in the rectangular coordinate system (nearest hundredth).

Possible Answers:

4.20

-1.05

1.05

1.82

-1.82

Correct answer:

1.05

Explanation:

Given the polar coordinates $(r,\theta )$ , the-coordinate is $x= r \cos \theta$ . We can find this coordinate by substituting $r = 2.1, \theta = \frac{7\pi }{3}$ :

$x = r \cos \theta = 2.1 \cdot \cos \frac{7\pi }{3} = 2.1 \cdot 0.5 = 1.05$

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Example Question #1 :Calculus Ii — Integrals

The polar coordinates of a point are $\left(1.2, \frac{2\pi }{5}\right )$ . Give its-coordinate in the rectangular coordinate system (nearest hundredth).

Possible Answers:

3.69

1.26

0.37

3.88

1.14

Correct answer:

1.14

Explanation:

Given the polar coordinates $(r,\theta )$ , the-coordinate is $y= r \sin \theta$ . We can find this coordinate by substituting $r = 1.2, \theta = \frac{2\pi }{5}$ :

$y = r \cos \theta = 1.2 \cdot \sin \frac{2\pi }{5} \approx 1.2 \cdot 0.9511 \approx 1.14$

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Example Question #1 :Understanding Polar Coordinates

The polar coordinates of a point are $\left(1.2, \frac{2\pi }{5}\right )$ . Give its-coordinate in the rectangular coordinate system (nearest hundredth).

Possible Answers:

3.69

3.88

0.37

1.14

1.26

Correct answer:

0.37

Explanation:

Given the polar coordinates $(r,\theta )$ , the-coordinate is $x= r \cos \theta$ . We can find this coordinate by substituting $r = 1.2, \theta = \frac{2\pi }{5}$ :

$x = r \cos \theta = 1.2 \cdot \cos \frac{2\pi }{5} \approx 1.2 \cdot 0.3090 \approx 0.37$

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Example Question #1 :Parametric, Polar, And Vector

Find the vector where its initial point is (-1,4) and its terminal point is (2,2) .

Possible Answers:

<1,6>

<1,-2>

<3,-2>

<1,2>

Correct answer:

<3,-2>

Explanation:

We need to subtract the-coordinate and the当考虑到坐标求解一个向量initial and terminal coordinates:

Initial pt: (-1,4)

Terminal pt: (2,2)

Vector: <2-(-1),2-4>

Vector:

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Example Question #1 :Understanding Vector Coordinates

Find the vector where its initial point is (-1,-5) and its terminal point is (2,3) .

Possible Answers:

$\left \langle3,8\right \rangle$

$\left \langle3,2\right \rangle$

$\left \langle1,2\right \rangle$

$\left \langle1,8\right \rangle$

Correct answer:

$\left \langle3,8\right \rangle$

Explanation:

We need to subtract the-coordinate and the-coordinate to solve for a vector when given its initial and terminal coordinates:

Initial pt: (-1,-5)

Terminal pt: (2,3)

Vector: $\left \langle 2-(-1),3-(-5)\right \rangle$

Vector: $\left \langle3,8\right \rangle$

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High School Math : Calculus II — Integrals

Study concepts, example questions & explanations for High School Math

All High School Math Resources

Example Questions

Example Question #1 :Using Limits With Continuity

Example Question #1 :Calculus Ii — Integrals

Example Question #2 :Using Limits With Continuity

Example Question #1 :Limits

Example Question #1 :Understanding Polar Coordinates

Example Question #1 :Calculus Ii — Integrals

Example Question #1 :Calculus Ii — Integrals

Example Question #1 :Understanding Polar Coordinates

Example Question #1 :Parametric, Polar, And Vector

Example Question #1 :Understanding Vector Coordinates

All High School Math Resources

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