28. Determine where [tex]$\frac{2 x-3}{5 x+4}$[/tex] is not continuous.
A. [tex]$\frac{5}{4}$[/tex]
B. [tex]$\frac{1}{5}$[/tex]
C. [tex]$-\frac{5}{4}$[/tex]
D. [tex]$-\frac{1}{5}$[/tex]
29. Evaluate [tex]$\lim _{x \rightarrow 0} \frac{\sin 3 x}{x}$[/tex].
A. 1
B. 3
C. [tex]$\frac{1}{3}$[/tex]
D. 0
30. The vertical asymptotes of [tex]$\frac{1 x^2+8 x}{x-4}$[/tex] is at [tex]$x=$[/tex] ?
A. 2
B. -2
C.
D. 0
31. Find the vertical asymptotes of [tex]$f(x)=\frac{2 x^3+9}{(x-4)^2}$[/tex].
A. [tex]$x=2$[/tex]
B. [tex]$x=-2$[/tex]
C. [tex]$x=4$[/tex]
D. [tex]$x=\frac{1}{2}$[/tex].
32. What is the vertical asymptote of [tex]$\frac{9}{x^2-9}$[/tex] ?
A. [tex]$x=9$[/tex]
B. [tex]$x=-3$[/tex] and [tex]$x=-3$[/tex]
C. [tex]$x=-3$[/tex] and [tex]$x=3$[/tex]
D. [tex]$x=9$[/tex].
33. Evaluate [tex]$\lim _{x \rightarrow \infty} \frac{3-4 x^2}{2 x^2-x-2}$[/tex]
A. -2
B. 1
C. [tex]$\frac{3}{2}$[/tex]
D. does not exists
34. The following are turning points except.
A. Maximum point.
B. Minimum point.
C. Inflexion point
D. Non-stationary point
35. Evaluate [tex]$\lim _{x \rightarrow 1^{+}} \frac{5}{x-1}$[/tex].
A. 0
B. [tex]$+\infty$[/tex]
C. [tex]$\frac{5}{2}$[/tex]
D. -5
36. Evaluate [tex]$\lim _{x \rightarrow 0} \frac{1-\cos x^2}{x}$[/tex].
A. 0
B. 1
C. [tex]$+\infty$[/tex]
D. -1
37. Compute [tex]$\lim _{x \rightarrow 0}\left(\frac{1}{x}-\frac{1}{x^2+x}\right)$[/tex]
A. Does not exists
B. 0
C. 1
D. 2
Answer (2)
Question 28: Identifies the point of discontinuity for a rational function by finding where the denominator equals zero, resulting in − 5 4 .
Question 29: Evaluates the limit using L'Hopital's rule, giving 3 .
Question 30: Determines the vertical asymptote of a rational function by setting the denominator to zero, yielding 4 .
Question 31: Finds the vertical asymptote of the function by setting the denominator to zero, resulting in 4 .
Question 32: Locates the vertical asymptotes by finding the zeros of the denominator, giving x = − 3 and x = 3 .
Question 33: Computes the limit at infinity by dividing by the highest power of x , resulting in − 2 .
Question 34: Recognizes that a non-stationary point is not necessarily a turning point.
Question 35: Evaluates the limit as x approaches 1 from the right, resulting in + ∞ .
Question 36: Applies L'Hopital's rule to find the limit, giving 0 .
Question 37: Simplifies the expression and evaluates the limit as x approaches 0, resulting in 1 .
Explanation
Introduction We will solve each multiple-choice question step-by-step, providing the correct answer and a brief explanation.
Question 28 Solution Question 28: The function 5 x + 4 2 x − 3 is not continuous where the denominator is zero. So, we solve 5 x + 4 = 0 , which gives x = − 5 4 . Therefore, the answer is C.
Question 29 Solution Question 29: To evaluate lim x → 0 x s i n 3 x , we can use L'Hopital's rule or the small angle approximation. Using L'Hopital's rule, we differentiate the numerator and denominator to get lim x → 0 1 3 c o s 3 x = 3 . Therefore, the answer is B.
Question 30 Solution Question 30: Vertical asymptotes of x − 4 x 2 + 8 x occur where the denominator is zero and the numerator is non-zero. Setting x − 4 = 0 , we get x = 4 . At x = 4 , the numerator is 4 2 + 8 ( 4 ) = 16 + 32 = 48 = 0 . Therefore, the answer is C.
Question 31 Solution Question 31: Vertical asymptotes of f ( x ) = ( x − 4 ) 2 2 x 3 + 9 occur where the denominator is zero and the numerator is non-zero. Setting ( x − 4 ) 2 = 0 , we get x = 4 . At x = 4 , the numerator is 2 ( 4 ) 3 + 9 = 2 ( 64 ) + 9 = 128 + 9 = 137 = 0 . Therefore, the answer is C.
Question 32 Solution Question 32: Vertical asymptotes of x 2 − 9 9 occur where the denominator is zero. Setting x 2 − 9 = 0 , we get x 2 = 9 , so x = ± 3 . Therefore, the answer is C.
Question 33 Solution Question 33: To evaluate lim x → ∞ 2 x 2 − x − 2 3 − 4 x 2 , we divide the numerator and denominator by x 2 to get lim x → ∞ 2 − x 1 − x 2 2 x 2 3 − 4 . As x → ∞ , the terms with x in the denominator go to zero, so the limit is 2 − 4 = − 2 . Therefore, the answer is A.
Question 34 Solution Question 34: Turning points are maximum points, minimum points, and inflection points. A non-stationary point is not necessarily a turning point. Therefore, the answer is D.
Question 35 Solution Question 35: To evaluate lim x → 1 + x − 1 5 , as x approaches 1 from the right, x − 1 approaches 0 from the positive side. Thus, x − 1 5 approaches + ∞ . Therefore, the answer is B.
Question 36 Solution Question 36: To evaluate lim x → 0 x 1 − c o s x 2 , we can use L'Hopital's rule. Differentiating the numerator and denominator, we get lim x → 0 1 2 x s i n x 2 = 0 . Therefore, the answer is A.
Question 37 Solution Question 37: To compute lim x → 0 ( x 1 − x 2 + x 1 ) , we first simplify the expression: x 1 − x ( x + 1 ) 1 = x ( x + 1 ) x + 1 − 1 = x ( x + 1 ) x = x + 1 1 . Then, we evaluate the limit as x approaches 0: lim x → 0 x + 1 1 = 1 . Therefore, the answer is C.
Final Answers Here are the final answers: 28: C, 29: B, 30: C, 31: C, 32: C, 33: A, 34: D, 35: B, 36: A, 37: C
Examples
Understanding continuity, limits, and asymptotes is crucial in various fields. For example, in physics, analyzing the motion of an object often involves understanding the limits of its velocity and acceleration. In economics, understanding limits can help predict market behavior as certain parameters approach extreme values. These concepts also form the foundation for more advanced topics in engineering and computer science.
The answers to the questions are as follows: 28 is C, 29 is B, 30 is C, 31 is C, 32 is C, 33 is A, 34 is D, 35 is B, 36 is A, and 37 is C.
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