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Continuity Ultimate Exam
Question 1. Which of the following best describes the meaning of (\displaystyle lim_{x\to c} f(x)=L)? A) (f(c)=L) for all (c) B) As (x) approaches (c), the values of (f(x)) get arbitrarily close to (L) C) The function (f) is continuous at (c) D) The derivative of (f) at (c) equals (L) Answer: B Explanation: The limit statement asserts that when (x) gets arbitrarily close to (c) (but not necessarily equal to (c)), the function values approach (L).
Question 2. In the limit notation (\displaystyle\lim_{x\to c^-} f(x)), the superscript “(-)” indicates: A) Approach from the right side of (c) B) Approach from the left side of (c) C) The limit is negative D) The function is decreasing near (c) Answer: B Explanation: The minus sign denotes the left-hand limit, i.e., (x) approaches (c) through values less than (c).
Question 3. Which of the following statements guarantees that (\displaystyle lim_{x\to c} f(x)) does not exist? A) (\displaystyle\lim_{x\to c^-} f(x)=\lim_{x\to c^+} f(x)=5) B) (\displaystyle\lim_{x\to c^-} f(x)=3) and (\displaystyle\lim_{x\to c^+} f(x)=3)
Continuity Ultimate Exam
C) (\displaystyle\lim_{x\to c^-} f(x)=2) and (\displaystyle\lim_{x\to c^+} f(x)=7) D) The function is undefined at (c) but both one-sided limits equal 4 Answer: C Explanation: A limit exists only when the left-hand and right-hand limits are equal. Different one-sided limits mean the overall limit does not exist.
Question 4. When evaluating (\displaystyle\lim_{x\to 2}\frac{x^2-4}{x-2}) by direct substitution, you obtain (0/0). Which algebraic technique should be applied first? A) Multiply numerator and denominator by the conjugate B) Factor the numerator and cancel the common factor C) Apply L’Hôpital’s Rule D) Use a table of values Answer: B Explanation: Factoring (x^2-4=(x-2)(x+2)) allows cancellation of the ((x-2)) factor, eliminating the indeterminate form.
Question 5. After simplifying (\displaystyle\frac{x^2-4}{x-2}) and evaluating the limit as (x\to2), the result is: A) 0 B) 2 C) 4 D) Does not exist
Continuity Ultimate Exam
Question 8. The limit (\displaystyle\lim_{x\to0}\frac{\sin x}{x}) equals: A) 0 B) 1 C) (\infty) D) Does not exist Answer: B Explanation: This is a fundamental trigonometric limit; the ratio approaches 1 as (x) (in radians) approaches 0.
Question 9. Which theorem justifies that (\displaystyle\lim_{x\to0}\frac{ sin(3x)}{x}=3)? A) Squeeze Theorem B) Intermediate Value Theorem C) Limit Law for constants D) Mean Value Theorem Answer: C Explanation: (\frac{\sin(3x)}{x}=3\cdot\frac{\sin(3x)}{3x}). The inner ratio tends to 1, so the overall limit is (3\cdot1=3).
Question 10. A function (f) is said to be continuous at (c) if which three conditions are satisfied? A) (f(c)) exists, (\displaystyle\lim_{x\to c}f(x)) exists, and the derivative at (c) exists
Continuity Ultimate Exam
B) (\displaystyle\lim_{x\to c^-}f(x)=\displaystyle\lim_{x\to c^+}f(x)) only C) (f(c)) exists, (\displaystyle\lim_{x\to c}f(x)) exists, and (\displaystyle\lim_{x\to c}f(x)=f(c)) D) The function is differentiable at (c) Answer: C Explanation: Continuity requires the function to be defined at the point, the limit to exist, and the limit to equal the function value.
Question 11. Which of the following points is a removable discontinuity of (g(x)=\frac{x^2-9}{x-3})? A) (x=3) B) (x=-3) C) (x=0) D) No discontinuities Answer: A Explanation: At (x=3) the denominator is zero, but the numerator also zero, allowing cancellation of ((x-3)). The “hole” can be filled, making it removable.
Question 12. After removing the removable discontinuity from (g(x)=\frac{x^2- 9}{x-3}), the resulting continuous function is: A) (x+3) B) (x-3) C) (\frac{x+3}{x-3}) D) (x^2-9)
Continuity Ultimate Exam
Question 15. Determine whether the function (h(x)=\frac{1}{x}) is continuous on the interval ((-\infty,0)). A) Yes, because it is differentiable there B) No, because it has a hole at (x=0) C) Yes, because the domain excludes the point of discontinuity D) No, because it is not defined for negative (x) Answer: C Explanation: Continuity is considered only on points where the function is defined. On ((-\infty,0)) the function has no breaks, so it is continuous there.
Question 16. Which of the following statements about polynomial functions is correct? A) They are continuous everywhere except at integer points B) They may have removable discontinuities C) They are continuous on their entire domain, which is (\mathbb{R}) D) They always have vertical asymptotes Answer: C Explanation: Polynomials are defined for all real numbers and have no breaks, holes, or asymptotes; thus they are continuous on (\mathbb{R}).
Continuity Ultimate Exam
Question 17. Evaluate (\displaystyle\lim_{x\to\infty}\frac{5x^3-2x} {2x^3+7}). A) (\frac{5}{2}) B) 0 C) (\infty) D) Does not exist Answer: A Explanation: Divide numerator and denominator by (x^3); the dominant terms give (\frac{5}{2}).
Question 18. For the rational function (r(x)=\frac{2x^2+3}{x^2-4}), the horizontal asymptote as (x\to\pm\infty) is: A) (y=0) B) (y=2) C) (y=1) D) No horizontal asymptote Answer: B Explanation: Degrees of numerator and denominator are equal; the asymptote is the ratio of leading coefficients (2/1=2).
Question 19. Which of the following functions grows the fastest as (x\to\infty)? A) (x^5) B) (e^{2x}) C) (\log x)
Continuity Ultimate Exam
Explanation: Since (-1\le\sin(1/x)\le1), multiplying by (x^2) gives (-x^2\le x^2\sin(1/x)\le x^2). Both bounding functions tend to 0, so the squeezed function also tends to 0.
Question 22. Which of the following is a necessary condition for the Intermediate Value Theorem (IVT) to apply on ([a,b])? A) The function must be differentiable on ([a,b]) B) The function must be continuous on ([a,b]) C) The function must be monotonic on ([a,b]) D) The function must have a derivative equal to zero somewhere in ((a,b)) Answer: B Explanation: IVT requires continuity on the closed interval; differentiability or monotonicity are not required.
Question 23. Suppose (f) is continuous on ([1,4]) and (f(1)=-2), (f(4)=5). Which of the following statements is guaranteed by the IVT? A) There exists (c\in(1,4)) such that (f(c)=0) B) (f) attains a maximum at (x=1) C) (f) is increasing on ([1,4]) D) The derivative (f'(c)=0) for some (c\in(1,4)) Answer: A Explanation: Since 0 lies between (-2) and (5), continuity ensures the function takes the value 0 somewhere in the interval.
Continuity Ultimate Exam
Question 24. Evaluate (\displaystyle\lim_{x\to0}\frac{1-\cos x}{x^2}). A) 0 B) (\frac{1}{2}) C) 1 D) (\infty) Answer: B Explanation: Using the standard limit (\displaystyle\lim_{x\to0}\frac{1-\cos x} {x^2}=\frac{1}{2}) (derived from the series or trig identity).
Question 25. Which algebraic manipulation is appropriate for (\displaystyle lim_{x\to4}\frac{\sqrt{x}-2}{x-4})? A) Multiply numerator and denominator by (\sqrt{x}+2) (conjugate) B) Factor denominator as ((x-2)(x+2)) C) Apply L’Hôpital’s Rule directly D) Substitute (x=4) Answer: A Explanation: The expression is a 0/0 indeterminate; rationalizing the numerator removes the square root.
Question 26. After rationalizing, the limit (\displaystyle\lim_{x\to4}\frac{ sqrt{x}-2}{x-4}) evaluates to: A) (\frac{1}{4}) B) (\frac{1}{8})
Continuity Ultimate Exam
Explanation: Polynomials are continuous everywhere; thus the limit exists at every real point.
Question 29. Consider the piecewise function [ p(x)=\begin{cases} x^2 & x<1\ 3 & x=1\ 2x-1 & x> \end{cases} ] Which statement correctly describes the continuity at (x=1)? A) Continuous, because left and right limits equal 3 B) Removable discontinuity, because the limit exists but differs from (p(1)) C) Jump discontinuity, because left and right limits are different D) Infinite discontinuity, because the function blows up Answer: C Explanation: (\displaystyle\lim_{x\to1^-}p(x)=1) and (\displaystyle\lim_{x to1^+}p(x)=1). Actually compute: left limit (=1^2=1); right limit (=2(1)-1=1); both equal 1, but (p(1)=3). So the limit exists (1) but does not equal the function value, giving a removable discontinuity. Correction: The correct answer is B. Answer: B Explanation: The one-sided limits both equal 1, so the overall limit exists and equals 1. Since (p(1)=3\neq1), the discontinuity is removable (a “hole” that could be filled by redefining (p(1)=1)).
Continuity Ultimate Exam
Question 30. Which of the following limits does not exist due to oscillation? A) (\displaystyle\lim_{x\to0}\frac{\sin x}{x}) B) (\displaystyle\lim_{x\to0}\sin\frac{1}{x}) C) (\displaystyle\lim_{x\to\infty}\frac{1}{x}) D) (\displaystyle\lim_{x\to2}\frac{x-2}{x-2}) Answer: B Explanation: As (x\to0), (\frac{1}{x}) swings infinitely fast, causing (\sin(1/x)) to oscillate between –1 and 1 without settling.
Question 31. The function (f(x)=\frac{x^2-1}{x-1}) has a hole at (x=1). After removing the hole, the simplified expression is: A) (x+1) B) (x-1) C) (\frac{x+1}{x-1}) D) (x^2-1) Answer: A Explanation: Factor numerator ((x-1)(x+1)) and cancel ((x-1)); the resulting function is (x+1) for (x\neq1).
Question 32. Which of the following statements about limits at infinity is true? A) (\displaystyle\lim_{x\to\infty}\frac{1}{x}=1) B) If the degrees of numerator and denominator are equal, the limit equals the ratio of leading coefficients
Continuity Ultimate Exam
Answer: C Explanation: As (x) approaches 0 from either side, (1/x^2) grows without bound, giving (+\infty).
Question 35. Which of the following limits can be evaluated directly by substitution without any algebraic manipulation? A) (\displaystyle\lim_{x\to2}\frac{x^2-4}{x-2}) B) (\displaystyle\lim_{x\to5}\sqrt{x+4}) C) (\displaystyle\lim_{x\to0}\frac{\sin x}{x}) D) (\displaystyle\lim_{x\to-1}\frac{x+1}{x^2-1}) Answer: B Explanation: Substituting (x=5) into (\sqrt{x+4}) gives (\sqrt{9}=3); no indeterminate form arises.
Question 36. The limit (\displaystyle\lim_{x\to0}\frac{e^{2x}-1}{x}) equals: A) 0 B) 1 C) 2 D) (\infty) Answer: C Explanation: Using the limit (\displaystyle\lim_{x\to0}\frac{e^{kx}-1}{x}=k). Here (k=2), so the limit is 2.
Continuity Ultimate Exam
Question 37. Which of the following functions is not continuous at (x=0)? A) (f(x)=\sin x) B) (g(x)=\frac{x}{|x|}) for (x\neq0) and (g(0)=0) C) (h(x)=\ln(x+1)) D) (p(x)=x^3) Answer: B Explanation: For (x>0), (g(x)=1); for (x<0), (g(x)=-1). The left-hand and right-hand limits differ, so continuity fails at 0.
Question 38. The limit (\displaystyle\lim_{x\to\infty}\left(1+\frac{1}{x} right)^x) equals: A) 0 B) 1 C) (e) D) (\infty) Answer: C Explanation: This is the classic definition of the constant (e).
Question 39. If (\displaystyle\lim_{x\to a} f(x)=L) and (\displaystyle\lim_{x\to a} g(x)=M), then (\displaystyle\lim_{x\to a} [f(x)+g(x)]) equals: A) (L+M) B) (LM) C) (\frac{L}{M})
Continuity Ultimate Exam
Question 42. Which of the following is a correct statement about vertical asymptotes? A) They occur where the numerator of a rational function is zero B) They are points where the limit of the function is infinite C) They exist only for polynomial functions D) They are always at (x=0) Answer: B Explanation: A vertical asymptote at (x=c) means (\displaystyle\lim_{x\to c^ pm} f(x)=\pm\infty).
Question 43. Evaluate (\displaystyle\lim_{x\to0}\frac{\ln(1+x)}{x}). A) 0 B) 1 C) (\infty) D) Does not exist Answer: B Explanation: Using the series (\ln(1+x)=x-\frac{x^2}{2}+...), the ratio approaches 1.
Question 44. For the function (f(x)=\frac{x^2}{\sqrt{x^2+1}}), what is ( displaystyle\lim_{x\to\infty} f(x))?
Continuity Ultimate Exam
A) 0
B) 1
C) (\infty) D) Does not exist Answer: B Explanation: Divide numerator and denominator by (|x|); the expression becomes (\frac{|x|}{\sqrt{1+1/x^2}}\to\frac{|x|}{|x|}=1).
Question 45. Which of the following limits is an example of an indeterminate form (0\cdot\infty)? A) (\displaystyle\lim_{x\to0^+} x\cdot\ln x) B) (\displaystyle\lim_{x\to\infty} \frac{1}{x}) C) (\displaystyle\lim_{x\to0} \frac{\sin x}{x}) D) (\displaystyle\lim_{x\to1} \frac{x-1}{x-1}) Answer: A Explanation: As (x\to0^+), (x\to0) and (\ln x\to -\infty); their product yields the (0\cdot\infty) indeterminate form.
Question 46. Transform the indeterminate form in Question 45 to apply L’Hôpital’s Rule. Which equivalent form should be used? A) (\displaystyle\lim_{x\to0^+} \frac{\ln x}{1/x}) B) (\displaystyle\lim_{x\to0^+} \frac{1}{x\ln x}) C) (\displaystyle\lim_{x\to0^+} \frac{x}{\ln x}) D) None of the above
