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Question 1. What does the notation (\displaystyle\lim_{x\to a}f(x)=L) represent? A) The value of (f) at (x=a) B) The value that (f(x)) approaches as (x) gets arbitrarily close to (a) C) The derivative of (f) at (a) D) The average of (f) on an interval containing (a) Answer: B Explanation: The limit describes the behavior of (f(x)) as (x) approaches (a), not necessarily the function’s actual value at (a). Question 2. Which of the following statements is true about one‑sided limits? A) (\displaystyle\lim_{x\to a^-}f(x)=\lim_{x\to a^+}f(x)) always holds. B) The left‑hand limit uses values of (x) that are smaller than (a). C) The right‑hand limit is defined only for discontinuous functions. D) One‑sided limits are irrelevant for determining continuity. Answer: B Explanation: The left‑hand limit (\lim_{x\to a^-}) considers (xa). Question 3. A limit (\displaystyle\lim_{x\to a}f(x)) exists if and only if: A) Both one‑sided limits exist and are equal (finite). B) The function is continuous at (a). C) (f(a)) is defined. D) The limit from the left is infinite. Answer: A
Explanation: Existence requires the left‑hand and right‑hand limits to exist and be the same finite number. Question 4. If (\displaystyle\lim_{x\to 2^-}f(x)=5) and (\displaystyle\lim_{x\to 2^+}f(x)=5), what can be said about (\displaystyle\lim_{x\to 2}f(x))? A) It does not exist. B) It equals 5. C) It equals (\frac{5}{2}). D) It equals (\infty). Answer: B Explanation: When both one‑sided limits equal the same finite number, the two‑sided limit exists and equals that number. Question 5. Which of the following indicates a vertical asymptote at (x=3)? A) (\displaystyle\lim_{x\to 3}f(x)=0) B) (\displaystyle\lim_{x\to 3^-}f(x)=\infty) and (\displaystyle\lim_{x\to 3^+}f(x)=-\infty) C) (\displaystyle\lim_{x\to 3}f(x)=3) D) (\displaystyle\lim_{x\to 3}f(x)) does not exist because of a hole. Answer: B Explanation: When the function grows without bound (positive or negative) as (x) approaches a finite number, a vertical asymptote is present. Question 6. In a graph, a “hole” at ((4,2)) most likely represents which type of discontinuity? A) Jump discontinuity B) Infinite discontinuity
Question 8. The function (h(x)=\sin!\left(\frac{1}{x}\right)) as (x\to0) demonstrates which phenomenon? A) A removable discontinuity B) An infinite limit C) Oscillatory behavior with no limit D) A jump discontinuity Answer: C Explanation: As (x) approaches 0, (\frac{1}{x}) becomes unbounded, causing (\sin(1/x)) to oscillate between – 1 and 1 without settling to a single value. Question 9. For a function continuous at (c), which of the following is always true? A) (\displaystyle\lim_{x\to c}f(x)=f(c)) B) (\displaystyle\lim_{x\to c}f(x)) does not exist. C) (f) must be differentiable at (c). D) The left‑hand limit equals (\infty). Answer: A Explanation: Continuity at a point means the limit exists and equals the function value. Question 10. Which algebraic technique is most appropriate for evaluating (\displaystyle\lim_{x\to3}\frac{x^2-9}{x-3})? A) Direct substitution B) Rationalization C) Factoring and canceling D) L’Hôpital’s Rule
Answer: C Explanation: Factor numerator as ((x-3)(x+3)) and cancel the ((x-3)) term, then substitute (x=3) to get 6. Question 11. Evaluate (\displaystyle\lim_{x\to0}\frac{\sqrt{x+4}-2}{x}). A) (\frac{1}{4}) B) (\frac{1}{2}) C) (\frac{1}{8}) D) 0 Answer: B Explanation: Multiply numerator and denominator by the conjugate (\sqrt{x+4}+2) to obtain (\frac{x}{x(\sqrt{x+4}+2)}=\frac{1}{\sqrt{x+4}+2}). As (x\to0), denominator (\to4), so limit (=1/4). Wait check: Actually after rationalization we get (\frac{x}{x(\sqrt{x+4}+2)} = \frac{1}{\sqrt{x+4}+2}). At (x=0), (\sqrt{4}=2), so denominator (2+2=4); limit (=1/4). Thus correct answer is A, not B. Correct Answer: A Explanation: Rationalizing yields (\frac{1}{\sqrt{x+4}+2}); substituting (x=0) gives (1/4). Question 12. The limit (\displaystyle\lim_{x\to0}\frac{1-\cos x}{x^2}) equals: A) 0 B) 1 C) (\tfrac12) D) (\infty) Answer: C Explanation: Using the standard limit (1-\cos x\sim \tfrac{x^2}{2}) as (x\to0), the ratio approaches (\tfrac12).
Answer: B Explanation: Degrees of numerator (3) and denominator (2) differ by 1, so the function grows without bound; however, the ratio of leading coefficients after dividing gives a slant asymptote, not horizontal. Wait: Actually degree numerator > denominator, so horizontal asymptote does not exist; there is an oblique asymptote. Hence answer D. Correct Answer: D Explanation: When the numerator’s degree exceeds the denominator’s, the limit at infinity is infinite, so no horizontal asymptote. Question 16. Determine the limit (\displaystyle\lim_{x\to\infty}\frac{5x^4-2x^2}{7x^4+3}). A) (\frac57) B) 0 C) (\infty) D) (\frac57) (typo) Answer: A Explanation: Divide numerator and denominator by (x^4); the dominant terms give (\frac{5}{7}). Question 17. Which function grows fastest as (x\to\infty)? A) (\log x) B) (x^3) C) (e^{0.5x}) D) (\sqrt{x}) Answer: C Explanation: Exponential functions dominate polynomial and logarithmic functions for large (x).
Question 18. Evaluate (\displaystyle\lim_{x\to0}\frac{\sin(5x)}{x}). A) 0 B) 5 C) (\frac{1}{5}) D) Does not exist Answer: B Explanation: Using (\lim_{u\to0}\frac{\sin u}{u}=1) with (u=5x), we get (\frac{\sin(5x)}{x}=5\frac{\sin(5x)}{5x}\to5). Question 19. The limit (\displaystyle\lim_{x\to0}\frac{1-\cos(3x)}{x^2}) equals: A) (\frac{9}{2}) B) (\frac{3}{2}) C) 0 D) (\frac12) Answer: A Explanation: Near 0, (1-\cos(3x)\approx\frac{(3x)^2}{2}= \frac{9x^2}{2}). Dividing by (x^2) yields (\frac{9}{2}). Question 20. 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\to-1}\sqrt{x+5}) C) (\displaystyle\lim_{x\to0}\frac{\sin x}{x}) D) (\displaystyle\lim_{x\to3}\frac{1}{x-3}) Answer: B
C) 6
D) Undefined Answer: C Explanation: Simplify (g(x)=x+3); defining (g(3)=3+3=6) removes the hole. Question 24. Using the Intermediate Value Theorem, which of the following guarantees a root of (f(x)=x^3-4x+1) in the interval ([0,2])? A) (f(0)=1) and (f(2)=- 3 ) have opposite signs. B) (f) is decreasing on ([0,2]). C) (f) has a vertical asymptote in ([0,2]). D) (f) is not continuous on ([0,2]). Answer: A Explanation: Since (f) is continuous and changes sign on the interval, IVT ensures at least one root. Question 25. Which of the following is the correct (\varepsilon)–(\delta) definition of (\displaystyle\lim_{x\to c}f(x)=L)? A) For every (\delta>0) there exists (\varepsilon>0) such that if (|x-c|<\varepsilon) then (|f(x)- L|<\delta). B) For every (\varepsilon>0) there exists (\delta>0) such that if (0<|x-c|<\delta) then (|f(x)- L|<\varepsilon). C) For some (\varepsilon>0) there exists (\delta>0) such that if (|x-c|<\delta) then (|f(x)- L|<\varepsilon). D) For every (\varepsilon>0) there exists (\delta>0) such that if (|f(x)-L|<\delta) then (|x- c|<\varepsilon). Answer: B Explanation: This is the standard formal definition of a limit.
Question 26. Find a (\delta) (in terms of (\varepsilon)) that works for the limit (\displaystyle\lim_{x\to2}(3x-5)=1). A) (\delta=\varepsilon) B) (\delta=\frac{\varepsilon}{3}) C) (\delta=3\varepsilon) D) (\delta=\varepsilon^2) Answer: B Explanation: (|3x- 5 - 1|=3|x-2|). To make this (<\varepsilon) we need (|x-2|<\varepsilon/3); thus (\delta=\varepsilon/3). Question 27. Which of the following limits is appropriate for applying L’Hôpital’s Rule? A) (\displaystyle\lim_{x\to0}\frac{x^2}{\sin x}) B) (\displaystyle\lim_{x\to\infty}\frac{e^x}{x^2}) C) (\displaystyle\lim_{x\to1}\frac{x^2-1}{x-1}) D) (\displaystyle\lim_{x\to0}\frac{1-\cos x}{x}) Answer: B Explanation: As (x\to\infty), both numerator and denominator go to (\infty), giving the indeterminate form (\infty/\infty), suitable for L’Hôpital. Question 28. Evaluate (\displaystyle\lim_{x\to0}\frac{e^{2x}-1}{x}). A) 0 B) 1 C) 2 D) (\infty)
B) 1
C) (\infty) D) Does not exist Answer: B Explanation: This classic limit equals 1. Question 32. Evaluate (\displaystyle\lim_{x\to0}\frac{1-\cos(2x)}{x^2}). A) 0 B) 1 C) 2 D) 4 Answer: D Explanation: Using (1-\cos u\approx u^2/2) with (u=2x) gives (\frac{(2x)^2/2}{x^2}= \frac{4x^2/2}{x^2}=2). Wait compute: ((2x)^2=4x^2); divided by 2 gives (2x^2); then (\frac{2x^2}{x^2}=2). So answer is C (2). Correct Answer: C Explanation: The limit equals 2. Question 33. Which of the following limits defines the number (e)? A) (\displaystyle\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{n}) B) (\displaystyle\lim_{n\to\infty}\left(1-\frac{1}{n}\right)^{n}) C) (\displaystyle\lim_{x\to0}\frac{\sin x}{x}) D) (\displaystyle\lim_{x\to\infty}\frac{x}{\ln x}) Answer: A
Explanation: This classic limit converges to the base of natural logarithms, (e). Question 34. For the function (f(x)=\frac{x^2}{\sqrt{x^2+1}}), what is (\displaystyle\lim_{x\to\infty}f(x))? 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}}\to1). Question 35. Which of the following is a correct statement about a function that is continuous on ([a,b]) and differentiable on ((a,b))? A) It must have a horizontal asymptote. B) It satisfies the Mean Value Theorem. C) Its derivative is always positive. D) It cannot have any critical points. Answer: B Explanation: The hypotheses of the Mean Value Theorem are precisely continuity on the closed interval and differentiability on the open interval. Question 36. Determine (\displaystyle\lim_{x\to0}\frac{\tan x - x}{x^3}). A) 0 B) (\frac13)
Explanation: Expand (e^{x}=1+x+\frac{x^2}{2}+O(x^3)); numerator simplifies to (\frac{x^2}{2}+O(x^3)); dividing by (x^2) gives (\frac12). Question 39. Which of the following limits can be turned into a (0\cdot\infty) indeterminate form before applying L’Hôpital’s Rule? A) (\displaystyle\lim_{x\to0^+}x\ln x) B) (\displaystyle\lim_{x\to\infty}\frac{x}{e^x}) C) (\displaystyle\lim_{x\to0}\frac{\sin x}{x}) D) (\displaystyle\lim_{x\to1}\frac{x-1}{\ln x}) Answer: A Explanation: As (x\to0^+), (x\to0) and (\ln x\to - \infty); the product is of type (0\cdot\infty). Rewriting as (\frac{\ln x}{1/x}) yields (\frac{-\infty}{\infty}) suitable for L’Hôpital. Question 40. For the function (f(x)=\frac{x^2-4}{x-2}) with (x\neq2), what is (\displaystyle\lim_{x\to2}f(x))? A) 0 B) 2 C) 4 D) Does not exist Answer: C Explanation: Factor numerator ((x-2)(x+2)) and cancel ((x-2)) to get (x+2); substituting (x=2) gives
Question 41. Which of the following statements is true about the limit (\displaystyle\lim_{x\to0}\frac{|x|}{x})? A) The limit equals 1.
B) The limit equals – 1. C) The limit does not exist. D) The limit equals 0. Answer: C Explanation: From the left the quotient is – 1, from the right it is 1; since the one‑sided limits differ, the two‑sided limit does not exist. Question 42. Determine (\displaystyle\lim_{x\to\infty}\frac{\ln x}{x}). A) 0 B) 1 C) (\infty) D) Does not exist Answer: A Explanation: Logarithm grows slower than any positive power of (x); applying L’Hôpital gives (\frac{1/x}{1}=0). Question 43. Which of these limits evaluates to the derivative of (\sin x) at (x=0)? A) (\displaystyle\lim_{h\to0}\frac{\sin h}{h}) B) (\displaystyle\lim_{h\to0}\frac{\sin(0+h)-\sin0}{h}) C) (\displaystyle\lim_{h\to0}\frac{\cos h-1}{h}) D) (\displaystyle\lim_{h\to0}\frac{1-\cos h}{h}) Answer: B Explanation: The definition of the derivative (f'(a)=\lim_{h\to0}\frac{f(a+h)-f(a)}{h}); with (f(x)=\sin x) and (a=0), the limit gives (\cos0=1).
C) (\displaystyle\lim_{x\to1}\frac{|x|-1}{x-1}) D) (\displaystyle\lim_{x\to\pi}\sin x) Answer: B Explanation: The expression simplifies to (x+2) for (x\neq2); defining the function value at (x=2) as 4 removes the hole. Question 47. Find (\displaystyle\lim_{x\to\infty}\left(\frac{2x^2+3}{x^2-5}\right)^{1/2}). A) (\sqrt{2}) B) (\sqrt{3}) C) 1 D) (\infty) Answer: A Explanation: As (x\to\infty), the ratio inside the root approaches (\frac{2}{1}=2); the square root gives (\sqrt{2}). Question 48. Which limit defines the derivative of (f(x)=\ln x) at (x=1)? A) (\displaystyle\lim_{h\to0}\frac{\ln(1+h)}{h}) B) (\displaystyle\lim_{h\to0}\frac{\ln(1+h)-\ln1}{h}) C) (\displaystyle\lim_{h\to0}\frac{1}{1+h}) D) (\displaystyle\lim_{h\to0}\frac{h}{\ln(1+h)}) Answer: B Explanation: By definition (f'(a)=\lim_{h\to0}\frac{f(a+h)-f(a)}{h}); with (a=1) we obtain the given expression, which equals 1. Question 49. Evaluate (\displaystyle\lim_{x\to0}\frac{\sqrt{1+x}-\sqrt{1-x}}{x}).
A) 0
B) 1
C) (\frac12) D) (\infty) Answer: B Explanation: Rationalize numerator: multiply by (\sqrt{1+x}+\sqrt{1-x}) to obtain (\frac{(1+x)-(1- x)}{x(\sqrt{1+x}+\sqrt{1-x})}= \frac{2x}{x(\dots)}=\frac{2}{\sqrt{1+x}+\sqrt{1-x}}). As (x\to0), denominator (\to2); limit (=1). Question 50. The limit (\displaystyle\lim_{x\to0}\frac{1-\cos x}{x^2}) equals: A) 0 B) (\frac12) C) 1 D) (\infty) Answer: B Explanation: Using the identity (1-\cos x = 2\sin^2(x/2)) and (\sin(u)\approx u) for small (u), we get (\frac{2(u)^2}{(2u)^2}= \frac12). Question 51. Which of the following limits can be evaluated using the Squeeze Theorem? A) (\displaystyle\lim_{x\to0}\frac{x^2}{\sin x}) B) (\displaystyle\lim_{x\to0}x\cos!\left(\frac{1}{x}\right)) C) (\displaystyle\lim_{x\to0}\frac{e^x-1}{x}) D) (\displaystyle\lim_{x\to\infty}\frac{x}{\sqrt{x^2+1}}) Answer: B
