3 free questions from this chapter (the full set has 30). Marking: Multi-correct: +4 all correct, +1 per correct option if none wrong, −2 if any wrong · Numerical: +4 / 0 · Matching: +3 / −1.
Question 1Level 2 · ExamSingle correct
The value of x→1lim415+x−237+x−3+x (two-sided limit) is ______. Exactly one option is correct.
A316
B−38
C−316
D−332
Show answer and solution
Answer: C
Write the numerator as (37+x−2)−(3+x−2) and divide everything by x−1.
Elementary route (no L'Hospital): with u=37+x, u3−8=x−1, so x−1u−2=u2+2u+41→121. With v=3+x, x−1v−2=v+21→41. With w=415+x, w4−16=x−1, so x−1w−2=(w+2)(w2+4)1→321.
Hence the limit is 321121−41=(−61)(32)=−316.
(L'Hospital gives the same: the derivatives at x=1 are 31⋅8−2/3=121, 21⋅4−1/2=41, 41⋅16−3/4=321.)
Difficulty: Hard
Question 2Level 2 · ExamOne or more correct
Let φ(x)=2π−tan−1(x1) for x=0, where tan−1 takes values in (−2π,2π). Which of the following statements is (are) TRUE? One or more options may be correct.
Ax→0+limxφ(x)=1
Bx→0limxφ(x) exists
Cx→0−limφ(x)=π
Dx→0−limxφ(x)−π=1
Show answer and solution
Answer: A, C, D
For x>0: tan−1(1/x)=2π−tan−1x, so φ(x)=tan−1x. For x<0: tan−1(1/x)=−2π−tan−1x, so φ(x)=π+tan−1x.
(A) x→0+: φ(x)/x=tan−1x/x→1. TRUE.
(B) From the left, φ(x)→π while x→0−, so φ(x)/x→−∞; the right-hand limit is 1. The two-sided limit does not exist. FALSE.
(C) limx→0−φ(x)=π+0=π. TRUE.
(D) x→0−: (φ(x)−π)/x=tan−1x/x→1. TRUE.
Answer: (A), (C), (D).
Difficulty: Hard
Question 3Level 2 · ExamSingle correct
The value of x→0limx(1−cos3x)tan2x−2sinx is ______. Exactly one option is correct.
A32
B94
C92
D31
Show answer and solution
Answer: A
Split the numerator: tan2x−2sinx=(tan2x−2tanx)+2(tanx−sinx).
First part: tan2x−2tanx=1−tan2x2tanx−2tanx=1−tan2x2tan3x, so divided by x3 it tends to 2.
Second part: 2(tanx−sinx)=2tanx(1−cosx), so divided by x3 it tends to 2⋅21=1.
So the numerator behaves like 3x3. The denominator: x(1−cos3x)=x⋅2sin223x, which behaves like 29x3.
Limit =9/23=32.
(Distractors: 94 keeps only the first part, 92 only the second.)
Difficulty: Hard
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