Calculus Limits and Continuity – Complete Study Guide (130+ Q&A)

Calculus Limits and Continuity –
Complete Exam Study Guide (130+
Questions & Answers) Guaranteed Pass
Description:
This comprehensive Calculus study guide focuses on limits and
continuity, essential concepts for mastering introductory
calculus. The document includes 130+ carefully structured
questions and answers with detailed explanations designed to
help students fully understand limit laws, evaluating limits
graphically and algebraically, infinite limits, and continuity
conditions. Each question is exam-focused and designed to
strengthen problem-solving skills. Perfect for college calculus
exams, quizzes, and practice tests. Students gain confidence
through step-by-step solutions and clear explanations, ensuring
strong preparation and a higher chance of passing calculus
exams successfully.



1. Q: Find the domain of the function f(x)=x+2x2−9f(x)=x2−9x+2.
A: The expression under the square root requires x+2≥0x+2≥0, so x≥−2x≥−2. The
denominator cannot be zero, so x2−9≠0x2−9 =0,
meaning x≠3x =3 and x≠−3x =−3. Combining these, the domain
is [−2,3)∪(3,∞)[−2,3)∪(3,∞).

2. Q: Determine whether the function f(x)=x3−2xf(x)=x3−2x is even, odd, or neither.
A: Evaluate f(−x)=(−x)3−2(−x)=−x3+2x=−(x3−2x)=−f(x)f(−x)=(−x)3−2(−x)=−x3
+2x=−(x3−2x)=−f(x). Since f(−x)=−f(x)f(−x)=−f(x), the function is odd.
3. Q: A ball is thrown into the air. Its height (in meters) after tt seconds is given
by h(t)=20t−5t2h(t)=20t−5t2. What is the practical domain of this function?

, A: The domain is the time the ball is in the air, from t=0t=0 until it hits the ground.
Solve h(t)=0h(t)=0: 20t−5t2=5t(4−t)=020t−5t2=5t(4−t)=0. The solutions
are t=0t=0 and t=4t=4. The practical domain is [0,4][0,4].

4. Q: For the function f(x)=xx−1f(x)=x−1x, find f−1(x)f−1(x).
A: Let y=xx−1y=x−1x. Swap xx and yy: x=yy−1x=y−1y. Solve
for yy: x(y−1)=y⇒xy−x=y⇒xy−y=x⇒y(x−1)=x⇒y=xx−1x(y−1)=y⇒xy−x=y⇒xy
−y=x⇒y(x−1)=x⇒y=x−1x. Thus, f−1(x)=xx−1f−1(x)=x−1x. (Note: The function is its
own inverse).

5. Q: Given f(x)=2x+1f(x)=2x+1 and g(x)=x2g(x)=x2,
find (f∘g)(x)(f∘g)(x) and (g∘f)(x)(g∘f)(x).
A:

 (f∘g)(x)=f(g(x))=f(x2)=2(x2)+1=2x2+1(f∘g)(x)=f(g(x))=f(x2)=2(x2)+1=2x2+1.
 (g∘f)(x)=g(f(x))=g(2x+1)=(2x+1)2=4x2+4x+1(g∘f)(x)=g(f(x))=g(2x+1)=(2x+1)2
=4x2+4x+1.
6. Q: Describe the transformation from the graph of y=xy=x to y=2x−1+3y=2x−1+3.
A: Starting from xx:

1. Horizontal shift right by 1 unit (inside the function: x−1x−1).
2. Vertical stretch by a factor of 2 (outside the function: 2⋯2⋯).
3. Vertical shift up by 3 units (+3+3).

7. Q: Write the piecewise function for the graph shown below (Description: A V-shaped
graph with vertex at (-1, 2), passing through (0, 4) on the right and (-2, 4) on the left).
A: The graph is an absolute value function. The slope on the right branch
(for x≥−1x≥−1) is 4−20−(−1)=20−(−1)4−2=2. The slope on the left branch
(for x≤−1x≤−1) is 4−2−2−(−1)=−2−2−(−1)4−2=−2. Using point-slope form with the
vertex (-1, 2):
f(x)={−2xif x≤−12x+4if x>−1f(x)={−2x2x+4if x≤−1if x>−1
8. Q: If ln⁡x=2.5lnx=2.5, find the value of ln⁡(ex2)ln(ex2).
A: Using logarithm
properties: ln⁡(ex2)=ln⁡(e)+ln⁡(x2)=1+2ln⁡x=1+2(2.5)=1+5=6ln(ex2)=ln(e)
+ln(x2)=1+2lnx=1+2(2.5)=1+5=6.
9. Q: Solve for xx: e2x−7ex+12=0e2x−7ex+12=0.
A: Let u=exu=ex. Then the equation becomes u2−7u+12=0u2−7u+12=0, which

, factors to (u−3)(u−4)=0(u−3)(u−4)=0. So u=3u=3 or u=4u=4. Substituting
back: ex=3⇒x=ln⁡3ex=3⇒x=ln3, and ex=4⇒x=ln⁡4ex=4⇒x=ln4.

10. Q: Find the average rate of change of f(x)=1xf(x)=x1 on the interval [1,4][1,4].
A: The average rate of change is f(4)−f(1)4−1=14−13=−343=−144−1f(4)−f(1)=341−1
=3−43=−41.
11. Q: For the function f(x)=x2−4x+3f(x)=x2−4x+3, find the difference
quotient f(x+h)−f(x)hhf(x+h)−f(x) and simplify.
A:
f(x+h)=(x+h)2−4(x+h)+3=x2+2xh+h2−4x−4h+3f(x+h)=(x+h)2−4(x+h)+3=x2+2
xh+h2−4x−4h+3.
f(x+h)−f(x)=(x2+2xh+h2−4x−4h+3)−(x2−4x+3)=2xh+h2−4h=h(2x+h−4)f(x+h)
−f(x)=(x2+2xh+h2−4x−4h+3)−(x2−4x+3)=2xh+h2−4h=h(2x+h−4).
Then, f(x+h)−f(x)h=h(2x+h−4)h=2x+h−4hf(x+h)−f(x)=hh(2x+h−4)=2x+h−4.

12. Q: Find all values of xx in the interval [0,2π][0,2π] that satisfy sin⁡x=32sinx=23.
A: Sine is positive in Quadrants I and II. The reference angle is π/3π/3 (60°).
Therefore, x=π/3x=π/3 and x=2π/3x=2π/3.

13. Q: Express log⁡b(x3yz2)logb(z2x3y) in terms of log⁡bxlogbx, log⁡bylogby,
and log⁡bzlogbz.
A: Using logarithm properties:
log⁡b(x3yz2)=log⁡b(x3y)−log⁡b(z2)=log⁡b(x3)+log⁡b(y1/2)−2log⁡bz=
3log⁡bx+12log⁡by−2log⁡bzlogb(z2x3y)=logb(x3y)−logb(z2)=logb(x3)+logb
(y1/2)−2logbz=3logbx+21logby−2logbz.
14. Q: Is the function f(x)=x4+2x2f(x)=x2x4+2 symmetric about the y-axis, the origin,
both, or neither?
A: Check f(−x)=(−x)4+2(−x)2=x4+2x2=f(x)f(−x)=(−x)2(−x)4+2=x2x4+2=f(x).
Since f(−x)=f(x)f(−x)=f(x), the function is even and symmetric about the y-axis.

15. Q: Find the point(s) of intersection of the
curves y=x2+1y=x2+1 and y=2x+1y=2x+1.
A: Set the equations
equal: x2+1=2x+1⇒x2−2x=0⇒x(x−2)=0x2+1=2x+1⇒x2−2x=0⇒x(x−2)=0.
So x=0x=0 or x=2x=2. The corresponding y-values are y=1y=1 and y=5y=5.
Intersection points are (0,1)(0,1) and (2,5)(2,5).