1. Solve for x in the equation: 2x = 8
Solution: We need to express both sides of the equation with the same base.
Since 8 = 2³, we can rewrite the equation as:
2x = 2³
Now that both sides have the same base, we set the exponents equal to each other:
x = 3
Result: The value of x is 3.

2. Solve for x in the equation: 5 • 3x = 45
Solution:
Start by dividing both sides by 5:
5•3x / 5 = 45 / 5
3x = 9
Next, express 9 as a power of 3:
9 = 3²
Now the equation becomes:
3x = 3²
Since the bases are the same, set the exponents equal to each other:
x = 2
Result: The value of x is 2.

3. The Fractional Exponent Limit (Radical Form)
Problem: lim 9^(x/2), x → 4
Solution: 9^(4/2) = 9²
9 × 9 = 81
Since the function is continuous at x = 4, we plug it in:
Result: The limit is 81.

4. The Power Function Limit
In this case, we have a variable in both the base and the exponent. As long as the base is positive at the limit point, direct substitution remains our best friend.
Problem: lim (x - 1)^(x + 1), x → 3
Solution: We substitute x = 3 into both the base and the exponent simultaneously:
Base: (3 - 1) = 2
Exponent: (3 + 1) = 4
Now, evaluate the resulting exponential expression:
2⁴ = 2 × 2 × 2 × 2 = 16
Result: The limit is 16.

1. The Basic Sine Limit
This example involves a simple trigonometric function. We can apply direct substitution here.
Problem: lim sin(x), as x → 30°
Solution: Substitute x = 30° directly into the sine function:
sin(30°) = 1/2
Result: The limit is 1/2.

2. The Basic Tangent Limit
This example uses a simple trigonometric function. We can directly substitute here as well.
Problem: lim tan(x) as x → 45°
Solution: Substitute x = 45° directly into the tangent function:
tan(45°) = 1
Result: The limit is 1.

3. The Composite "Root-Trig" Limit
This example nests a trigonometric function inside a square root. For direct substitution to work, the value inside the square root must be non-negative.
Problem: lim √(2cos(x)), x → π/3
Solution: Substitute x = π/3 directly into the cosine function:
cos(π/3) = 1/2
Multiply by the constant: 2 • 1/2 = 1
Apply the root: √1 = 1
Result: The limit is 1.

4. The Reciprocal Product Limit
This example uses sec(x) (which is 1/cos(x)) and a squared term. Track multiple operations at once.
Problem: lim sec²(x) • sin(x), x → π/6
Solution: Substitute x = π/6 into both parts:
Find sec(π/6): cos(π/6) = √3/2 → sec(π/6) = 2/√3
Square it: (2/√3)² = 4/3
Find sin(π/6): sin(π/6) = 1/2
Multiply them: 4/3 • 1/2 = 4/6 = 2/3
Result: The limit is 2/3.

1. Problem: lim log₂(x) = 5
Solution: Rewrite the given equation in exponential form:
log₂(x) = 5 → 2⁵ = x
Result: x = 32

2. Problem: lim log₃(x) = 4
Solution: Rewrite the given equation in exponential form:
log₃(x) = 4 → 3⁴ = x
Result: x = 81

3. Problem: lim log₂(2x - 5), as x → 3
Solution: Substitute x = 3 into the expression and simplify:
log₂[2(3) - 5]
= log₂(6 - 5)
= log₂(1)
Since 2⁰ = 1, log₂(1) = 0
Result: The limit is 0

4. Problem: lim log₁₀((2x)/(x - 4)), as x → 5
Solution: Substitute x = 5 into the fraction:
Numerator: 2(5) = 10
Denominator: 5 - 4 = 1
Fraction: 10/1 = 10
Apply the logarithm: log₁₀(10)
Since 10¹ = 10, log₁₀(10) = 1
Result: The limit is 1

Problem no. 1: f(x) = (3x - 1)/x, x → 1
Solution: Substitute x = 1 directly into the expression:
(3(1) - 1)/1 = 2
Make a table for x approaching 1 from the left and right:

Left-hand approach (x → 1⁻):

x	f(x)
0.5	1.464
0.8	1.746
0.9	1.848
0.99	1.977
0.999	1.997

Right-hand approach (x → 1⁺):

x	f(x)
1.001	2.004
1.01	2.02
1.1	2.05
1.2	2.045
1.5	2.131

Result: The limit of f(x) as x → 1 is 2 because both sides approach 2.

Problem no. 2: f(x) = (4x - 3)/x, x → 4
Solution: Substitute x = 4 directly into the expression:
(4(4) - 3)/4 = 1
Make a table for x approaching 4 from the left and right:

Left-hand approach (x → 4⁻):

x	f(x)
3	1
3.5	2
3.9	3.48220
3.99	3.94493
3.999	3.99446

Right-hand approach (x → 4⁺):

x	f(x)
4.001	4.00555
4.01	4.05584
4.1	4.59479
4.5	8
5	16

Result: From both the left and right, f(x) approaches 4. Therefore, the overall limit of f(x) as x → 4 is 4.

Problem 3: lim_{x → 0} (x / sin x)

Solution: Using a table of values to estimate the limit from both sides.

Left-hand approach (x → 0⁻):

x	f(x)
-1	57.2987
-0.5	57.2965
-0.1	57.2958
-0.01	57.2958
-0.001	57.2958

Right-hand approach (x → 0⁺):

x	f(x)
1	57.2987
0.5	57.2965
0.1	57.2958
0.01	57.2958
0.001	57.2958

Result: The limit of f(x) = x / sin(x) as x → 0 is approximately 57.2957 because both sides approach the same value.

Problem 4: lim_{x → π/2} cos x

Solution: Using a table of values to estimate the limit from both sides.

Left-hand approach (x → π/2⁻):

x	f(x)
1.5	-0.9991
1.57	-1
1.570	-1
1.5707	-1
1.57079	-1

Right-hand approach (x → π/2⁺):

x	f(x)
1.6	-0.9985
1.58	-1
1.570	-1
1.5708	-1
1.57079	-1

Result: The limit of cos(x) as x → π/2 is -1 because both sides approach -1.

1. Problem: Evaluate the limit:
Solution:
Using L'Hopital's Rule:
Since the direct substitution gives 0/0, we apply L'Hopital's Rule, which involves differentiating the numerator and denominator:

d/dx[sin(x)] = cos(x)
d/dx[x] = 1

Now, substitute the derivatives into the limit:
lim_x→0 cos(x)/1
Substitute x = 0:
cos(0) = 1
Result: The limit is 1.

2. Problem: Evaluate the limit:
lim_x→1 (x² - 1)/(x - 1)
Solution:
Direct substitution gives 0/0, so we apply factoring to simplify the expression.
Factor the numerator:
x² - 1 = (x - 1)(x + 1)
Now, the expression becomes:
(x - 1)(x + 1)/(x - 1)
Cancel out the common factor of (x - 1):
x + 1
Substitute x = 1:
1 + 1
Result: The limit is 2.

3. The Algebraic Factoring Method
Problem: Find the limit of (x² - 9)/(x - 3) as x → 3
Solution: simplify the given expression
(x - 3)(x + 3)/(x - 3)
The (x - 3) terms cancel out, leaving us with lim (x + 3) as x → 3
Solve: x + 3 = 3 + 3 = 6
Result: The limit is 6.

4. Problem: Find the limit of the following function as x approaches 3:
lim_x→3 (x² - 2x - 3)/(x - 3)
Solution: Find the factors of the numerator
We need two numbers that multiply to -3 and add to -2. Those numbers are -3 and 1.
x² - 2x - 3 = (x - 3)(x + 1)
Now the expression becomes:
(x - 3)(x + 1)/(x - 3)
Simplify:
As long as x ≠ 3 (which is true for the "approach" of a limit), we can cancel out the common (x - 3) term:
(x - 3)(x + 1)/(x - 3) = x + 1
Re-evaluate the limit:
x + 1 = 3 + 1 = 4
Result: The limit is 4.

1. Evaluate the one-sided limit:
lim_x→0+ 1/x
Solution: As x→0+, 1/x approaches infinity:
lim_x→0+ 1/x = +∞

2. Evaluate the one-sided limit:
lim_x→0- 1/x
Solution: As x→0-, 1/x approaches negative infinity:
lim_x→0- 1/x = -∞

3. Problem: Given the function below, find lim f(x) as x→1- and lim f(x) as x→1+
f(x) = x + 2 if x < 1
f(x) = x² - 4 if x ≥ 1
Solution: Find the value approaching from the left and right of 1
Approach from the left (x→1-):
When x is less than 1, we use the top piece of the function: f(x) = x + 2
Plug x = 1 into that equation:
1 + 2 = 3

Approach from the right (x→1+):
When x is greater than or equal to 1, we use the bottom piece: f(x) = x² - 4
Plug x = 1 into that equation:
(1)² - 4 = 1 - 4 = -3

Result: Left-hand limit (x→1-) = 3
Right-hand limit (x→1+) = -3
Overall limit: DNE

4. Problem: Find the one-sided limits and overall limit of f(x) as x approaches 2
f(x) = 3x - 4 if x < 2
x² - 2 if x ≥ 2
Solution: Find the value approaching 2 from the left and right

Evaluate the Left-Hand Limit (x → 2⁻):
When x approaches from the left, x < 2, so we use the top piece: f(x) = 3x - 4
Plug in x = 2:
3(2) - 4 = 6 - 4 = 2

Evaluate the Right-Hand Limit (x → 2⁺):
When x approaches from the right, x ≥ 2, so we use the bottom piece: f(x) = x² - 2
Plug in x = 2:
(2)² - 2 = 4 - 2 = 2

Result: Left-Hand Limit = 2
Right-Hand Limit = 2
Overall Limit = 2

1. Problem: Find the limit of f(x) = 1/x² as x → 0
Solution:
Look at the right-hand side (x → 0⁺)
Pick a tiny positive number like 0.1
0.1² = 0.01
1 / 0.01 = 100
As x gets even smaller (like 0.001), the result becomes 1,000,000. It is heading towards +∞.
Look at the left-hand side (x → 0⁻)
Pick a tiny negative number like -0.01
When you square a negative, it becomes positive: (-0.01)² = 0.0001
1 / 0.0001 = 10,000
Even though x started negative, x² is positive. So this also heads toward +∞
Result: Left side (x → 0⁻) = +∞
Right side (x → 0⁺) = +∞
Overall limit: Since both sides go to +∞, the overall limit is ∞

2. Problem: Find the limit of f(x) = 2x / (x - 1) as x → 1
Solution:
Find the right-hand limit (x → 1⁺)
Imagine a number slightly bigger than 1, like 1.01
Numerator: 2(1.01) ≈ 2 (positive number)
Denominator: 1.01 - 1 = 0.01 (a tiny positive number)
lim 2x / (x - 1) → +∞
Find the left-hand limit (x → 1⁻)
Imagine a number slightly smaller than 1, like 0.99
Numerator: 2(0.99) ≈ 2 (positive number)
Denominator: 0.99 - 1 = -0.01 (a negative number)
lim 2x / (x - 1) → -∞
Result: Right-Hand Limit (x → 1⁺) = +∞
Left-Hand Limit (x → 1⁻) = -∞
Overall Limit: The two sides are heading in opposite directions, therefore the limit is DNE

1. Problem: Find the lim (2x² - 4)/(x - 2) as x → 2 using the following graph:
Solution: Check the value at 2 from the left, and 2 from the right
The graph values are both approaching 2, so the limit is 2.
Note: Limit is found on the y-axis.
Result: The lim (2x² - 4)/(x - 2) as x → 2 is 2.

2. Problem: Find the lim |x - 2| / (x - 2) as x → 2 using the following graph:
Solution: Check the value at 2 from the left, and 2 from the right
To the left of 2 = -1
To the right of 2 = 1
Note: Limit is found on the y-axis
Result: Both sides are not approaching the same number, so the limit is DNE.

Problem 3:
lim x → π/2 of y = cot(x)
As x gets close to π/2, the graph of y = cot(x) approaches 0.
The curve is smooth near π/2 and crosses the x-axis at that point, indicating that the limit is 0.
Result: The limit is 0.

Problem 4:
lim x → π/2 of sin(x) cos(x - π/6)
When evaluating the limit of a trigonometric function, we start by substituting the value that the variable approaches.
As long as this substitution does not make the expression undefined, the limit equals the value of the function.
Substitute x = π/2:
sin(π/2) cos(π/2 - π/6)
sin(π/2) cos(π/3)
1 × 1/2 = 1/2
Result: The limit is 1/2.