1) Introduction to Limit Laws: Streamlining Limit Calculations
Welcome to the practical side of limits! In Topic 2, we built a strong intuitive understanding of what limits mean. Now, we move to limit laws, which are a set of rules that allow us to calculate limits more efficiently and systematically, rather than relying solely on intuition and numerical approximation.
Think of limit laws as the "algebra of limits." Just as algebraic laws simplify complex arithmetic expressions, limit laws break down complex limit problems into simpler, manageable parts. They are based on the fundamental properties of limits and provide a toolbox for evaluating limits of combinations of functions.
Why Limit Laws are Essential
Without limit laws, we would have to painstakingly examine function values getting closer and closer to a point for every single limit problem. Limit laws provide shortcuts and a framework, making limit calculations much more efficient and less prone to error, especially for complicated functions built from simpler ones.
2) The Basic Limit Laws - Building Blocks of Limit Evaluation
These laws apply to limits of sums, differences, products, quotients, powers, and roots of functions, provided that the individual limits exist. Let's explore each one in detail.
2.1) Limit Law 1: Limit of a Constant
The limit of a constant function is simply the constant itself.
Formal Statement: If \( c \) is a constant, then for any number \( a \),
\[ \lim_{x \to a} c = c \]
Example 1: Limit of a Constant Function
Find \( \lim_{x \to 5} 7 \).
Since \( f(x) = 7 \) is a constant function (always equals 7, regardless of \( x \)), its limit as \( x \) approaches any value is 7. \[ \lim_{x \to 5} 7 = 7 \]
2.2) Limit Law 2: Limit of \( x \)
The limit of the identity function \( f(x) = x \) as \( x \) approaches \( a \) is simply \( a \).
Formal Statement: For any number \( a \),
\[ \lim_{x \to a} x = a \]
Example 2: Limit of the Identity Function
Find \( \lim_{x \to -2} x \).
For the function \( f(x) = x \), as \( x \) approaches -2, the function value also approaches -2. \[ \lim_{x \to -2} x = -2 \]
2.3) Limit Law 3: Limit of a Sum
The limit of a sum of two functions is the sum of their limits (provided the individual limits exist).
Formal Statement: If \( \lim_{x \to a} f(x) \) and \( \lim_{x \to a} g(x) \) exist, then
\[ \lim_{x \to a} [f(x) + g(x)] = \lim_{x \to a} f(x) + \lim_{x \to a} g(x) \]
Example 3: Limit of a Sum
Find \( \lim_{x \to 3} (x + 4) \).
We can break this limit into the sum of limits: \[ \lim_{x \to 3} (x + 4) = \lim_{x \to 3} x + \lim_{x \to 3} 4 \] Using Law 2 and Law 1: \[ = 3 + 4 = 7 \] Thus, \( \lim_{x \to 3} (x + 4) = 7 \).
2.4) Limit Law 4: Limit of a Difference
The limit of a difference of two functions is the difference of their limits (provided the individual limits exist).
Formal Statement: If \( \lim_{x \to a} f(x) \) and \( \lim_{x \to a} g(x) \) exist, then
\[ \lim_{x \to a} [f(x) - g(x)] = \lim_{x \to a} f(x) - \lim_{x \to a} g(x) \]
Example 4: Limit of a Difference
Find \( \lim_{x \to 2} (x^2 - 3x) \).
We can use the difference and constant multiple laws along with Law 2: \[ \lim_{x \to 2} (x^2 - 3x) = \lim_{x \to 2} x^2 - \lim_{x \to 2} 3x = \lim_{x \to 2} x^2 - 3\lim_{x \to 2} x \] We will use the power law (Law 7) for \( \lim_{x \to 2} x^2 \) later, but for now, let's intuitively evaluate \( \lim_{x \to 2} x^2 = 2^2 = 4 \) and \( \lim_{x \to 2} x = 2 \). \[ = 4 - 3(2) = 4 - 6 = -2 \] Thus, \( \lim_{x \to 2} (x^2 - 3x) = -2 \).
2.5) Limit Law 5: Limit of a Constant Multiple
The limit of a constant times a function is the constant times the limit of the function (provided the limit of the function exists).
Formal Statement: If \( c \) is a constant and \( \lim_{x \to a} f(x) \) exists, then
\[ \lim_{x \to a} [c \cdot f(x)] = c \cdot \lim_{x \to a} f(x) \]
Example 5: Limit of a Constant Multiple
Find \( \lim_{x \to 4} (5x^3) \).
We can pull the constant 5 out of the limit: \[ \lim_{x \to 4} (5x^3) = 5 \lim_{x \to 4} x^3 \] Again, we'll use the power rule (Law 7) later, but intuitively, \( \lim_{x \to 4} x^3 = 4^3 = 64 \). \[ = 5 \cdot 64 = 320 \] Thus, \( \lim_{x \to 4} (5x^3) = 320 \).
2.6) Limit Law 6: Limit of a Product
The limit of a product of two functions is the product of their limits (provided the individual limits exist).
Formal Statement: If \( \lim_{x \to a} f(x) \) and \( \lim_{x \to a} g(x) \) exist, then
\[ \lim_{x \to a} [f(x) \cdot g(x)] = \left[ \lim_{x \to a} f(x) \right] \cdot \left[ \lim_{x \to a} g(x) \right] \]
Example 6: Limit of a Product
Find \( \lim_{x \to -1} (x \cdot (x + 2)) \).
We can separate this limit into the product of two limits: \[ \lim_{x \to -1} (x \cdot (x + 2)) = \left[ \lim_{x \to -1} x \right] \cdot \left[ \lim_{x \to -1} (x + 2) \right] \] Using Law 2 and the sum law (Law 3) for the second limit: \[ = (-1) \cdot \left[ \lim_{x \to -1} x + \lim_{x \to -1} 2 \right] = (-1) \cdot [-1 + 2] = (-1) \cdot (1) = -1 \] Thus, \( \lim_{x \to -1} (x \cdot (x + 2)) = -1 \).
2.7) Limit Law 7: Limit of a Quotient
The limit of a quotient of two functions is the quotient of their limits, provided the limits of both functions exist and the limit of the denominator is not zero.
Formal Statement: If \( \lim_{x \to a} f(x) \) and \( \lim_{x \to a} g(x) \) exist, and \( \lim_{x \to a} g(x) \neq 0 \), then
\[ \lim_{x \to a} \frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)} \]
Example 7: Limit of a Quotient
Find \( \lim_{x \to 2} \frac{x^2 + 1}{x - 3} \).
We can take the limit of the numerator and the denominator separately: \[ \lim_{x \to 2} \frac{x^2 + 1}{x - 3} = \frac{\lim_{x \to 2} (x^2 + 1)}{\lim_{x \to 2} (x - 3)} \] Using sum/difference and power laws (and direct substitution for polynomials): \[ = \frac{\lim_{x \to 2} x^2 + \lim_{x \to 2} 1}{\lim_{x \to 2} x - \lim_{x \to 2} 3} = \frac{(2)^2 + 1}{2 - 3} = \frac{4 + 1}{-1} = \frac{5}{-1} = -5 \] Thus, \( \lim_{x \to 2} \frac{x^2 + 1}{x - 3} = -5 \).
Important Condition for Quotient Law
Remember, the Quotient Law only applies if the limit of the denominator is NOT zero. If the denominator's limit is zero, we cannot use this law directly. We might need to use other techniques, which we will discuss shortly.
2.8) Limit Law 8: Limit of a Power
The limit of a function raised to a power is the limit of the function raised to that power (provided the limit of the base function exists).
Formal Statement: If \( \lim_{x \to a} f(x) \) exists and \( n \) is a positive integer, then
\[ \lim_{x \to a} [f(x)]^n = \left[ \lim_{x \to a} f(x) \right]^n \]
Example 8: Limit of a Power
Find \( \lim_{x \to 2} (x^2 + 1)^3 \).
Using the power law, we can move the exponent outside the limit: \[ \lim_{x \to 2} (x^2 + 1)^3 = \left[ \lim_{x \to 2} (x^2 + 1) \right]^3 \] Now apply sum and power laws inside the brackets: \[ = \left[ \lim_{x \to 2} x^2 + \lim_{x \to 2} 1 \right]^3 = \left[ (2)^2 + 1 \right]^3 = [4 + 1]^3 = [5]^3 = 125 \] Thus, \( \lim_{x \to 2} (x^2 + 1)^3 = 125 \).
2.9) Limit Law 9: Limit of a Root
The limit of the \( n \)-th root of a function is the \( n \)-th root of the limit of the function (provided the limit of the function exists, and for even roots, the limit is non-negative).
Formal Statement: If \( \lim_{x \to a} f(x) \) exists and \( n \) is a positive integer, then
\[ \lim_{x \to a} \sqrt[n]{f(x)} = \sqrt[n]{\lim_{x \to a} f(x)} \]
(If \( n \) is even, we assume \( \lim_{x \to a} f(x) \geq 0 \).)
Example 9: Limit of a Root
Find \( \lim_{x \to 1} \sqrt{3x + 1} \).
Using the root law, we can move the square root outside the limit: \[ \lim_{x \to 1} \sqrt{3x + 1} = \sqrt{\lim_{x \to 1} (3x + 1)} \] Now apply sum and constant multiple laws inside the square root: \[ = \sqrt{\lim_{x \to 1} 3x + \lim_{x \to 1} 1} = \sqrt{3 \lim_{x \to 1} x + \lim_{x \to 1} 1} = \sqrt{3(1) + 1} = \sqrt{3 + 1} = \sqrt{4} = 2 \] Thus, \( \lim_{x \to 1} \sqrt{3x + 1} = 2 \).
3) Techniques for Evaluating Limits - Going Beyond Direct Substitution
While limit laws are powerful, sometimes direct substitution or simple application of laws isn't enough. We need to employ specific techniques to handle different types of limit problems.
3.1) Technique 1: Direct Substitution - The First Approach
For many "well-behaved" functions (polynomials, rational functions, trigonometric functions, exponential, logarithmic functions) at points within their domain, you can find the limit by simply substituting the value \( a \) into the function.
When to Use: For polynomials, rational functions where the denominator is not zero at \( x = a \), and other continuous functions at points within their domain.
Example 10: Direct Substitution
Find \( \lim_{x \to \pi/2} \sin(x) \).
Since \( \sin(x) \) is continuous at \( x = \pi/2 \), we can directly substitute: \[ \lim_{x \to \pi/2} \sin(x) = \sin(\pi/2) = 1 \] Thus, \( \lim_{x \to \pi/2} \sin(x) = 1 \).
3.2) Technique 2: Factoring and Cancellation - Removing Removable Discontinuities
When direct substitution leads to an indeterminate form like \( \frac{0}{0} \) (which means both numerator and denominator become zero at \( x = a \)), it often indicates a "removable discontinuity". Factoring and cancelling common factors can simplify the expression and allow us to find the limit.
When to Use: When direct substitution gives \( \frac{0}{0} \) for rational functions.
Example 11: Factoring and Cancellation
Find \( \lim_{x \to -3} \frac{x^2 + x - 6}{x + 3} \).
Direct substitution gives \( \frac{(-3)^2 + (-3) - 6}{-3 + 3} = \frac{9 - 3 - 6}{0} = \frac{0}{0} \). Indeterminate form! Factor the numerator: \( x^2 + x - 6 = (x + 3)(x - 2) \). \[ \lim_{x \to -3} \frac{x^2 + x - 6}{x + 3} = \lim_{x \to -3} \frac{(x + 3)(x - 2)}{x + 3} \] Cancel the common factor \( (x + 3) \) for \( x \neq -3 \): \[ = \lim_{x \to -3} (x - 2) \] Now, use direct substitution: \[ = -3 - 2 = -5 \] Thus, \( \lim_{x \to -3} \frac{x^2 + x - 6}{x + 3} = -5 \).
3.3) Technique 3: Rationalizing Numerator or Denominator - Dealing with Square Roots
When dealing with limits involving square roots, especially when direct substitution leads to \( \frac{0}{0} \) or involves differences of square roots, rationalizing (multiplying by the conjugate) can simplify the expression.
When to Use: Limits involving square roots and indeterminate forms, especially of the form \( \frac{0}{0} \) or \( \infty - \infty \).
Example 12: Rationalizing the Numerator
Find \( \lim_{x \to 0} \frac{\sqrt{x + 4} - 2}{x} \).
Direct substitution gives \( \frac{\sqrt{0 + 4} - 2}{0} = \frac{\sqrt{4} - 2}{0} = \frac{2 - 2}{0} = \frac{0}{0} \). Indeterminate form! Rationalize the numerator by multiplying by the conjugate \( \sqrt{x + 4} + 2 \): \[ \lim_{x \to 0} \frac{\sqrt{x + 4} - 2}{x} \cdot \frac{\sqrt{x + 4} + 2}{\sqrt{x + 4} + 2} = \lim_{x \to 0} \frac{(x + 4) - 4}{x(\sqrt{x + 4} + 2)} = \lim_{x \to 0} \frac{x}{x(\sqrt{x + 4} + 2)} \] Cancel the common factor \( x \) for \( x \neq 0 \): \[ = \lim_{x \to 0} \frac{1}{\sqrt{x + 4} + 2} \] Now, use direct substitution: \[ = \frac{1}{\sqrt{0 + 4} + 2} = \frac{1}{\sqrt{4} + 2} = \frac{1}{2 + 2} = \frac{1}{4} \] Thus, \( \lim_{x \to 0} \frac{\sqrt{x + 4} - 2}{x} = \frac{1}{4} \).
3.4) Technique 4: Dividing by the Highest Power (Limits at Infinity)
When evaluating limits as \( x \) approaches \( \infty \) or \( -\infty \) for rational functions, dividing both the numerator and the denominator by the highest power of \( x \) in the denominator can simplify the expression and help determine the limit.
When to Use: Limits of rational functions as \( x \to \infty \) or \( x \to -\infty \).
Example 13: Limit at Infinity using Dividing by Highest Power
Find \( \lim_{x \to \infty} \frac{3x^2 - 5x + 7}{6x^2 + 2x - 1} \).
Divide numerator and denominator by \( x^2 \) (the highest power of \( x \) in the denominator): \[ \lim_{x \to \infty} \frac{3x^2 - 5x + 7}{6x^2 + 2x - 1} = \lim_{x \to \infty} \frac{\frac{3x^2}{x^2} - \frac{5x}{x^2} + \frac{7}{x^2}}{\frac{6x^2}{x^2} + \frac{2x}{x^2} - \frac{1}{x^2}} = \lim_{x \to \infty} \frac{3 - \frac{5}{x} + \frac{7}{x^2}}{6 + \frac{2}{x} - \frac{1}{x^2}} \] As \( x \to \infty \), terms like \( \frac{5}{x}, \frac{7}{x^2}, \frac{2}{x}, \frac{1}{x^2} \) approach 0 (since \( \lim_{x \to \infty} \frac{c}{x^n} = 0 \) for \( n > 0 \)). \[ = \frac{3 - 0 + 0}{6 + 0 - 0} = \frac{3}{6} = \frac{1}{2} \] Thus, \( \lim_{x \to \infty} \frac{3x^2 - 5x + 7}{6x^2 + 2x - 1} = \frac{1}{2} \).
4) Combining Limit Laws and Techniques - More Complex Examples
Often, solving a limit problem requires using a combination of limit laws and techniques. Let's look at some more complex examples.
Example 14: Combining Laws and Factoring
Find \( \lim_{x \to 1} \frac{x^3 - 1}{x^2 - 1} \).
Direct substitution gives \( \frac{1^3 - 1}{1^2 - 1} = \frac{0}{0} \). Indeterminate form! Factor both numerator and denominator. Numerator is a difference of cubes: \( x^3 - 1 = (x - 1)(x^2 + x + 1) \). Denominator is a difference of squares: \( x^2 - 1 = (x - 1)(x + 1) \). \[ \lim_{x \to 1} \frac{x^3 - 1}{x^2 - 1} = \lim_{x \to 1} \frac{(x - 1)(x^2 + x + 1)}{(x - 1)(x + 1)} \] Cancel the common factor \( (x - 1) \) for \( x \neq 1 \): \[ = \lim_{x \to 1} \frac{x^2 + x + 1}{x + 1} \] Now use the quotient law and direct substitution (for polynomials): \[ = \frac{\lim_{x \to 1} (x^2 + x + 1)}{\lim_{x \to 1} (x + 1)} = \frac{(1)^2 + 1 + 1}{1 + 1} = \frac{3}{2} \] Thus, \( \lim_{x \to 1} \frac{x^3 - 1}{x^2 - 1} = \frac{3}{2} \).
Example 15: Combining Laws and Rationalization with Square Roots
Find \( \lim_{x \to 5} \frac{\sqrt{x + 4} - 3}{x - 5} \).
Direct substitution gives \( \frac{\sqrt{5 + 4} - 3}{5 - 5} = \frac{\sqrt{9} - 3}{0} = \frac{3 - 3}{0} = \frac{0}{0} \). Indeterminate form! Rationalize the numerator by multiplying by \( \sqrt{x + 4} + 3 \): \[ \lim_{x \to 5} \frac{\sqrt{x + 4} - 3}{x - 5} \cdot \frac{\sqrt{x + 4} + 3}{\sqrt{x + 4} + 3} = \lim_{x \to 5} \frac{(x + 4) - 9}{(x - 5)(\sqrt{x + 4} + 3)} = \lim_{x \to 5} \frac{x - 5}{(x - 5)(\sqrt{x + 4} + 3)} \] Cancel the common factor \( (x - 5) \) for \( x \neq 5 \): \[ = \lim_{x \to 5} \frac{1}{\sqrt{x + 4} + 3} \] Now use the quotient, sum, root laws and direct substitution: \[ = \frac{1}{\sqrt{\lim_{x \to 5} (x + 4)} + \lim_{x \to 5} 3} = \frac{1}{\sqrt{5 + 4} + 3} = \frac{1}{\sqrt{9} + 3} = \frac{1}{3 + 3} = \frac{1}{6} \] Thus, \( \lim_{x \to 5} \frac{\sqrt{x + 4} - 3}{x - 5} = \frac{1}{6} \).
5) Practice Questions - Applying Limit Laws and Techniques
It's practice time! These questions will test your ability to apply limit laws and techniques to evaluate various limits.
Fundamental Practice Questions
Instructions: Evaluate each limit using limit laws and techniques discussed. Show your steps.
- Q1. Evaluate \( \lim_{x \to -2} (3x^2 - 4x + 5) \).
- Q2. Find \( \lim_{x \to 4} \frac{2x + 1}{x - 3} \).
- Q3. Evaluate \( \lim_{x \to 0} \frac{\sqrt{x + 9} - 3}{x} \). (Rationalize numerator).
- Q4. Find \( \lim_{x \to 2} \frac{x^2 - 4}{x - 2} \). (Factor and cancel).
- Q5. Evaluate \( \lim_{x \to \infty} \frac{4x + 3}{2x - 1} \). (Divide by highest power of x).
- Q6. Find \( \lim_{x \to -1} (x + 1)^2 \cdot (2x - 3) \).
- Q7. Evaluate \( \lim_{x \to 3} \sqrt{\frac{x^2 - 9}{x - 3}} \). (Simplify inside the root first).
- Q8. Find \( \lim_{x \to \infty} \frac{x^3 + 2x}{2x^3 - x^2 + 5} \). (Divide by highest power of x).
- Q9. Evaluate \( \lim_{x \to 1} \frac{x^2 + 2x - 3}{x - 1} \). (Factor and cancel).
- Q10. Find \( \lim_{x \to 0} \frac{\sqrt{x + 16} - 4}{x} \). (Rationalize numerator).
Challenging Practice Questions
Instructions: These questions may require more creative application of limit laws and techniques. Show all steps clearly.
- Q1. Evaluate \( \lim_{h \to 0} \frac{(x + h)^2 - x^2}{h} \). (Treat \( x \) as a constant, and \( h \) as the variable approaching 0).
- Q2. Find \( \lim_{x \to 0} \frac{\frac{1}{x+2} - \frac{1}{2}}{x} \). (Simplify the complex fraction).
- Q3. Evaluate \( \lim_{x \to \infty} (\sqrt{x^2 + 3x} - x) \). (Rationalize, thinking of it as \( \infty - \infty \) form).
- Q4. Find \( \lim_{x \to -2} \frac{x^3 + 8}{x + 2} \). (Factor sum of cubes: \( a^3 + b^3 = (a+b)(a^2 - ab + b^2) \)).
- Q5. Evaluate \( \lim_{x \to 7} \frac{\sqrt{x + 2} - \sqrt{9}}{x - 7} \). (Rationalize numerator, involving two square roots).
6) Summary & Cheat Sheet - Limit Laws and Evaluation Techniques
Let's summarize the key limit laws and techniques we've covered. Use this as your quick reference guide for limit evaluation.
6.1) Core Limit Laws
- Limit of a Constant: \( \lim_{x \to a} c = c \)
- Limit of \( x \): \( \lim_{x \to a} x = a \)
- Sum Law: \( \lim_{x \to a} [f(x) + g(x)] = \lim_{x \to a} f(x) + \lim_{x \to a} g(x) \)
- Difference Law: \( \lim_{x \to a} [f(x) - g(x)] = \lim_{x \to a} f(x) - \lim_{x \to a} g(x) \)
- Constant Multiple Law: \( \lim_{x \to a} [c \cdot f(x)] = c \cdot \lim_{x \to a} f(x) \)
- Product Law: \( \lim_{x \to a} [f(x) \cdot g(x)] = \left[ \lim_{x \to a} f(x) \right] \cdot \left[ \lim_{x \to a} g(x) \right] \)
- Quotient Law: \( \lim_{x \to a} \frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)} \) (if \( \lim_{x \to a} g(x) \neq 0 \))
- Power Law: \( \lim_{x \to a} [f(x)]^n = \left[ \lim_{x \to a} f(x) \right]^n \)
- Root Law: \( \lim_{x \to a} \sqrt[n]{f(x)} = \sqrt[n]{\lim_{x \to a} f(x)} \) (for valid roots)
6.2) Key Techniques for Limit Evaluation
- Direct Substitution: For "well-behaved" functions, try substituting \( x = a \) directly.
- Factoring and Cancellation: For \( \frac{0}{0} \) form in rational functions, factor and cancel common factors.
- Rationalization: For square roots and \( \frac{0}{0} \) or \( \infty - \infty \) forms, multiply by the conjugate.
- Dividing by Highest Power: For limits of rational functions as \( x \to \infty \) or \( -\infty \), divide numerator and denominator by the highest power of \( x \) in the denominator.
Explore more advanced limit problems and techniques in further calculus studies!