Lesson 3: Finding Limits Analytically
Finding Limits Analytically
To find \(\lim_{x \to c} f(x)\) analytically, we consider \(f(c)\), that is, we plug \(x = c\) into \(f(x)\). The result of this substitution determines which approach we must use to evaluate the limit. We will examine three possible scenarios based on what happens when we substitute \(x = c\) into the function.
Case 1: Direct Substitution Yields a Number
Case 1: If \(f(c)\) returns a number (note that 0 is acceptable), then \(\lim_{x \to c} f(x) = f(c)\).
This is the simplest case. When direct substitution produces a real number, that number is the limit. This property holds for all continuous functions at points where they are defined.
Example 1: Direct Substitution
Evaluate the following limit analytically:
\[\lim_{x \to 3} (2x^2 + 7)\]Solution:
We substitute \(x = 3\) directly into the function \(2x^2+7\):
\[2\cdot 3^2 + 7 = 2\cdot 9 + 7 = 18 + 7 = 25\]Since this produces a real number, we have:
\[\lim_{x \to 3} (2x^2 + 7) = 25\]Case 2: Nonzero Number Over Zero
Case 2: If \(f(c) = \frac{\text{nonzero number}}{0}\), then \(f(x)\) has a vertical asymptote at \(x = c\). Therefore, \(\lim_{x \to c} f(x)\) can be equal to \(\infty\), \(-\infty\), or DNE. We need to evaluate the left-sided and right-sided limits to determine \(\lim_{x \to c} f(x)\).
When substitution yields a nonzero number divided by zero, the function has a vertical asymptote at that point. The behavior of the function as it approaches the asymptote from each side determines whether the limit is infinite or does not exist.
Example 2: Vertical Asymptote with Same-Signed Limits
Evaluate the following limit analytically:
\[\lim_{x \to -1} \frac{10}{(x + 1)^2}\]Solution:
First, we attempt direct substitution:
\[\frac{10}{(-1 + 1)^2} = \frac{10}{0}\]This is a nonzero number divided by zero, indicating a vertical asymptote at \(x = -1\). We must examine the one-sided limits.
For \(x\) slightly less than \(-1\) (approaching from the left): \(x + 1\) is a small negative number, so \((x + 1)^2\) is a small positive number. Therefore, \(\frac{10}{(x + 1)^2}\) approaches \(+\infty\).
For \(x\) slightly greater than \(-1\) (approaching from the right): \(x + 1\) is a small positive number, so \((x + 1)^2\) is a small positive number. Therefore, \(\frac{10}{(x + 1)^2}\) approaches \(+\infty\).
Since both one-sided limits are equal:
\[\lim_{x \to -1} \frac{10}{(x + 1)^2} = \infty\]Example 3: Vertical Asymptote with Different-Signed Limits
Evaluate the following limit analytically:
\[\lim_{x \to 3} \frac{1}{x - 3}\]Solution:
First, we attempt direct substitution:
\[\frac{1}{3 - 3} = \frac{1}{0}\]This is a nonzero number divided by zero, indicating a vertical asymptote at \(x = 3\). We examine the one-sided limits.
For \(x\) slightly less than 3 (approaching from the left): \(x - 3\) is a small negative number. Therefore, \(\frac{1}{x - 3}\) approaches \(-\infty\).
For \(x\) slightly greater than 3 (approaching from the right): \(x - 3\) is a small positive number. Therefore, \(\frac{1}{x - 3}\) approaches \(+\infty\).
Since the one-sided limits are different:
\[\lim_{x \to 3} \frac{1}{x - 3} = \text{DNE}\]Case 3: Zero Over Zero (Indeterminate Form)
Case 3: If \(f(c) = \frac{0}{0}\), then \(f(x)\) has a hole or vertical asymptote at \(x = c\). We need to manipulate \(f(x)\) using algebra so that \(f(c)\) returns a number or \(\frac{\text{nonzero number}}{0}\), that is, case 1 or case 2.
The indeterminate form \(\frac{0}{0}\) indicates that both the numerator and denominator have a factor that becomes zero at \(x = c\). We must simplify the expression algebraically, typically by factoring and canceling common factors, to reveal the true behavior of the limit.
Example 4: Removable Discontinuity (Hole)
Evaluate the following limit analytically:
\[\lim_{x \to 0} \frac{x^3 - 6x^2}{x^3 + 3x^2}\]Solution:
First, we attempt direct substitution:
\[\frac{0^3 - 6\cdot 0^2}{0^3 + 3\cdot 0^2} = \frac{0}{0}\]This is the indeterminate form, so we must factor and simplify. We factor out the greatest common factor from both the numerator and denominator:
\[\frac{x^3 - 6x^2}{x^3 + 3x^2} = \frac{x^2(x - 6)}{x^2(x + 3)}\]Cancel the common factor \(x^2\) (valid for \(x \neq 0\)):
\[\frac{x^2(x - 6)}{x^2(x + 3)} = \frac{x - 6}{x + 3}\]Now substitute \(x = 0\) into the simplified expression:
\[\frac{0 - 6}{0 + 3} = \frac{-6}{3} = -2\]Therefore:
\[\lim_{x \to 0} \frac{x^3 - 6x^2}{x^3 + 3x^2} = -2\]Example 5: Indeterminate Form Leading to Vertical Asymptote
Evaluate the following limit analytically:
\[\lim_{x \to 1} \frac{x^2 - x}{(x - 1)^2}\]Solution:
First, we attempt direct substitution:
\[\frac{1^2 - 1}{(1 - 1)^2} = \frac{0}{0}\]This is the indeterminate form, so we factor the numerator:
\[\frac{x^2 - x}{(x - 1)^2} = \frac{x(x - 1)}{(x - 1)^2}\]Cancel the common factor \((x - 1)\) (valid for \(x \neq 1\)):
\[\frac{x(x - 1)}{(x - 1)^2} = \frac{x}{x - 1}\]Now substitute \(x = 1\) into the simplified expression:
\[\frac{1}{1 - 1} = \frac{1}{0}\]This is a nonzero number over zero (case 2), indicating a vertical asymptote. We examine the one-sided limits.
For \(x\) slightly less than 1: \(x\) is close to 1 (positive), and \(x - 1\) is a small negative number. Therefore, \(\frac{x}{x - 1}\) approaches \(-\infty\).
For \(x\) slightly greater than 1: \(x\) is close to 1 (positive), and \(x - 1\) is a small positive number. Therefore, \(\frac{x}{x - 1}\) approaches \(+\infty\).
Since the one-sided limits are different:
\[\lim_{x \to 1} \frac{x^2 - x}{(x - 1)^2} = \text{DNE}\]Limits of Piecewise Functions
We can also find limits of piecewise functions analytically. For piecewise functions, we must carefully consider which piece of the function applies as we approach the point from each direction.
Example 6: Limits of a Piecewise Function
Suppose
\[f(x) = \begin{cases} x + 3 & \text{if } x \leq 0, \\ x^2 + 3 & \text{if } 0 < x < 1, \\ 2x^2 + 1 & \text{if } x \geq 1. \end{cases}\]Evaluate the following limits analytically:
\[\begin{align}&\lim_{x \to 0^-}f(x) \qquad &\lim_{x \to 1^-}f(x)\\ &\lim_{x \to 0^+}f(x) \qquad &\lim_{x \to 1^+}f(x)\\ &\lim_{x \to 0}f(x) \qquad &\lim_{x \to 1}f(x)\end{align}\]Solution:
For \(x \to 0\):
Left-sided limit: As \(x\) approaches 0 from the left, we use the first piece \(f(x) = x + 3\).
\[\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (x + 3) = 0 + 3 = 3\]Right-sided limit: As \(x\) approaches 0 from the right, we use the second piece \(f(x) = x^2 + 3\).
\[\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (x^2+3) = 0^2 + 3 = 3\]Two-sided limit: Since the one-sided limits are equal,
\[\lim_{x \to 0} f(x) = 3.\]For \(x \to 1\):
Left-sided limit: As \(x\) approaches 1 from the left, we use the second piece \(f(x) = x^2 + 3\).
\[\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x^2 + 3) = 1^2 + 3 = 4\]Right-sided limit: As \(x\) approaches 1 from the right, we use the third piece \(f(x) = 2x^2 + 1\).
\[\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (2x^2 + 1) = 2\cdot 1^2 + 1 = 2 + 1 = 3\]Two-sided limit: Since the one-sided limits are different,
\[\lim_{x \to 1} f(x) = \text{DNE}.\]Limit Properties
The following properties allow us to break down complex limit problems into simpler components. These properties are fundamental tools for analytical limit computation.
Limit Properties:
Let \(c\), \(k\), \(L\), and \(M\) be real numbers, and let \(n\) be a positive integer. If \(\lim_{x \to c} f(x) = L\) and \(\lim_{x \to c} g(x) = M\), then the following properties hold:
\[\lim_{x \to c} [kf(x)] = kL\] \[\lim_{x \to c} [f(x) \pm g(x)] = L \pm M\] \[\lim_{x \to c} [f(x)g(x)] = LM\] \[\lim_{x \to c} \frac{f(x)}{g(x)} = \frac{L}{M}, \quad \text{if \(M \neq 0\)}\] \[\lim_{x \to c} [f(x)]^n = L^n\]These properties allow us to evaluate limits of combinations of functions by working with the limits of individual components. The constant multiple rule states that we can factor constants out of limits. The sum and difference rules allow us to split limits across addition and subtraction. The product and quotient rules extend these operations to multiplication and division, with the quotient rule requiring that the denominator's limit is nonzero. The power rule extends limits through exponentiation.
Example 7: Applying Limit Properties
Suppose \(\lim_{x \to 1} f(x) = 2\) and \(\lim_{x \to 1} g(x) = 4\). Evaluate
\[\lim_{x \to 1} \left[5f(x) - g(x)^2 + 7\right]\]using the limit properties above.
Solution:
We apply the limit properties step by step. First, use the sum/difference rule to separate the limit into three parts:
\[\lim_{x \to 1} \left[5f(x) - g(x)^2 + 7\right] = \lim_{x \to 1} [5f(x)] - \lim_{x \to 1} [g(x)^2] + \lim_{x \to 1} 7\]Apply the constant multiple rule to the first term, the power rule to the second term, and recognize that the limit of a constant is that constant:
\[\lim_{x \to 1} [5f(x)] - \lim_{x \to 1} [g(x)^2] + \lim_{x \to 1} 7 = 5 \cdot \lim_{x \to 1} f(x) - \left[\lim_{x \to 1} g(x)\right]^2 + 7\]Substitute the given limit values:
\[5 \cdot \lim_{x \to 1} f(x) - \left[\lim_{x \to 1} g(x)\right]^2 + 7= 5\cdot 2 - 4^2 + 7= 10 -16+7=1\]Therefore:
\[\lim_{x \to 1} \left[5f(x) - \frac{g(x)}{2} + 7\right] = 1\]Key Takeaway: When evaluating limits analytically, always begin by attempting direct substitution. The result of this substitution determines which case applies and guides the appropriate method for finding the limit. Understanding these three cases and the limit properties provides a systematic approach to analytical limit evaluation.