Chapter 10: Introduction to Calculus (Limits Preview)
This final chapter of precalculus serves as a bridge to calculus. You have spent the preceding chapters building a toolkit of functions, graphs, and algebraic techniques. Now we turn to the two central questions that motivated the invention of calculus: How do we find the slope of a curve at a single point? And how do we find the area of a region with curved boundaries? Both questions lead to the same fundamental concept: the limit.
10.1 The Intuitive Idea of a Limit
In precalculus, you have evaluated functions by plugging in values: given \(f(x) = x^2\), you know that \(f(3) = 9\). But what happens when you cannot simply plug in a value? Consider the function
\[f(x) = \frac{x^2 - 4}{x - 2}\]At \(x = 2\), we get \(\frac{0}{0}\), which is undefined. The function has no output at \(x = 2\). Yet if we ask what value \(f(x)\) is approaching as \(x\) gets closer and closer to \(2\), we discover something meaningful. This idea of "approaching" is the heart of a limit.
Definition: Limit (Informal)
We write
$$\lim_{x \to a} f(x) = L$$
and say "the limit of \(f(x)\) as \(x\) approaches \(a\) equals \(L\)" if the values of \(f(x)\) get arbitrarily close to \(L\) whenever \(x\) is sufficiently close to \(a\) (but not equal to \(a\)).
The key insight is that a limit describes behavior near a point, not at the point itself. The function does not need to be defined at \(x = a\) for the limit to exist. Think of it as asking: "Where is the function heading?" rather than "Where is the function right now?"
Worked Example 10.1
Investigate \(\displaystyle\lim_{x \to 2} \frac{x^2 - 4}{x - 2}\).
Step 1 Notice that \(f(2)\) is undefined since the denominator is zero. But we can simplify for \(x \neq 2\):
$$\frac{x^2 - 4}{x - 2} = \frac{(x-2)(x+2)}{x-2} = x + 2$$
Step 2 The simplified function \(g(x) = x + 2\) agrees with \(f(x)\) everywhere except at \(x = 2\). As \(x\) approaches \(2\), the expression \(x + 2\) approaches \(4\).
Step 3 Therefore, \(\displaystyle\lim_{x \to 2} \frac{x^2 - 4}{x - 2} = 4\).
The graph of \(f\) is the line \(y = x + 2\) with a hole at the point \((2, 4)\). The limit tells us the \(y\)-value the function is heading toward, even though there is a hole there.
10.2 Evaluating Limits Numerically and Graphically
Numerical Approach: Tables of Values
One way to estimate a limit is to build a table of function values for inputs approaching the target from both sides. Let us revisit the function \(f(x) = \dfrac{x^2 - 4}{x - 2}\) near \(x = 2\):
| \(x\) | 1.9 | 1.99 | 1.999 | 2 | 2.001 | 2.01 | 2.1 |
|---|---|---|---|---|---|---|---|
| \(f(x)\) | 3.9 | 3.99 | 3.999 | undef. | 4.001 | 4.01 | 4.1 |
From both the left and the right, the output values converge toward \(4\). This numerical evidence strongly suggests that the limit is \(4\).
Worked Example 10.2a — Numerical Limit Estimation
Use a table of values to estimate \(\displaystyle\lim_{x \to 0} \frac{\sin x}{x}\) (with \(x\) in radians).
| \(x\) | \(-0.1\) | \(-0.01\) | \(-0.001\) | 0 | 0.001 | 0.01 | 0.1 |
|---|---|---|---|---|---|---|---|
| \(\frac{\sin x}{x}\) | 0.9983 | 0.99998 | 0.9999998 | undef. | 0.9999998 | 0.99998 | 0.9983 |
The values approach \(1\) from both sides. Therefore \(\displaystyle\lim_{x \to 0} \frac{\sin x}{x} = 1\). This is one of the most important limits in all of calculus.
Graphical Approach
A graph provides visual confirmation of a limit. To find \(\displaystyle\lim_{x \to a} f(x)\) graphically, trace the curve from both sides toward \(x = a\). If both sides approach the same \(y\)-value, that value is the limit. Use the interactive graph below to explore how function values behave as \(x\) approaches a target value.
Drag the slider for \(a\) to explore the limit at different points. Blue points approach from the left; green from the right. The open circle marks the target point.
10.3 Limit Laws and Algebraic Techniques
While tables and graphs give useful estimates, we need algebraic methods for exact answers. The following laws allow us to compute limits systematically.
Theorem: Basic Limit Laws
If \(\displaystyle\lim_{x \to a} f(x) = L\) and \(\displaystyle\lim_{x \to a} g(x) = M\), then:
$$\lim_{x \to a} [f(x) + g(x)] = L + M \qquad \text{(Sum Law)}$$
$$\lim_{x \to a} [f(x) - g(x)] = L - M \qquad \text{(Difference Law)}$$
$$\lim_{x \to a} [c \cdot f(x)] = cL \qquad \text{(Constant Multiple Law)}$$
$$\lim_{x \to a} [f(x) \cdot g(x)] = L \cdot M \qquad \text{(Product Law)}$$
$$\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{L}{M}, \quad M \neq 0 \qquad \text{(Quotient Law)}$$
Technique 1: Direct Substitution
The simplest approach: if the function is "well-behaved" at \(x = a\) (specifically, if it is continuous there), then \(\displaystyle\lim_{x \to a} f(x) = f(a)\). All polynomials and rational functions (where the denominator is nonzero) allow direct substitution.
Worked Example 10.3a — Direct Substitution
Evaluate \(\displaystyle\lim_{x \to 3} (x^3 - 4x + 7)\).
Solution Since this is a polynomial, substitute directly:
$$\lim_{x \to 3} (x^3 - 4x + 7) = 27 - 12 + 7 = 22$$
Technique 2: Factoring
When direct substitution produces the indeterminate form \(\frac{0}{0}\), factor the numerator and denominator to cancel the common factor that is causing both to be zero.
Worked Example 10.3b — Factoring
Evaluate \(\displaystyle\lim_{x \to -3} \frac{x^2 + 5x + 6}{x + 3}\).
Step 1 Direct substitution: \(\dfrac{(-3)^2 + 5(-3) + 6}{-3 + 3} = \dfrac{9 - 15 + 6}{0} = \dfrac{0}{0}\). Indeterminate.
Step 2 Factor the numerator: \(x^2 + 5x + 6 = (x + 2)(x + 3)\).
Step 3 Cancel the common factor (valid for \(x \neq -3\)):
$$\frac{(x+2)(x+3)}{x+3} = x + 2$$
Step 4 Substitute: \(\displaystyle\lim_{x \to -3} (x + 2) = -3 + 2 = -1\).
Technique 3: Rationalizing
When the indeterminate form involves square roots, multiply numerator and denominator by the conjugate of the radical expression.
Worked Example 10.3c — Rationalizing
Evaluate \(\displaystyle\lim_{x \to 0} \frac{\sqrt{x + 4} - 2}{x}\).
Step 1 Direct substitution: \(\dfrac{\sqrt{4} - 2}{0} = \dfrac{0}{0}\). Indeterminate.
Step 2 Multiply by the conjugate:
$$\frac{\sqrt{x+4} - 2}{x} \cdot \frac{\sqrt{x+4} + 2}{\sqrt{x+4} + 2} = \frac{(x+4) - 4}{x(\sqrt{x+4} + 2)} = \frac{x}{x(\sqrt{x+4} + 2)}$$
Step 3 Cancel \(x\) (valid for \(x \neq 0\)):
$$= \frac{1}{\sqrt{x+4} + 2}$$
Step 4 Substitute \(x = 0\):
$$\lim_{x \to 0} \frac{1}{\sqrt{x+4} + 2} = \frac{1}{\sqrt{4} + 2} = \frac{1}{4}$$
10.4 One-Sided Limits
Sometimes a function behaves differently depending on which direction \(x\) approaches the target value. We formalize this with one-sided limits.
Definition: One-Sided Limits
Left-hand limit: \(\displaystyle\lim_{x \to a^-} f(x) = L\) means \(f(x) \to L\) as \(x\) approaches \(a\) from values less than \(a\).
Right-hand limit: \(\displaystyle\lim_{x \to a^+} f(x) = L\) means \(f(x) \to L\) as \(x\) approaches \(a\) from values greater than \(a\).
The two-sided limit exists if and only if both one-sided limits exist and are equal:
$$\lim_{x \to a} f(x) = L \quad \Longleftrightarrow \quad \lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = L$$
Worked Example 10.4
Let \(f(x) = \begin{cases} x^2 + 1, & x < 1 \\ 3x - 1, & x \geq 1 \end{cases}\). Does \(\displaystyle\lim_{x \to 1} f(x)\) exist?
Step 1 Left-hand limit: as \(x \to 1^-\), use \(f(x) = x^2 + 1\):
$$\lim_{x \to 1^-} (x^2 + 1) = 1 + 1 = 2$$
Step 2 Right-hand limit: as \(x \to 1^+\), use \(f(x) = 3x - 1\):
$$\lim_{x \to 1^+} (3x - 1) = 3 - 1 = 2$$
Step 3 Since both one-sided limits equal \(2\), the two-sided limit exists: \(\displaystyle\lim_{x \to 1} f(x) = 2\). Moreover, \(f(1) = 3(1) - 1 = 2\), so the function is also continuous at \(x = 1\).
10.5 Limits at Infinity and Horizontal Asymptotes
We can also ask what happens to \(f(x)\) as \(x\) grows without bound. This is the concept of a limit at infinity, which connects directly to the horizontal asymptotes you have already studied in precalculus.
Definition: Limit at Infinity
We write \(\displaystyle\lim_{x \to \infty} f(x) = L\) if the values of \(f(x)\) can be made arbitrarily close to \(L\) by taking \(x\) sufficiently large. If this limit equals \(L\), the line \(y = L\) is a horizontal asymptote of the graph.
Theorem: End Behavior of Rational Functions
For the rational function \(R(x) = \dfrac{a_n x^n + \cdots}{b_m x^m + \cdots}\) with leading coefficients \(a_n\) and \(b_m\):
$$\lim_{x \to \infty} R(x) = \begin{cases} 0, & n < m \\ \dfrac{a_n}{b_m}, & n = m \\ \pm\infty, & n > m \end{cases}$$
Worked Example 10.5
Find all horizontal asymptotes of \(f(x) = \dfrac{3x^2 + 1}{x^2 - 5x + 2}\).
Step 1 Both numerator and denominator have degree 2. The leading coefficients are \(3\) and \(1\).
Step 2 Divide every term by \(x^2\):
$$\frac{3x^2 + 1}{x^2 - 5x + 2} = \frac{3 + \frac{1}{x^2}}{1 - \frac{5}{x} + \frac{2}{x^2}}$$
Step 3 As \(x \to \pm\infty\), the fractional terms vanish:
$$\lim_{x \to \pm\infty} f(x) = \frac{3 + 0}{1 - 0 + 0} = 3$$
The horizontal asymptote is \(y = 3\).
10.6 The Tangent Line Problem
One of the great motivating problems of calculus is this: given a curve, how do we find the slope of the line tangent to the curve at a single point? A tangent line touches the curve at exactly one point (locally) and represents the direction the curve is heading at that instant.
The strategy is to approximate the tangent line using secant lines. A secant line passes through two points on the curve. As we bring the second point closer to the first, the secant line rotates toward the tangent line.
Definition: Slope of the Secant Line
Given a function \(f\) and two points \(x = a\) and \(x = a + h\) on its graph, the slope of the secant line through \((a, f(a))\) and \((a+h, f(a+h))\) is:
$$m_{\text{sec}} = \frac{f(a+h) - f(a)}{h}$$
This expression is called a difference quotient.
As \(h \to 0\), the second point slides along the curve toward the first point, and the secant line approaches the tangent line. The slope of the tangent line is:
$$m_{\text{tan}} = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$$
Use the interactive graph below to see this convergence in action. Drag the slider for \(h\) toward zero and watch the red secant line rotate to match the tangent line.
Drag \(h\) toward 0 to watch the secant line (red) approach the tangent line. The green point is fixed at \(x = a\); the red point slides toward it.
Worked Example 10.6 — Finding the Tangent Slope
Find the slope of the tangent line to \(f(x) = x^2\) at \(x = 3\).
Step 1 Write the difference quotient:
$$\frac{f(3+h) - f(3)}{h} = \frac{(3+h)^2 - 9}{h}$$
Step 2 Expand the numerator:
$$\frac{9 + 6h + h^2 - 9}{h} = \frac{6h + h^2}{h} = 6 + h$$
Step 3 Take the limit as \(h \to 0\):
$$m_{\text{tan}} = \lim_{h \to 0} (6 + h) = 6$$
The slope of the tangent line at \(x = 3\) is \(6\). The tangent line equation is \(y - 9 = 6(x - 3)\), or \(y = 6x - 9\).
10.7 Average vs. Instantaneous Rate of Change
The difference quotient from the previous section has a physical interpretation. If \(f(t)\) represents position at time \(t\), then the difference quotient gives the average rate of change (average velocity) over a time interval, while the limit gives the instantaneous rate of change (instantaneous velocity) at a single moment.
Definition: Average Rate of Change
The average rate of change of \(f\) on the interval \([a, b]\) is:
$$\text{Average rate} = \frac{f(b) - f(a)}{b - a}$$
Geometrically, this is the slope of the secant line through \((a, f(a))\) and \((b, f(b))\).
Definition: Instantaneous Rate of Change
The instantaneous rate of change of \(f\) at \(x = a\) is:
$$\text{Instantaneous rate} = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$$
Geometrically, this is the slope of the tangent line at \(x = a\). In calculus, this limit is called the derivative of \(f\) at \(a\), written \(f'(a)\).
Worked Example 10.7 — Average vs. Instantaneous Velocity
A ball dropped from a building has height \(s(t) = 100 - 4.9t^2\) meters after \(t\) seconds.
(a) Find the average velocity on \([1, 3]\).
Solution
$$\frac{s(3) - s(1)}{3 - 1} = \frac{(100 - 44.1) - (100 - 4.9)}{2} = \frac{55.9 - 95.1}{2} = \frac{-39.2}{2} = -19.6 \text{ m/s}$$
(b) Find the instantaneous velocity at \(t = 2\).
Solution
$$\lim_{h \to 0} \frac{s(2+h) - s(2)}{h} = \lim_{h \to 0} \frac{[100 - 4.9(2+h)^2] - [100 - 4.9(4)]}{h}$$
$$= \lim_{h \to 0} \frac{-4.9(4 + 4h + h^2) + 19.6}{h} = \lim_{h \to 0} \frac{-19.6 - 19.6h - 4.9h^2 + 19.6}{h}$$
$$= \lim_{h \to 0} \frac{-19.6h - 4.9h^2}{h} = \lim_{h \to 0} (-19.6 - 4.9h) = -19.6 \text{ m/s}$$
The negative sign indicates the ball is moving downward at \(19.6\) m/s at the instant \(t = 2\).
Introduction to the Derivative Concept
The limit of the difference quotient appears so frequently in calculus that it receives its own name and notation.
Definition: The Derivative (Preview)
The derivative of \(f\) at \(x = a\) is defined as:
$$f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$$
provided this limit exists. The derivative gives the slope of the tangent line to the graph of \(f\) at the point \((a, f(a))\), and represents the instantaneous rate of change of \(f\) at \(x = a\).
You will study derivatives in depth in calculus. For now, the important takeaway is that the derivative is simply a limit, and you already have the tools to compute it for polynomial functions using the difference quotient.
10.8 Area Under a Curve Preview
The second great problem of calculus is computing the area of a region bounded by a curve. For rectangles, triangles, and circles, you already know area formulas. But what about the area under a general curve like \(y = x^2\) from \(x = 0\) to \(x = 1\)?
The key idea is approximation by rectangles. Divide the interval into \(n\) equal subintervals, build a rectangle on each subinterval whose height matches the function value, and add up the areas. As \(n\) increases, the approximation improves. The exact area is the limit as \(n \to \infty\).
Definition: Right Riemann Sum
To approximate the area under \(f(x)\) from \(x = a\) to \(x = b\) using \(n\) rectangles of equal width:
$$\Delta x = \frac{b - a}{n}, \qquad R_n = \sum_{i=1}^{n} f\!\left(a + i\,\Delta x\right) \cdot \Delta x$$
The exact area is \(\displaystyle A = \lim_{n \to \infty} R_n\).
Worked Example 10.8 — Riemann Sum Approximation
Estimate the area under \(f(x) = x^2\) from \(x = 0\) to \(x = 1\) using \(n = 4\) right-endpoint rectangles.
Step 1 Compute \(\Delta x = \frac{1 - 0}{4} = 0.25\). The right endpoints are \(x_1 = 0.25,\ x_2 = 0.5,\ x_3 = 0.75,\ x_4 = 1\).
Step 2 Evaluate the function at each right endpoint:
- \(f(0.25) = 0.0625\)
- \(f(0.5) = 0.25\)
- \(f(0.75) = 0.5625\)
- \(f(1) = 1\)
Step 3 Sum the rectangle areas:
$$R_4 = (0.0625 + 0.25 + 0.5625 + 1)(0.25) = 1.875 \times 0.25 = 0.46875$$
This is an overestimate because \(f(x) = x^2\) is increasing on \([0, 1]\) and we used right endpoints. The exact area (which you will learn to compute using integration in calculus) is \(\frac{1}{3} \approx 0.3333\). With more rectangles, the approximation converges to this exact value.
Key Insight: Limits Connect Everything
Both the tangent line problem and the area problem are solved using limits. The derivative is the limit of a difference quotient, and the definite integral is the limit of a Riemann sum. This unifying role of limits is why they are the starting point of calculus.
10.9 Practice Problems
Test your understanding with the following problems. Attempt each one before revealing the solution.
Problem 1
Evaluate \(\displaystyle\lim_{x \to 5} \frac{x^2 - 25}{x - 5}\).
Show Solution
Direct substitution gives \(\frac{0}{0}\). Factor the numerator:
$$\frac{x^2 - 25}{x - 5} = \frac{(x-5)(x+5)}{x-5} = x + 5 \quad (x \neq 5)$$
$$\lim_{x \to 5} (x + 5) = 10$$
Problem 2
Evaluate \(\displaystyle\lim_{x \to 0} \frac{\sqrt{x + 1} - 1}{x}\).
Show Solution
Direct substitution gives \(\frac{0}{0}\). Multiply by the conjugate:
$$\frac{\sqrt{x+1} - 1}{x} \cdot \frac{\sqrt{x+1} + 1}{\sqrt{x+1} + 1} = \frac{(x+1) - 1}{x(\sqrt{x+1} + 1)} = \frac{1}{\sqrt{x+1} + 1}$$
$$\lim_{x \to 0} \frac{1}{\sqrt{x+1} + 1} = \frac{1}{1 + 1} = \frac{1}{2}$$
Problem 3
Evaluate \(\displaystyle\lim_{x \to \infty} \frac{4x^3 + x}{2x^3 - 5x^2 + 1}\).
Show Solution
Both numerator and denominator have degree 3. Divide every term by \(x^3\):
$$\frac{4 + \frac{1}{x^2}}{2 - \frac{5}{x} + \frac{1}{x^3}}$$
As \(x \to \infty\), all fractional terms vanish:
$$\lim_{x \to \infty} = \frac{4}{2} = 2$$
The horizontal asymptote is \(y = 2\).
Problem 4
Let \(f(x) = \begin{cases} 2x + 3, & x < 2 \\ x^2, & x \geq 2 \end{cases}\). Does \(\displaystyle\lim_{x \to 2} f(x)\) exist? Is \(f\) continuous at \(x = 2\)?
Show Solution
Left-hand limit: \(\displaystyle\lim_{x \to 2^-} (2x + 3) = 4 + 3 = 7\)
Right-hand limit: \(\displaystyle\lim_{x \to 2^+} x^2 = 4\)
Since \(7 \neq 4\), the one-sided limits are not equal. The two-sided limit does not exist, and therefore \(f\) is not continuous at \(x = 2\). This is a jump discontinuity.
Problem 5
Use the difference quotient to find the slope of the tangent line to \(f(x) = x^2 + 3x\) at \(x = 2\).
Show Solution
Compute the difference quotient:
$$\frac{f(2+h) - f(2)}{h} = \frac{[(2+h)^2 + 3(2+h)] - [4 + 6]}{h}$$
$$= \frac{[4 + 4h + h^2 + 6 + 3h] - 10}{h} = \frac{7h + h^2}{h} = 7 + h$$
$$f'(2) = \lim_{h \to 0} (7 + h) = 7$$
The slope of the tangent line at \(x = 2\) is \(7\).
Problem 6
Use the difference quotient to find the slope of the tangent line to \(f(x) = \dfrac{1}{x}\) at \(x = 3\).
Show Solution
Compute the difference quotient:
$$\frac{f(3+h) - f(3)}{h} = \frac{\frac{1}{3+h} - \frac{1}{3}}{h} = \frac{\frac{3 - (3+h)}{3(3+h)}}{h} = \frac{-h}{3h(3+h)} = \frac{-1}{3(3+h)}$$
$$f'(3) = \lim_{h \to 0} \frac{-1}{3(3+h)} = \frac{-1}{3 \cdot 3} = -\frac{1}{9}$$
The slope of the tangent line at \(x = 3\) is \(-\frac{1}{9}\).
Problem 7
A car's position is given by \(s(t) = t^3 - 2t\) meters at time \(t\) seconds. Find the average velocity on \([1, 4]\) and the instantaneous velocity at \(t = 2\).
Show Solution
Average velocity on \([1, 4]\):
$$\frac{s(4) - s(1)}{4 - 1} = \frac{(64 - 8) - (1 - 2)}{3} = \frac{56 - (-1)}{3} = \frac{57}{3} = 19 \text{ m/s}$$
Instantaneous velocity at \(t = 2\):
$$\lim_{h \to 0} \frac{s(2+h) - s(2)}{h} = \lim_{h \to 0} \frac{[(2+h)^3 - 2(2+h)] - [8 - 4]}{h}$$
$$= \lim_{h \to 0} \frac{[8 + 12h + 6h^2 + h^3 - 4 - 2h] - 4}{h} = \lim_{h \to 0} \frac{10h + 6h^2 + h^3}{h}$$
$$= \lim_{h \to 0} (10 + 6h + h^2) = 10 \text{ m/s}$$
Problem 8
Evaluate \(\displaystyle\lim_{x \to 1} \frac{x^3 - 1}{x - 1}\).
Show Solution
Direct substitution gives \(\frac{0}{0}\). Factor using the difference of cubes formula:
$$x^3 - 1 = (x - 1)(x^2 + x + 1)$$
$$\frac{(x-1)(x^2 + x + 1)}{x - 1} = x^2 + x + 1 \quad (x \neq 1)$$
$$\lim_{x \to 1} (x^2 + x + 1) = 1 + 1 + 1 = 3$$
Problem 9
Estimate the area under \(f(x) = x^3\) from \(x = 0\) to \(x = 2\) using a right Riemann sum with \(n = 4\) rectangles.
Show Solution
\(\Delta x = \frac{2 - 0}{4} = 0.5\). The right endpoints are \(0.5,\ 1,\ 1.5,\ 2\).
Function values:
- \(f(0.5) = 0.125\)
- \(f(1) = 1\)
- \(f(1.5) = 3.375\)
- \(f(2) = 8\)
$$R_4 = (0.125 + 1 + 3.375 + 8)(0.5) = 12.5 \times 0.5 = 6.25$$
The exact area (computed in calculus using integration) is \(\frac{2^4}{4} = 4\). The right Riemann sum overestimates because \(f(x) = x^3\) is increasing on \([0, 2]\).
Problem 10
Evaluate \(\displaystyle\lim_{x \to \infty} \frac{2x + 1}{5x - 3}\).
Show Solution
Both numerator and denominator are degree 1. Divide every term by \(x\):
$$\frac{2 + \frac{1}{x}}{5 - \frac{3}{x}}$$
As \(x \to \infty\):
$$\lim_{x \to \infty} \frac{2 + 0}{5 - 0} = \frac{2}{5}$$
The horizontal asymptote is \(y = \frac{2}{5}\).
Problem 11
Let \(f(x) = \begin{cases} ax + 1, & x \leq 2 \\ x^2 - 3, & x > 2 \end{cases}\). Find the value of \(a\) that makes \(f\) continuous at \(x = 2\).
Show Solution
For continuity at \(x = 2\), we need \(\displaystyle\lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) = f(2)\).
Left-hand limit and function value: \(\displaystyle\lim_{x \to 2^-} (ax + 1) = 2a + 1\), and \(f(2) = 2a + 1\).
Right-hand limit: \(\displaystyle\lim_{x \to 2^+} (x^2 - 3) = 4 - 3 = 1\).
Set them equal: \(2a + 1 = 1\), so \(a = 0\).
Problem 12
Find the equation of the tangent line to \(f(x) = x^2 - x\) at the point where \(x = 4\).
Show Solution
Find the point: \(f(4) = 16 - 4 = 12\). The point is \((4, 12)\).
Find the slope using the difference quotient:
$$\frac{f(4+h) - f(4)}{h} = \frac{[(4+h)^2 - (4+h)] - 12}{h} = \frac{16 + 8h + h^2 - 4 - h - 12}{h} = \frac{7h + h^2}{h} = 7 + h$$
$$m = \lim_{h \to 0} (7 + h) = 7$$
Write the equation: Using point-slope form:
$$y - 12 = 7(x - 4) \quad \Longrightarrow \quad y = 7x - 16$$