Find the Tangent Line Slope With the Limit Definition
Find the slope of a tangent line from first principles: lim h->0 [f(a+h) - f(a)]/h. Worked examples for x², 1/x and √x, plus the secant-line picture.
Correcting a Common Misconception About Tangent Lines
Most textbooks say a tangent line touches a curve at exactly one point and never crosses it. This is wrong. At an inflection point, the tangent line crosses the curve. For a cubic like f(x)=x³ at x=0, the tangent line y=0 cuts through the curve. The honest definition is the limit of secant lines as the second point approaches the first. The slope of that limiting line is the derivative. The definition comes from OpenStax Calculus Vol. 1 section 3.1: the tangent line through P(c, f(c)) has slope m_tan = lim_{x→c} (f(x)-f(c))/(x-c), provided the limit exists.
From Secant Slope to Tangent Slope
A secant line goes through two distinct points on a curve. Its slope is the average rate of change between those points. If those points are at x=a and x=x, the secant line slope is (f(x)-f(a))/(x-a). As the second point gets closer to the first, the secant slopes approach a single number: the instantaneous rate of change, which is the slope of the tangent line. That process is the limit definition of the derivative. If the two-sided limit does not exist, the function is not differentiable at that point. A corner, cusp, or vertical tangent means no unique tangent line exists. The TI-84 Plus nDeriv( function uses a symmetric difference quotient with h=0.001 and does not check differentiability, so it can give a false slope at a corner. Always verify the limit by hand.
Two Forms of the Limit Definition
The limit definition appears in two interchangeable forms. The first is f'(a) = lim_{x→a} (f(x)-f(a))/(x-a). The second is f'(a) = lim_{h→0} (f(a+h)-f(a))/h. Both give the same slope. Use the first when the problem gives specific x-values. Use the second when the algebra is simpler with a small increment h. OpenStax Calculus Vol. 1 section 3.1 states both forms. The point is always the same: take the difference quotient of a secant line and let the gap shrink to zero.
Example: f(x) = x² at x = 3
Find the slope of the tangent line to f(x)=x² at x=3 using the limit definition. Use the h-form: f'(3) = lim_{h→0} ((3+h)² - 3²)/h = lim_{h→0} (9 + 6h + h² - 9)/h = lim_{h→0} (6h + h²)/h = lim_{h→0} (6 + h) = 6. The slope is 6. The point of tangency is (3, 9). The tangent line equation in point-slope form is y - 9 = 6(x - 3), or y = 6x - 9. The normal line slope is -1/6. Horizontal tangents occur where f'(x)=0, which for x² is at x=0. Vertical tangents do not occur for polynomial functions.
Example: f(x) = 1/x (Common-Denominator Algebra)
Find the slope of the tangent line to f(x)=1/x at x=2. Use the x-form: f'(2) = lim_{x→2} (1/x - 1/2)/(x - 2). Combine the numerator: (2 - x)/(2x) over (x - 2) gives (2 - x)/(2x(x - 2)). Factor -1 from the numerator: -(x - 2)/(2x(x - 2)) = -1/(2x). As x→2, the limit is -1/4. The slope is -1/4. The point is (2, 1/2). The tangent line equation is y - 1/2 = (-1/4)(x - 2), or y = -x/4 + 1. The normal line slope is 4, giving the line y - 1/2 = 4(x - 2). OpenStax Calculus Vol. 1 Exercise 59 confirms the normal line to 1/x at x=2 is y = 4x - 15/2.
Example: f(x) = √x (Conjugate Trick)
Find the slope of the tangent line to f(x)=√x at x=4. Use the h-form: f'(4) = lim_{h→0} (√(4+h) - √4)/h = lim_{h→0} (√(4+h) - 2)/h. Multiply numerator and denominator by the conjugate (√(4+h) + 2): ( (4+h) - 4 )/(h(√(4+h) + 2)) = h/(h(√(4+h) + 2)) = 1/(√(4+h) + 2). As h→0, the limit is 1/(√4 + 2) = 1/4. The slope is 1/4. The point is (4, 2). The tangent line equation is y - 2 = (1/4)(x - 4), or y = x/4 + 1. The normal line slope is -4, giving y - 2 = -4(x - 4). OpenStax Calculus Vol. 1 Exercise 61 gives the normal line to √x at x=4 as y = -4x + 18. The conjugate trick is essential for square root functions; without it, the denominator goes to zero and the limit is indeterminate.
Writing the Tangent Line Equation from the Slope
Once you have the slope f'(a) and the point (a, f(a)), write the tangent line equation using point-slope form: y - f(a) = f'(a)(x - a). That is the standard output for homework and exams. Slope-intercept form y = mx + b is fine for graphing. The linearization L(x) = f(a) + f'(a)(x - a) is the same line used for approximation. The error when using linearization is bounded by (M/2)(x-a)², where M is the maximum of |f''| on the interval between a and x. For f(x)=√x at x=4, L(x) = 2 + (1/4)(x-4). At x=4.1, the approximation gives 2.025; the actual value is √4.1 ≈ 2.02485, an error of about 0.00015.
| h | Secant Slope ( (3+h)² - 9 ) / h | Distance from Tangent Slope |
|---|---|---|
| 1 | (16 - 9)/1 = 7 | 1 |
| 0.1 | (9.61 - 9)/0.1 = 6.1 | 0.1 |
| 0.01 | (9.0601 - 9)/0.01 = 6.01 | 0.01 |
| 0.001 | (9.006001 - 9)/0.001 = 6.001 | 0.001 |
| -0.1 | (8.41 - 9)/(-0.1) = 5.9 | 0.1 |
| -0.01 | (8.9401 - 9)/(-0.01) = 5.99 | 0.01 |
| -0.001 | (8.994001 - 9)/(-0.001) = 5.999 | 0.001 |
Graph of Secants Approaching the Tangent
Numerical Convergence
The secant lines through (3, 9) and (3+h, (3+h)²) rotate toward the tangent line y=6x-9 as h shrinks. For h=1, the secant slope is 7. For h=0.1, it is 6.1. For h=0.01, it is 6.01. From the negative side, h=-0.1 gives 5.9, and h=-0.01 gives 5.99. The secant slopes converge to 6 from both sides. This numeric convergence is what the limit definition captures.
Calculator Pitfalls
If you graph these lines on a calculator or software, you see the secants cluster around the tangent. The TI-84 Plus DRAW Tangent( command draws the tangent at the cursor point automatically, but it relies on nDeriv( internally. For f(x)=x² at x=3, nDeriv( gives 6 correctly, but for |x| at x=0 it gives 0, which is wrong because the function is not differentiable there.
What to Do Next
Your next homework problem will ask you to find the slope of a tangent line using the limit definition for a function like f(x)=x³ or f(x)=1/(x+1). Write down the difference quotient, simplify algebraically (expand, factor, conjugate, or find a common denominator), cancel the h or x-a term, then evaluate the limit. The single thing that most often goes wrong is failing to simplify before substituting. Always check that the denominator eliminates the zero before you take the limit. Use the three examples above as templates: polynomial, rational, and radical cover 90% of first-semester problems. If the function is trigonometric, remember the special limit sin(θ)/θ → 1.
Common Questions
What is the difference between a secant line and a tangent line?
A secant line goes through two distinct points on a curve. A tangent line is the limit of secant lines as the second point approaches the first. The secant always has two intersection points; the tangent has a double intersection at the point of tangency. The secant line slope is the difference quotient; the tangent line slope is the derivative.
When does the tangent line cross the curve instead of just touching it?
At an inflection point, the tangent line crosses the curve. For f(x)=x³ at x=0, the tangent line is y=0, which crosses the curve. This happens because the second derivative changes sign at the inflection point. The old rule that a tangent never crosses is false.
What should I do if my calculator's nDeriv( gives a slope but the function has a corner?
Check differentiability by computing the left-hand and right-hand limits of the difference quotient. For |x| at x=0, the left-hand limit is -1 and the right-hand limit is +1, so no derivative exists. nDeriv( returns 0, which is meaningless. Use the limit definition by hand to verify.
How do I find the tangent line when the derivative is undefined?
If the limit of secant slopes goes to +∞ or -∞, the tangent is vertical. Its equation is x = a. For example, f(x)=x^(1/3) has a vertical tangent at x=0. No finite slope exists, and the line cannot be written in y=mx+b form.
What is the most common mistake when using the limit definition?
Forgetting to simplify the difference quotient before taking the limit. For f(x)=x², students often write lim_{h→0} ( (3+h)² - 9 )/h and try to evaluate by plugging h=0, getting 0/0. You must expand and cancel h first. The same applies to square roots: use the conjugate to eliminate the radical in the numerator before letting h go to zero.