From a Limit to a Tangent Line
A derivative at a point gives the slope of the tangent line there. To write the line, however, you also need a point on it. That point comes from the function value: if the tangent is at \(x=a\), the point on the curve is \((a,f(a))\).
This tutorial brings those pieces together. First compute \(f'(a)\) from the limit definition, then use the point \((a,f(a))\) and slope \(f'(a)\) to write the tangent line. The order matters: the input \(a\) is not generally the output \(f(a)\).
The limit compares the function values at \(a\) and \(a+h\). For \(h\ne 0\), the quotient is the slope of a secant line through those two points. As \(h\) approaches zero, the secant slope approaches the tangent slope, when that limit exists.
Writing the Equation
A line with slope \(m\) passing through \((x_1,y_1)\) can be written in point-slope form as \(y-y_1=m(x-x_1)\). For a tangent to \(y=f(x)\) at \(x=a\), the point is \((a,f(a))\) and the slope is \(f'(a)\). Therefore the tangent-line equation is:
This form keeps the two required pieces visible: \(f(a)\) gives the vertical coordinate of the point, and \(f'(a)\) gives the slope. You may simplify the equation into slope-intercept form afterward, but the point-slope form is often the clearest way to show how the tangent was obtained.
Why the Formula Works
The tangent-line formula follows from the meaning of the derivative and the point-slope equation. Here is the full connection. Assume \(f(a)\) is defined and the derivative limit at \(a\) exists as a finite number.
At input \(a\), the graph has output \(f(a)\), so the point is \((a,f(a))\).
For \(h\ne 0\), the secant through \((a,f(a))\) and \((a+h,f(a+h))\) has slope \(\frac{f(a+h)-f(a)}{(a+h)-a}=\frac{f(a+h)-f(a)}{h}\). Its limit as \(h\to0\) is \(f'(a)\).
Any line of slope \(f'(a)\) passing through \((a,f(a))\) satisfies \(y-f(a)=f'(a)(x-a)\). By the derivative's tangent-slope interpretation, this is the tangent line.
This also explains the condition on the derivative. If the limit does not exist as a finite real number, it does not supply a finite slope for this point-slope equation. The calculation below is specifically for points where the stated limit exists.
Worked Examples
Worked Example: Tangent to a Quadratic at \(x=2\)
Find the tangent line to \(f(x)=x^2\) at \(x=2\), using the limit definition.
First find the function value:
Thus the point on the curve is \((2,4)\). Now compute the derivative at \(2\):
The tangent slope is \(4\). Use point-slope form with \((2,4)\):
Check: At \(x=2\), the line gives \(y=4(2)-4=4\), so it passes through \((2,4)\). Its slope is the coefficient \(4\), matching the limit. The tangent line is \(y=4x-4\).
Worked Example: Tangent to a Linear Function at \(x=-1\)
Find the tangent line to \(f(x)=3x+1\) at \(x=-1\), computing the derivative from its limit.
The function value is:
So the point is \((-1,-2)\). The difference quotient is:
Substitute the point and slope into point-slope form:
Check: Substituting \(x=-1\) into the line gives \(y=3(-1)+1=-2\), the required point. The line's slope is \(3\), which agrees with the limit. In this case the tangent line is the original line itself.
Worked Example: Tangent to a Reciprocal Function at \(x=2\)
Find the tangent line to \(f(x)=\frac{1}{x}\) at \(x=2\), using the limit definition.
First, the point on the graph is found from:
Thus the point is \(\left(2,\frac12\right)\). Calculate the slope:
In the cancellation, \(h\ne0\) because the difference quotient is defined for nonzero \(h\); the limit then evaluates the simplified expression as \(h\) approaches zero. Now use slope \(-\frac14\) and point \(\left(2,\frac12\right)\):
Check: At \(x=2\), the line gives \(y=-\frac14(2)+1=-\frac12+1=\frac12\), as required. The line's slope is \(-\frac14\), equal to the derivative limit. Therefore the tangent line is \(y=-\frac14x+1\).
A Common Point-of-Tangency Mistake
A frequent error is to use \((a,a)\) as the point of tangency. But \(a\) is the input, while \(f(a)\) is the output. The graph point is \((a,f(a))\), and these coordinates are only the same in a coordinate when the function value happens to equal the input.
For example, for \(f(x)=x^2\) at \(x=2\), an incorrect setup might use the slope \(4\) but write \(y-2=4(x-2)\). That equation passes through \((2,2)\), not the actual graph point \((2,f(2))=(2,4)\). The corrected setup is:
Checking the result at \(x=2\) catches the error: the incorrect line gives \(y=2\), but the graph has \(y=4\). A useful final check is therefore to substitute \(x=a\) into the proposed tangent line and confirm that it returns \(f(a)\). Also check that the line's slope equals the limit-defined derivative.
Using the Result
The tangent line packages the local information supplied by the derivative into a familiar linear equation. It identifies the curve point and describes the line whose slope matches the curve's instantaneous rate of change there. In an AP Calculus problem, the derivative may be given or may need to be calculated from its definition; either way, the tangent-line step still requires both the derivative value and the function value at the specified input.
Before finalizing an equation, make sure that you can identify each piece:
- The input \(a\) is the \(x\)-coordinate of the point.
- The function value \(f(a)\) is the \(y\)-coordinate of the point.
- The limit \(f'(a)\) is the slope of the tangent line.
- The line equation is \(y-f(a)=f'(a)(x-a)\).
Keeping the point and slope distinct prevents the most common setup error. Writing point-slope form first also makes the construction easy to inspect before you simplify.
Check Your Understanding
For each question, identify the point and slope before writing or assessing a tangent line.
- For \(f(x)=x^2\), compute \(f(3)\) and \(f'(3)\) using the limit definition, then write the tangent line at \(x=3\).
- For \(g(x)=2x-5\), find \(g(-2)\) and use the limit definition to find \(g'(-2)\). What is the tangent-line equation there?
- A function has \(f(4)=7\) and \(f'(4)=-3\). Write the tangent line at \(x=4\) in point-slope form and slope-intercept form.
- For \(p(x)=\frac1x\), find the point on the graph at \(x=1\). Which information is needed in addition to the derivative to write the tangent line?
- A proposed tangent line at \(x=2\) has slope \(5\) and passes through \((2,2)\), while the function value is \(f(2)=6\). What part of the proposed setup must be corrected?
