1-4 Using Limits to Find Tangents
Using limits, we are now able to find the slope of a tangent without continually finding the
slopes of secants with the points of the secant line being ever closer together.
Remember that the slope of any secant of a function y = f ( x ) between the points,
f ( x) − f (a )
. If we are interested in finding
x− a
the slope at (a, f ( a )) , we could simply set x to be as close to a as we want, or, using the
( x, f ( x)) and (a, f ( a )) , is given by m =
theory of limits we could say that the slope of the tangent line at (a, f ( a )) is given by
f ( x) − f (a )
∆y
m = lim
, or just as m = lim
x→ a
∆x ∆ x
x− a
Example 1: Find the equation of the tangent line to the curve y = − 2 x 2 + 3 x + 3 at the
point (2,1) . The slope at that point is given by
f ( x) − f (2)
x→ 2
x− 2
(− 2 x 2 + 3 x + 3) − (− 2(2) 2 + 3(2) + 3)
= lim
x→ 2
x− 2
2
− 2 x + 3x + 2
= lim
x→ 2
x− 2
− 1(2 x 2 − 3x − 2)
= lim
x→ 2
x− 2
− (2 x + 1)( x − 2)
= lim
x→ 2
x− 2
= lim(− 2 x − 1)
m = lim
x→ 2
= −5
To get the equation of the tangent line, we write
that
y − 1 = − 5( x − 2)
y = − 5 x + 11
Often, another definition for the slope of a tangent is more useful. By this definition, the
slope at a point ( x, f ( x )) is given by the definition
m = lim
h→ 0
f ( x + h) − f ( x )
. This definition really amounts to the same thing.
h
Example 2: Find the equation of the tangent line to the hyperbola xy = 12 at the point
(− 3, − 4) .
Solution:
f ( x + h) − f ( x )
h→ 0
h
f (− 3 + h) − f (− 3)
lim
h→ 0
h
12
12
−
lim − 3 + h − 3
h→ 0
h
− 3(12) − (− 3 + h)(12)
− 3(− 3 + h)
lim
h→ 0
h
− 3(12) + 3(12) − 12h
lim
h→ 0
− 3h(− 3 + h)
− 12h
lim
h → 0 − 3h( − 3 + h)
− 12
lim
h → 0 − 3( − 3 + h)
− 12
9
4
−
3
m = lim
=
=
=
=
=
=
=
=
4
( x + 3)
3
4
y = − x− 8
3
y+ 4= −
The equation for the line is
Complete the assignment starting on page 35 - 36 #1i, 2ii, 6, 7, 8b, c, d