Limits and the Derivative
I. Quiz
II. Homework
III. Limits infinities and zeros.
It is useful to have the following symbolic fractions when dealing with limits.
A) (infinity)/(finite) = infinity
B) 0/(finite nonzero) = infinity
C) (finite)/(infinity) = 0
D) (finite nonzero)/0 = infinity
E) (infinity)/(infinity) = Do Algebra
F) 0/0 = Do Algebra
Example:
1) lim x -> 0 (3x + 5)/(2x2 + x) = 5/0 = DNE
2) lim x -> 1 of (x2 - 1)/(x + 3) = 0/4 = 0
3) lim x -> 2 of (x2 - 4)/(x - 2) = 0/0
We factor to get: ((x - 2)(x + 2))/(x - 2) = x + 2 = 4
4) Try lim x -> -1 of (x2 - 2x - 3)/(x2 + 8x + 7)
5) Try lim x -> 3/2 of (6x2 + x - 15)/(8x2 - 6x - 9)
6) lim x -> 2 of (sqrt(x - 1) - 1)/(x2 - 4) = 0/0
we rationalize the denominator by multiplying by the conjugate root: sqrt(x - 1) + 1 to obtain
7) Find lim x -> 0 of (sqrt(3 + x) - sqrt(3))/x
IV. Limits and Trigonometry
Use your calculator to graph (sinx)/x and discover that
lim x -> 0 of (sinx)/x = 1
Corollary:
lim x -> 0 of (1 - cosx)/x = 0
proof: (1 - cosx)/x = (1 - sqrt(1 - sin2x))/x =
((1 + sqrt(1 - sin2x)) (1 - sqrt(1 - sin2x)))/((x)(1 + sqrt(1 - sin2x)))
= (1 - (1 - sin2x))/((x)(1 + sqrt(1 - sin2x)))
= sin2x/((x)(1 + sqrt(1 - sin2x)))
= ((sinx)/x) (sinx)/((1 + sqrt(1 - sin2x))
= (1) (0/2) = 0
Applications:
1) lim x -> 0 of (tanx)/x = ((sinx)/(cosx))/x = ((sinx)/x)(1/cosx) = (1)(1) = 1
2) lim x -> 0 of (1 - cos(2x))/x
We set u = 2x then as x goes to zero so does u. we have x = u/2 substituting we get
(1 - cosu)/(u/2) = 2((1 - cosu)/u) = 2(0) = 0
Try (sin(5x))/(2x)
V. The Squeeze Theorem
The squeeze theorem says that if a function f is between two functions that have the same limit, then f has that limit also.
Let h < f < g in an open interval containing c, except possibly at c. then if
lim as x approaches c of h = lim x -> c of g = L, then
lim as x -> c of f = L.
Application: Show that lim as x -> 0 of x sin(1/x) = 0
Proof: We have
-x < xsin(1/x) < x for all x,
Since lim x -> 0 (-x) = lim x -> 0 (x) = 0 the squeeze theorem tells us that
lim x-> 0 x sin(1/x) = 0.