Math 1710
Topics for first exam
Chapter 1: Limits and Continuity
 § 1:

Rates of change and limits
Calculus = Precalculus + (limits)
Limit of a function f at a point x_{0} = the value the function `should' take at the point
= the value that the points `near' x_{0} tell you f should have at x_{0}
lim_{x®x0}f(x) = L means f(x) is close to L when x is close to (but not
equal
to) x_{0}
Idea: slopes of tangent lines
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lim_{x®x0}f(x) = L does
not care what f(x_{0})
is; it ignores it
lim_{x®x0}f(x) need not exist! (function can't make up it's mind?)
 § 2:

Rules for finding limits
If two functions f(x) and g(x) agree (are equal) for every x near a
(but maybe not
at a), then
lim_{x® a}f(x)lim_{x® a}g(x)
Ex.: lim_{x®2} [(x^{2}3x+2)/(x^{2}4)] = lim_{x®2} [(x1)/(x+2)]
If f(x)®L and g(x)®M as x®x_{0} (and c is a constant), then
f(x)+g(x)®L+M ; f(x)g(x)®LM ; cf(x)®cL ;
f(x)g(x)®LM ;
and f(x)/g(x)®L/M
provided M ¹ 0
If f(x) is a polynomial, then lim_{x® x0}f(x)= f(x_{0})
Basic principle: to solve lim_{x® x0} , plug in x = x_{0} !
If (and when) you get 0/0 , try something else! (Factor (xx_{0}) out of top and bottom...)
If a function has something like Öx  Öa in it, try multiplying (top and bottom)
with Öx + Öa
Sandwich Theorem: If f(x) £ g(x) £ h(x) , for all x near a (but not
at a), and
lim_{x® a}f(x)lim_{x® a}h(x)L , then lim_{x® a}g(x)L .
 § 4:

Extensions of the limit concept
Motivation: the Heaviside function
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Limit from the right: lim_{x® a+}f(x)L means f(x) is close to L
when x is close to, and
bigger than, a
Limit from the left: lim_{x® a}f(x)M means f(x) is close to M
when x is close to, and
smaller than, a
lim_{x® a}f(x)L then means lim_{x® a+}f(x)lim_{x® a}f(x)L
Infinite limits: ¥ represents something bigger than any number we can think of
lim_{x® a}f(x)¥ means f(x) gets really large as x gets close to a
Also have lim_{x® a}f(x)¥ ; lim_{x® a+}f(x)¥ ;
lim_{x® a}f(x)¥ ; etc....
Typically, an infinite limit occurs where the denomenator of f(x) is zero
(although not always)
 § 5:

Continuity
A function f is
continuous (cts)
at
a if lim_{x® a}f(x)f(a)
This means: (1) lim_{x® a}f(x) exists ; (2) f(a) exists ; and
(3) they're equal.
Limit theorems say (sum, difference, product, quotient) of cts functions are cts.
Polynomials are continuous at every point;
rational functions are continuous except
where denom=0.
Points where a function is not continuous are called
discontinuities
Four flavors:
removable: both onesided limits are the same
jump: onesided limts exist, not the same
infinite: one or both onesided limits is ¥ or ¥
oscillating: one or both onesided limits DNE
Intermediate Value Theorem:
If f(x) is cts at every point in an interval [a,b], and M is between f(a) and f(b),
then there is (at least one) c between a and b so that f(c) = M.
Application: finding roots of polynomials
 § 6:

Tangent lines
Slope of tangent line = limit of slopes of secant lines; at (x_{0},f(x_{0}) :
lim_{x® x0}[(f(x)f(x_{0}))/(xx_{0})] Notation: call this limit f^{¢}(x_{0})derivative of f at x_{0}
Different formulation: h = xx_{0}, x = x_{0}+h
f^{¢}(x_{0})lim_{h® 0}[(f(x_{0}+h)f(x_{0}))/h]
Chapter 2: Derivatives
 § 1:

The derivative of a function
derivative = limit of difference quotient (two flavors)
f^{¢}(x_{0}) exists, say f is
differentiable at x_{0}
Fact: f differentiable (diff'ble) at x_{0}, then f cts at x_{0}
h® 0 notation: replace x_{0} with x (= variable), get f^{¢}(x)
new
function
f^{¢}(x)the derivative of f = function whose vaules are the slopes of the tangent
lines to the graph of y=f(x) . Domain = every point where the limit exists
Notation:
f^{¢}(x)[dy/dx][d/dx](f(x)) = [df/dx] = y^{¢} = D_{x} f = Df = (f(x))^{¢}
 § 2:

Differentiation rules
[d/dx](constant) = 0
[d/dx](x) = 1
(f(x)+g(x))^{¢} = (f(x))^{¢}+ (g(x))^{¢}
(f(x)g(x))^{¢}= (f(x))^{¢} (g(x))^{¢}
(cf(x))^{¢}= c(f(x))^{¢}
(f(x)g(x))^{¢}= (f(x))^{¢}g(x)+ f(x)(g(x))^{¢}
([f(x)/g(x)])^{¢}= [(f^{¢}(x)g(x)f(x)g^{¢}(x))/(g^{2}(x))]
(x^{n})^{¢}= nx^{n1} , for n=0,1,1,2,2,3,.......
(( (1/g(x))^{¢}= ggp/(g(x))^{2} ))
f^{¢}(x) is `just' a
function, so we can take its derivative!
(f^{¢}(x))^{¢}= f^{¢¢}(x)(= y^{¢¢} = [(d^{2}y)/(dx^{2})] = [(d^{2}f)/(dx^{2})])
= second derivative of f (=rate of change of rate of change of f !)
Keep going! f^{¢¢¢}(x) = 3rd derivative, f^{(n)}(x) = nth derivative
 § 3:

Rates of change
Physical interpretation:
f(t)= position at time t
f^{¢}(t)= rate of change of position = velocity
f^{¢¢}(t)= rate of change of velocity = acceleration
f^{¢}(t) = speed
Basic principle: for object to change direction (velocity changes sign),
f^{¢}(t)= 0 somewhere (IVT!)
Examples:
Freefall: object falling near earth; s(t) = s_{0}+v_{0} t[g/2] t^{2}
s_{0} = s(0) = initial position; v_{0} = initial velocity; g= acceleration due to gravity
Economics:
C(x) = cost of making x objects; R(x) = revenue from selling x objects;
P = RC = profit
C^{¢}(x) = marginal cost = cost of making `one more' object
R^{¢}(x) = marginal revenue ; profit is maximized when P^{¢}(x) = 0 ;
i.e., R^{¢}(x) = C^{¢}(x)
 § 4:

Derivatives of trigonometric functions
Basic limit: lim_{x®0}[sinx/x] = 1 ; everything else comes from this!
Note: this uses radian measure! lim_{x®0}[sin(bx)/x] = b
Then we get:
(sinx)^{¢}= cosx (cosx)^{¢}= sinx
(tanx)^{¢}= sec^{2} x (cotx)^{¢}= = csc^{2} x
(secx)^{¢}= secx tanx (cscx)^{¢}= = cscxcotx
 § 5:

The Chain Rule
Composition (g°f)(x_{0}) = g(f(x_{0})) ; (note: we
don't know what g(x_{0}) is.)
(g°f)^{¢} ought to have something to do with g^{¢}(x) and f^{¢}(x)
in particular, (g°f)^{¢}(x_{0}) should depend on f^{¢}(x_{0}) and g^{¢}(f(x_{0}))
Chain Rule: (g°f)^{¢}(x_{0}) = g^{¢}(f(x_{0}))f^{¢}(x_{0})
= (d(outside) eval'd at inside fcn)·(d(inside))
Ex: ((x^{3}+x1)^{4})^{¢}= (4(x^{3}+11)^{3})(3x^{2}+1)
Different notiation:
y = g(f(x)) = g(u), where u = f(x), then [dy/dx] = [dy/du][du/dx]
 § 6:

Implicit differentiation
We can differentiate functions; what about equations? (e.g., x^{2}+y^{2} = 1)
graph
looks
like it has tangent lines
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Idea:
Pretend equation defines y as a function of x : x^{2}+(f(x))^{2} = 1
and differentiate!
2x+2f(x) f^{¢}(x) = 0 ; so
f^{¢}(x) = [(x)/f(x)] = [(x)/y]
Different notation:
x^{2}+xy^{2}y^{3} = 6 ; then 2x+(y^{2}+x(2y[dy/dx])3y^{2}[dy/dx] = 0
[dy/dx] = [(2xy^{2})/(2xy3y^{2})]
Application: extend the power rule
[d/dx](x^{r}) = rx^{r1} works for any rational number r
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