1. CONTINUITY
In an informal way, we can say that a function f is continuous on an interval if its graph can be drawn without taking the pencil off of the paper ( f doesn’t cut or jump).
More precisely f is continuous at x = c if and only if: I) f(c) is defined, and
II) lim f(x) = f(c)
x c
(meaning you can get to the point (c,f(c) ) by walking on the curve from left or from the right of it)
A function f is said to be continuous on an interval I (domain) if and only if f is continuous at any point in I.
Note that if f is continuous on I, then lim f(x) coincides with f(c)
x c
Ex 1. Find where f(x) = (2x-4) / (x2 –x –6) is not continuous ( points of discontinuity ).
Since f(x) is not defined at x where its denominator is zero. The discontinuity of f occurs when x2 –x –6 = 0. Solving x2 –x –6 = 0, we get x=-2 or x= 3. So the values for which f is discontinuous are x=-2 or x=3.
Ex 2. The line f(x) = ax+b is continuous in its domain R.
f(c) = ac+b exists and lim x c
(ax+b) =ac+b = f( c ), so f is continuous at any real c
Ex 3. Polynomials p(x) are continuous on R, since lim p(x) = p(a) and p(x)| x=a= p(a) xa

2. C0NTINUOUS FUNTIONS V/S DISCONTINUOUS FUNCTIONS
Analyze continuity for each of the four functions illustrated with the graphs shown below.
1) y = f(x) is continuous, since f has no disconnecting places or jumps. Y
y=h(x)
y= f(x)
2) y = g(x) is continuous, since g has no disconnecting places or jumps.
y= g(x)
y=r(x)
3) y = h(x) is discontinuous , since h has a jump at x = a.
Jump X
4) y = r(x) is discontinuous, since r is disconnected at x=c. c
There’s a hole. A point is missing!

3. MORE PROPERTIES OF CONTINUOUS FUNCTIONS
Theorem: (Combination of continuous functions). If f and g are continuous at x=c then the following combinations of them are continuous at x=c too.
1) Scalar multiple: kf is continuous, for any constant k
2) Sum and Difference: f +g and f - g are continuous
3) Product: fg is continuous
4) Quotient: f/g is continuous if g(c) ≠ 0
y = 3x
5) Composition: fog is continuous (provides that f is continuous at g(c) )
y = x2 ` Y
Example 1: We know f(x) = x is continuous.
The following functions are continuous too See figure on the right.
y=x2 - 4x+ 5
y = x
y=x-1
1) 3f(x) = 3x
2) ( ff)(x) = f(x)f(x) = xx = x2 X
3) g((x) =( f•f)(x)+4 f(x)+5 = x2 - 4x + 5 2 2 2
4) h(x) = (x2 - 3x + 4 ) / (x-2)
Note that x =2 is a Vertical Asymptote for h(x) So h(x) is continuous for any x  2 .
And the oblique asymptote is y=x-1.

4. The above examples show us that polynomial functions are continuous functions on R and that the rational functions q(x) = p1(x) / p2(x) (where p1 & p2 are polynomials) are continuous on R – { the zeros of p2(x) }.
Example 1: Study continuity for f(x) = (x+1)/ (x2 – x –12)
We look for the zeros of denominator, i.e. x2–x–12 =0 (x +3)(x – 4)=0. So x=4 or x= - 3.
So f(x) is continuous on R – {-3,4}
Example 2: Study continuity for g(x) = sin2 x /(x4 + 1). (Assume that sinx is continuous)
Since sin x is continuous its, square is also continuous.
Now not only that the denominator is a continuous function (it’s a polynomial ! ).
Furthermore x cannot be zero, since x4 + 1 is always  1 so . So g(x) is continuous on R.
Example 3: Study continuity for sin x & cos x
Since sin(x+h) = sin x cos h + cos x sinh we have 1
lim sin(x+h) = lim (sin x cos h + cos x sinh) h h0
= sin x lim cos h + cos x lim sin h h h0
But we know that lim sin = 0 & lim cos h = h h0
So lim sin(x+h) = sin x and this means that sin x is a continuous function h 0
Similarly cos x is a continuous function using cos(x+h) = cos x cos h - sinx sinh

5. Intermediate Value Theorem
Theorem: If f(x) is a continuous function on a closed interval [a,b] , then f(x) must take all values between f(a) and f(b). Y
f(a)
f(b)
The figure on the right shows a continuous function on [a,b] c
The Theorem assures us that if we choose any number c between f(a) and f(b) , the line y=c will have to intersect at least one point on the graph. In our case it cuts the graph at three points (x1, c) , (x2 , c) & (x3 , c). X
a b
x x2 x3
Example: Prove that f(x) = x3+5x – 7 has a zero between 1 and 2. Y
(2,7) 7
We know that f(x) = x3+5x – 7 is a continuous function. X
Since f(1) = = - 5 & f(2) = = 7. We have f(1)<0<f(2).
- 5
(1,-5)
The Intermediate Theorem assures us that the line y=0 intersects the graph of f(x) at a value of x between 1 and 2.

6. Example 1: Find lim f(x) , . x 16
Theorem: If f is continuous at g(x) then lim f( g(x) ) = f( lim (g(x) ) xa xa
Example 1: Find lim f(x) , x 16
for f(x) = x3/2 . (Assume that f(x) is continuous)
Interchanging lim and f we get
lim f(x) = [ lim x ]3/2 =163/2 = x x 16
Example 2: Find lim x 
ln [(3x+1)/(x+9)] (Assume that ln x is continuous for x>0)
Interchanging lim and ln, we get ln [ lim ((3x+1)/(x+9))] x 
= ln(3)
Example 3: Find lim f(x) , x-1
for f(x) = (sin [(x2 – 1)/(x2+4x+3) ] )2
Interchanging lim and ( )2 …. {lim sin[(x2 – 1)/(x2+4x+3) } x -1
Interchanging lim and sin ….. {sin lim (x2 – 1)/(x2+4x+3) } x-1
But lim (x2 – 1)/(x2+4x+3) is type 0/0 , so we try to simplify by (x+1) … x-1 2 2
Factoring we get lim (x2 – 1)/(x2+4x+3) = lim (x+1)(x-1) / (x+1)(x+3) x x -1
= -
So lim f(x) = (sin (-))2 = x-1