# Definition of Continuous Function

Definition. A function f is continuous at a if lim_(x->a)=f(a).

Continuity implies three things:

1. f(a) is defined (i.e. a is in domain of f)
2. lim_(x->a)f(x) exists
3. lim_(x->a)f(x)=f(a)

Geometrically continuity means that you can draw function without removing pen from the paper.

Also continuity means that small changes in x produce small changes in y, i.e. function is continuous at a if for any epsilon>0 there exists delta>0 such that |f(x)-f(a)|<epsilon when |x-a|<delta.

If function is not continuous at a then we say that f is discontinuous at a, or f has a discontinuity at a.

Definition. Function y=f(x) is continuous on interval if it is continuous at every point of the interval.

Example 1. Show that f(x)=sqrt(4-x^2)+1 is continuous on the interval (-2,2).

If -2<a<2 then using properties of limits we obtain that

lim_(x->a)f(x)=lim_(x->a)(sqrt(4-x^2)+1)=

=lim_(x->a)sqrt(4-x^2)+lim_(x->a)1= by law 2

=sqrt(lim_(x->a)(4-x^2))+1= by laws 7 and 8

=sqrt(4-a^2)+1= by laws 2, 7 and 8

=f(a).

Thus, f is continuous on (-2, 2).

Actually, we can generalize above example for any elementary function.

Continuity of Combinations of Functions. Suppose that f(x) and g(x) are both continuous at a, and c is some arbitrary constant, then following functions are also continuous at a:

• f(x)+g(x)
• f(x)-g(x)
• cf(x)
• f(x)g(x)
• (f(x))/(g(x)), if g(a)!=0.

Fact. Inverse of continuous function is also continuous.

Continuity of Elementary Functions. All elementary functions, i.e. linear, polynomial,rational, power, exponential, logarithmic, trigonometric, inverse trigonometric, hyperbolic, inverse hyperbolic, are continuous at every point of their domain.

Actually we used this fact in previous notes. For example, since y=cos(x) is continuous at 0 then lim_(x->0)cos(x)=cos(0)=1.

Fact. If f(x) is continuous at b and lim_(x->a)g(x)=b then lim_(x->a)f(g(x))=f(b). In other words lim_(x->a)f(g(x))=f(lim_(x->a)g(x)).

This fact seems reasonable because if x is close to a, then g(x) is close to b, and since f is continuous at b, if g(x) is close to b, then f(g(x)) is close to f(b).

From this fact we obtain following fact.

Continuity of Composite Function. Suppose that g(x) is continuous at a and f(x) is continuous at g(a), then (f@g)(x)=f(g(x)) is continuous at a.

All above facts allow us to easily calulate limits of functions that are continuous at point a.

Example 2. Find lim_(x->1)(x^2+3x+6)/(2x-1).

Since f(x) is rational function then it is continuous on its domain, which in this case is {x|x!=1/2}. Thus, it is continuous at point 1.

Therefore, lim_(x->1)(x^2+3x+6)/(2x-1)=(1^2+3*1+6)/(2*1-1)=10.

Example 3. Where is (ln(x)+sqrt(4-x^2))/(x^2-1) continuous?

We know that ln(x) is continuous for x>0 (its domain), sqrt(4-x^2) is continuous on interval [-2,2] (its domain) and denominator is polynomial, so it is continuous everywhere.

Thus, function is continuous on interval (0,2] where denominator not equal 0 or x!=+-1. Therefore, function is continuous on (0,1)uu(1,2].

Example 4. Find lim_(x->pi)(cos(x))/(1+sin(x)).

In numerator we have continuous function cos(x). In denominator we have sum of two continuous functions, so denominator is also continuous.

Thus, ratio is continuous everywhere except where denominator equals 0, i.e. where sin(x)=-1.

But sin(pi)=0, so at this point function is defined.

Thus, lim_(x->pi) cos(x)/(1+sin(x))=cos(pi)/(1+sin(pi))=-1.

Example 5. Find lim_(x->1)sin((x^2-1)/(x-1)).

Since sine is a continuous function then lim_(x->1)sin((x^2-1)/(x-1))=sin(lim_(x->1)(x^2-1)/(x-1))=sin(lim_(x->1)((x-1)(x+1))/(x-1))=sin(lim_(x->1)(x+1))=sin(2).

Example 6. Where is function h(x)=ln(sqrt(4-x^2)) continuous?

Here we have composite function h(x)=f(g(x)), where f(x)=ln(x) and g(x)=sqrt(4-x^2).

We know that ln(x) is continuous for x>0 and sqrt(4-x^2) is continuous on [-2,2]. Therefore, h(x) is continuous on interval (0,2].