# Definition of Continuous Function

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

Continuity implies three things:

- `f(a)` is defined (i.e. `a` is in domain of `f`)
- `lim_(x->a)f(x)` exists
- `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]`.