In calculus, limit and continuity is one of the most crucial topics. A limit of a function is a number that a function approaches as the non dependent variable approaches a given value. We define limit as the value the function is approaching.

A function can have either of the following properties:

**1)**It can be continuous

**2)**It can be discontinuous

A continuous function is one which can be graphed without lifting the pen from the paper easily. If the function cannot be graphed this way then it is discontinuous.

The common properties of a continuous function are:

The common properties of a continuous function are:

**i)**A continuous function neither jumps nor blast to infinity.

**ii)**The graph of such functions can be drawn easily without lifting pencil up from the paper.

Limits help us to check on the continuity of a function. If $lim_{x \rightarrow c}$$ f (x) = f (c)$, then we say that the function is continuous, that is, the function f(x) is said to be “everywhere continuous” or simply continuous if for all real numbers ‘c’, f (x) is continuous at x = c. This is very helpful at times when it is not easy to make graph of a function

Mathematically we say that a function ‘h’ is continuous at x = c if

**1)**h (c) is defined.

**2)**$lim_{x \rightarrow c}$$ h (x)$ exists

**3)**$lim_{x \rightarrow c}$$ h (x)$ = h (c)

If either of the conditions above is not fulfilled then we say that the function ’h’ is discontinuous at x = c.

## Intermediate Value Theorem

If a function h (x) is continuous in some interval [c, d], then there exist some ‘a’ belonging to [c, d] such that y = h (a) where h (a) lies between h (c) and h (d). This implies that the function ‘h’ attains all the values between h (c) and h (d) as ‘x’ reaches from c to d.

This theorem in general clears the fact that the functions that are continuous do not have jumps.

Boundedness Theorem:

If a function h (x) is continuous in some interval [c, d], then real numbers p and P exist such that p <= h (x) <= P for every ‘x’ belonging to [c, d]. in other words we can say that the function ‘h’ is bounded from both sides above and below in the interval [c, d].

This theorem in general clears the fact that function that are continuous do not blast to infinity since they are bounded in the interval.

## Examples

**Example 1:**For what values of c and d the given function is continuous.

h(y) = $\left\{\begin{matrix}

y^2& y < 2 \\

y + c& 2 \leq y \leq 3 \\

d - y^2& 3 < y

\end{matrix}\right.$

**Solution:**It is clear that the function is continuous at (-$\infty$, 2), (2, 3), (3, $\infty$). This is because ‘h’ is defined in them separately. The problem arises at points 2 and 3 only.

Consider $lim_{y \rightarrow 2^-}$ h (y) = $2^2$ = 4 …(i)

$lim_{y \rightarrow 2^+}$ h (y) = 2 + c …(ii)

For function to be continuous (i) and (ii) should be equal.

$\rightarrow$ 2 + c = 4

$\rightarrow$ c = 2.

Now consider $lim_{y \rightarrow 3^-}$ h (y) = 3 + c = 3 + 2 = 5 …(iii)

$lim_{y \rightarrow 3^+}$ h (y) = d - $3^2$ = d - 9 …(iv)

Again (iii) and (iv) needs to be same for function to be continuous.

$\rightarrow$ d - 9 = 5

$\rightarrow$ d = 14

Thus when c = 5 and d = 14, then the function ‘h’ here will be continuous.

**Example 2:**Determine if the following function is continuous at x = 3.

F(x) = $\left\{\begin{matrix}

x^2+3 & ;\ x \leq 3\\

x^3 + 4x & ;\ x > 3

\end{matrix}\right.$

**Solution:**Given function is F(x) = $\left\{\begin{matrix}

x^2+3 & x \leq 3\\

x^3 + 4x & x > 3

\end{matrix}\right.$

Check the value of function at x = 3

F(3) = (3)$^2$ + 3 = 12

Left hand limit = Lim$_{x \rightarrow 3^-}$ F(x) = Lim$_{x \rightarrow 3^-}$ (x$^2$+3) = (3)$^2$ + 3 = 12

Right hand limit = Lim$_{x \rightarrow 3^+}$ F(x) = Lim$_{x \rightarrow 3^+}$ (x$^3$+4x) = (3)$^3$ + 4.3 = 27 + 12 = 39

From above Lim$_{x \rightarrow 3^-}$ F(x) $\neq$ Lim$_{x \rightarrow 3^+}$ F(x)

Thus, function f is not continuous at x = 3.