# limit-continuity-4

**L’ Hospital’s Rule: **

L’ Hospital’s Rule is applied only to the two forms _{ } .The rule states that if _{ } is of the form_{ } then _{ } provided later limit exists. But if still it takes the form _{ } then we differentiate numerator and denominator again.

i.e. _{ } and this process continues till we find that _{ } form is removed.

Illustration 13:

Evaluate _{ }

Solution:

Let A = _{ }

_{ }

__ __

**Continuity at point: **** **

A function ‘f’ is said to be continuous at a point ‘a’ in domain of f, if following conditions are met.

1) f (a) exists.

2) _{ } exists finitely which essentially means that _{ } and _{ } both exist finitely and are equal.

3)_{ } . If anyone or more conditions are not met than function f is said to be discontinuous at point ‘a’. Geometrically graph of function exhibit a break at point x=a.

Illustration 14: Discuss the continuity of function y = f(x) as shown by graph below at the point x=1.

Fig (3)

Solution:

Now clearly the function exhibits a break at point x=1 and is thus discontinuous at x=1. But if we go by the rules mentioned in above article we observe that.

_{ } And _{ }

So _{ } does not exists and thus the function is discontinuous at x=1.

__Continuity of function in open interval: __

A function f is said to be continuous in (a, b) if f is continuous at each and every point belonging to interval (a, b).

__Continuity of function in closed interval: __

A function f is said to be continuous in closed interval [a, b] if

1) f is continuous in open interval (a, b) and

2) _{ } exists finitely and _{ } and

3) _{ } exists finitely and_{ } .

Illustration 15:

Consider a function f(x) which is continuous in [1, 2] and given that f (1) = -1 and f (1) = 2, prove that there exists a point x such that f(x) =1.

Solution:

Let us arbitrarily choose some graph for f(x)

Fig (4)

For that matter function f(x) can take any shape. But since the function f is continuous between 1 and 2 it must be defined on all values lying between 1 and 2 and the function’s curve will go from y = -1 to y = 2 through some path. Whatever path be chosen y =1 will lie in between (Remember there cannot be a sudden jump because function is continuous)

Hence there exists a point x-where f(x) = 1.

__Properties of continuous functions: __

Let f(x) and g(x) are continuous functions at x=a then

1) K f(x) is continuous where K is constant

2) f(x) ± g(x) is continuous at x=a.

3) f(x) ´g(x) is continuous at x=a.

4) f(x)/g(x) is continuous at x=0 provided g (a) ¹ 0.

5) If m = f(x) is continuous at x=x_{0} and f (m) is continuous at the point m_{0} = f(x_{0}), then composite function f (f(x)) is continuous at point x_{0}.

6) If f(x) is continuous on [a, b] such that f (a) and f (b) are of opposite signs. Then there exists at least one solution of equation f(x) = 0 in the open interval (a, b).

7) Functions like sinx, cosx, tanx, cotx, secx, cosecx, logx, e^{x} etc are continuous in their domain.

Dumb Question:

1) How can function like tanx be continuous?

Ans: Please read the point mentioned above carefully. It is written that tanx is continuous in its domain. The points (2n+1)p/2 when tanx® do not lie in the domain of function f.

Illustration 16:

Let function _{ } Prove that there is a solution for equation f(x) =0 in the interval [0, 1].

Solution:

We observe that e^{x}, cosx, sinx, _{ } all are continuous functions in interval [0, 1]. So the function f is also continuous in [0, 1].

_{ }

Since 0<1< (p/2) so cos1 and sin1 are positive quantities and e>_{ } .

So, f (1)>0.

Now f (0) and f (1) are of opposite signs. So f(x) =0 has a solution in interval [0, 1].

__Classification of discontinuities: __

1) Removable discontinuity:

Here at a point x=a of discontinuity, _{ } exists but f (a) is either undefined or if defined it does not tally with_{ } .

In this case the removal of discontinuity is achieved by modifying definition of function suitably so that f (a) tallies with _{ } which is found existing.

2) Irremovable discontinuity:

Here _{ } does not exist even though f (a) exists or additionally f (a) also does not exist.

Here, since _{ } does not exist, no modification of definition at x=a can succeed to make _{ } exist. Hence discontinuity remains irremovable.

Illustration 17:

Consider the function

_{ }

Discuss the continuity of function at x=0. If discontinuous suggest how to remove discontinuity.

Solution:

Let us plot the function f(x)

Fig (5)

Now _{ }

And _{ }

So, _{ } = _{ } = -1

But f (0) =_{ } .

\f (x) has removable discontinuity and f(x) can be made continuous by taking f (0) = -1.

_{ }