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=x₀ and f (m) is continuous at the point m₀ = f(x₀), then composite function f (f(x)) is continuous at point x₀.

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