5.12 Continuous function  (Page 6/6)

For x ≠1, the function is :

$$f\left(x\right)=\frac{x^3+x^2-3x+3}{{\left(x-1\right)}^2}$$

We guess here that one of the roots of numerator is 1. It is true as numerator is zero for x=1. Thus, (x-1) is a factor of cubic expression in the numerator.

$$\Rightarrow f\left(x\right)=\frac{\left(x-1\right)\left(x^2+2x-3\right)}{{\left(x-1\right)}^2}==\frac{\left(x-1\right)\left(x-1\right)\left(x+3\right)}{{\left(x-1\right)}^2}$$

$$\Rightarrow f\left(x\right)=x+3$$

Clearly,

$$\Rightarrow \underset{x\to 1}{\overset{}{\lim }}f\left(x\right)=\underset{x\to 1}{\overset{}{\lim }}x+3=4$$

For function to be continuous, this limit should be equal to value of function at x=1. Hence,

$$\Rightarrow f\left(1\right)=k=4$$ $$\Rightarrow k=4$$

For x ≠ 0, the function is of standard form whose limit is 1 when x approaches 0. Thus, limit is not equal to function value at x=0. Clearly, function is discontinuous at x=0. It is a removable discontinuity.

In order to evaluate limit at x=0, we determine left and right limits at x=0. We see here that as x approaches zero 1/x approaches negative infinity from left, whereas it approaches to positive infinity from the right. Using these facts, left hand limit is :

$$\Rightarrow \underset{x\to 0-}{\overset{}{\lim }}\frac{e^{-\infty }}{1+e^{-\infty }}=\frac{0}{1+0}=0$$

Now right limit is :

$$\Rightarrow \underset{x\to 0+}{\overset{}{\lim }}\frac{e^{1/x}}{1+e^{1/x}}\left[\frac{\infty }{\infty }\text{form}\right]$$

Dividing by $e^{1/x}$ , we have :

$$\Rightarrow \underset{x\to 0+}{\overset{}{\lim }}\frac{1}{\frac{1}{e^{1/x}}+1}=\frac{1}{\frac{1}{e^{\infty }}+1}=1$$

Clearly, $L_l\ne L_r$ . Hence, function is discontinuous at x=0.

In order to determine continuity, we consider set of rational and irrational numbers separately. Let us first work with rational set. We determine limit when as x approaches any real value point “a” in real domain. Note that x approaches real number “a” assuming only rational number in its set. We say that x approaches "a" through rational numbers. Then,

$$\Rightarrow \underset{x\to a}{\overset{}{\lim }}f\left(x\right)=\underset{x\to a}{\overset{}{\lim }}x=a$$

Now, we work with irrational set. Here, x approaches real number “a” assuming only irrational numbers in the domain. Then,

$$\Rightarrow \underset{x\to a}{\overset{}{\lim }}f\left(x\right)=\underset{x\to a}{\overset{}{\lim }}2-x=2-a$$

Let us consider that function is continuous at x=a. In that case,

$$L_l=L_r$$ $$\Rightarrow a=2-a$$ $$\Rightarrow a=1$$

We have taken any arbitrary point x=a and we find that function is continuous only for a single value – not an interval or union of intervals. Thus, we conclude given function is continuous only at one point and discontinuous at all other points.