5.10 Limits  (Page 4/5)

Let us now consider reciprocal function :

$$f\left(x\right)=\frac{1}{x}$$

Its singularity is obtained by setting denominator to zero. Thus, singularity exists x=0 for the reciprocal function. As such, domain of function is R-{0}. In order to know the function, we need to know nature of function in the vicinity of undefined point. We can do this by evaluating limit on either side of the singularity.

$$\underset{x\to 0-}{\overset{}{\lim }}\frac{1}{x}=-\infty$$

$$\underset{x\to 0+}{\overset{}{\lim }}\frac{1}{x}=\infty$$

Important point to underscore here is that limiting values of x or f(x) as infinity is a valid estimates. To be more explicit, value of function can approach infinity as limiting value. In this case, left and right limits are not same. Therefore, limit of function does not exist at exception point x=0. In order to explore limit at exception point, we consider the case of modulus of reciprocal function. In this case also, function is not defined at x=0. But, for x<0; |x| = -x and for x>0; |x| = x. Hence, left and right limits are :

$$\underset{x\to 0-}{\overset{}{\lim }}\left|\frac{1}{x}\right|=\underset{x\to 0-}{\overset{}{\lim }}-\frac{1}{x}=\infty$$

$$\underset{x\to 0+}{\overset{}{\lim }}\left|\frac{1}{x}\right|=\underset{x\to 0+}{\overset{}{\lim }}\frac{1}{x}=\infty$$

In this case, left and right limits are equal. Therefore, limit of function exist at exception point x=0 and it is given as :

$$\underset{x\to 0}{\overset{}{\lim }}\left|\frac{1}{x}\right|=\infty$$

Limits and infinity

We have noted that limit of function can be positive or negative infinity to reflect the estimate that function value is expected to be either very large positive or negative value. It happens when a finite value is divided by a value which is exceedingly small. If the divisor is a exceedingly small negative value, then function approaches negative infinity and if the divisor is a exceedingly small positive value, then function approaches positive infinity. Similar intuitive limiting values involving infinity are given here :

(1) Let “a” be a finite real number.

$$\underset{x\to \infty }{\overset{}{\lim }}\frac{a}{x}=0$$

$$\underset{x\to -\infty }{\overset{}{\lim }}\frac{a}{x}=0$$

(2) Let “a” be a finite real number.

$$\underset{x\to 0-}{\overset{}{\lim }}\frac{a}{x}=-\infty$$ $$\underset{x\to 0+}{\overset{}{\lim }}\frac{a}{x}=\infty$$

(3) Let “a” be a finite real number.

$$\underset{x\to \infty }{\overset{}{\lim }}ax=\infty ;a>0$$ $$=0;a=0$$ $$=-\infty ;a<0$$

(4) Let “a” be a finite real number.

$$\underset{x\to \infty }{\overset{}{\lim }}{a}^x=\infty ;a>1$$ $$=1;a>0$$ $$=-\infty ;0<a<1$$

Indeterminate limit forms

The indeterminate limit form is also called meaningless form. There are seven such forms in total. We, however, need to be careful in interpreting these forms. The interpretation is most important part of evaluation of limit. For example, if we say that 0/0 is indeterminate limit form, then we mean that both numerator and denominator of function approach zero, but none are equal to zero. In the example below, both numerator and denominator approach to zero as x approaches 2 :

$$\underset{x\to 2}{\overset{}{\lim }}\frac{\left(x^2-4\right)}{\left(x-2\right)}$$

As x approaches 2, both numerator and denominator approaches to zero. Therefore, the function expression is an indeterminate form 0/0. However, following is not an indeterminate form :

$$\underset{x\to 0}{\overset{}{\lim }}\frac{0}{x}=\infty$$

In the limit given above, the numerator is zero (not approaches to zero), whereas denominator approaches to zero. Thus, the rational form is determinate form and approaches infinity and the limit is also infinity.

In addition to 0/0 indeterminate form, there are other indeterminate forms which needs to be converted to determinate form so that limit can be evaluated. The seven indeterminate forms are :