5.10 Limits  (Page 3/5)

| x ; x>0 f(x) = | 0 ; x=0| -x ; x<0

Clearly, at test point x=0,

$$L=f\left(0\right)=0$$

This is an important result which gives us a method to determine limit of a function. If function is continuous, simply put the test value into function definition. The value of function is limit of function at that point.

Limit and discontinuity

If function rule changes exactly at the test point, then limit of the function, L, and value of function, f(a), are not same. In order to clearly understand the implication of the statement about inequality of limit and function value, we consider a modification to the modulus function :

| x; x>0 f(x) = | 1; x=0| -x; x<0

At test point x=0,

$$L=0$$ $$f\left(0\right)=1$$ $$L\ne f\left(0\right)$$

Therefore, limit of function exists at x=0 even though it is not equal to function value. This is an important result which gives us a method to determine limit of a piece-wise defined functions. We need to evaluate function from both left and right side. If limits are equal from both sides, then limit of function at test point is equal to either limit. However, if left and right limits are not equal then limit of function does not exist at the test point.

Limit and singularity

Singularity or exception point is a point where function is not defined. It is outside definition of function. However, function can point (or tend or approach) to a value at a point where it is not defined. Limit as we know estimate value from a close point where function exits and can project a value based on function definition at points very close to exception point. Consider limit of a rational function :

$$\underset{x\to 1}{\overset{}{\lim }}\frac{\left(x-1\right)\left(x+3\right)}{\left(x-1\right)}$$

The singularity of function is obtained by setting denominator to zero. Thus, singularity exists at x=1. We want to know nature of function about this point. In other words, we want to know what would have been the value of function at this point had the function been defined there. For this, we need to evaluate left and right limit at this point. Graphically, there is a hole in the graph of the function. How can we estimate value of function at a point if it is not defined there ?

We keep linear factor in the denominator to know singularity. Extrapolating value at the singularity is a reverse process. We need to calculate function value in the neighborhood, where function is defined. For this, we require to remove linear factor from the denominator. Canceling out (x-1) from both numerator and denominator, we have :

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

Thus, function is not defined at x=1, but its limit at the point is 4. This means that nature of function in its immediate vicinity is such that the function should have attained a value of 4 had it been estimated on the basis of nature of function in the neighborhood.

This is again an important result which gives us a method to determine limit of a function, when function is not defined at certain point or has other indeterminate or meaningless forms. We need to simplify function expression till we get a form which can be evaluated.