 maths > calculus-limits

Indeterminate value in functions

what you'll learn...

Indeterminate Value in Functions

»  Functions evaluate to 00$\frac{0}{0}$
→  eg: x24x2x=2=00$\frac{{x}^{2} - 4}{x - 2} {|}_{x = 2} = \frac{0}{0}$

Though the function evaluates to 00$\frac{0}{0}$, it may take a value.

evaluating a function

Consider the value of function f(x)=x21x1$f \left(x\right) = \frac{\textcolor{\mathrm{de} e p s k y b l u e}{{x}^{2} - 1}}{\textcolor{c \mathmr{and} a l}{x - 1}}$ when x=2.2$x = 2.2$.

On substituting $x = 2.2$, we get $f \left(2.2\right) = \frac{\textcolor{\mathrm{de} e p s k y b l u e}{{2.2}^{2} - 1}}{\textcolor{c \mathmr{and} a l}{2.2 - 1}}$.
$f \left(2.2\right) = 3.2$

function evaluating to 0/0

Consider the value of function $f \left(x\right) = \frac{\textcolor{\mathrm{de} e p s k y b l u e}{{x}^{2} - 1}}{\textcolor{c \mathmr{and} a l}{x - 1}}$ when $x = 1$

On substituting $x = 1$, we get
$f \left(1\right)$
$\quad \quad = \frac{\textcolor{\mathrm{de} e p s k y b l u e}{{1}^{2} - 1}}{\textcolor{c \mathmr{and} a l}{1 - 1}}$
$\quad \quad = \frac{\textcolor{\mathrm{de} e p s k y b l u e}{0}}{\textcolor{c \mathmr{and} a l}{0}}$

That is, the function evaluates to indeterminate value when $x = 1$.

Let us closely examine the function $f \left(x\right) = \frac{\textcolor{\mathrm{de} e p s k y b l u e}{{x}^{2} - 1}}{\textcolor{c \mathmr{and} a l}{x - 1}}$.

The numerator be factorized as ${x}^{2} - 1 = \left(x + 1\right) \left(x - 1\right)$

Rewriting the function $f \left(x\right) = \frac{\textcolor{\mathrm{de} e p s k y b l u e}{{x}^{2} - 1}}{\textcolor{c \mathmr{and} a l}{x - 1}}$ as the function $f \left(x\right) = \frac{\textcolor{\mathrm{de} e p s k y b l u e}{\left(x + 1\right) \left(x - 1\right)}}{\textcolor{c \mathmr{and} a l}{x - 1}}$.

It is noted that the function can be simplified to $f \left(x\right) = x + 1$ when $x \ne 1$. Note that $0$ cannot be canceled out in expressions or equations.

So the given function
$f \left(x\right)$
$\quad \quad = x + 1$ when $x \ne 1$
$\quad \quad = \frac{{x}^{2} - 1}{x - 1}$ when $x = 1$

By this it is concluded that $f \left(x\right) {|}_{x = 1}$ is indeterminate value $\frac{0}{0}$.

Many students wrongly understand that the algebraic simplification (like canceling $x - 1$ in the example above) solves the indeterminate value. It is not so -- the function remains indeterminate at that input value $x = 1$.

another example

Given $f \left(x\right) = \frac{{x}^{3} - 8}{{x}^{2} - 4}$ What is $f \left(2\right)$?

The answer is '$\frac{0}{0}$'

summary

Function evaluates to indeterminate value: Function $f \left(x\right)$ evaluates to indeterminate value for $x = a$ if $f \left(a\right) = \frac{0}{0}$.

Outline