 maths > calculus-limits

Limit of Ratios

what you'll learn...

Limit of Ratios

»  for defined polynomials, algebra of limits apply.

»  For polynomials evaluating to 00$\frac{0}{0}$ value
→  factorize numerator and denominator

→  cancel common factors

→  organize to standard results

→  apply limit to modified expressions

trivial for limit

To find limits of function f(x)=3x2+5x2x2+x2$f \left(x\right) = \frac{\textcolor{\mathrm{de} e p s k y b l u e}{3 {x}^{2} + 5 x - 2}}{\textcolor{c \mathmr{and} a l}{{x}^{2} + x - 2}}$ at x=2$x = 2$, first we should check if function evaluate to 00$\frac{0}{0}$.

limits of function $f \left(x\right) = \frac{\textcolor{\mathrm{de} e p s k y b l u e}{3 {x}^{2} + 5 x - 2}}{\textcolor{c \mathmr{and} a l}{{x}^{2} + x - 2}}$ at $x = 2$

By Substitution :
$f \left(x\right) {|}_{x = 2}$
$\quad \quad = \frac{\textcolor{\mathrm{de} e p s k y b l u e}{3 {\left(2\right)}^{2} + 5 \times 2 - 2}}{\textcolor{c \mathmr{and} a l}{{\left(2\right)}^{2} + 2 - 2}}$
$\quad \quad = \frac{\textcolor{\mathrm{de} e p s k y b l u e}{12 + 10 - 2}}{\textcolor{c \mathmr{and} a l}{4 + 2 - 2}}$
$\quad \quad = \frac{\textcolor{\mathrm{de} e p s k y b l u e}{20}}{\textcolor{c \mathmr{and} a l}{4}}$
$\quad \quad = 5$

For left-hand-limit and right-hand-limit substitute $x = 2 - \delta$ and $x = 2 + \delta$. You are required to work out this once.
It is generalized that, the left-hand-limit and right-hand-limit need not be computed. A standard result is specified as follows.

When the function does not evaluate to indeterminate value or undefined large, the limits of the function will be equal to the value got by substitution.
This generalization is not applicable when the function is given in piecewise form.

Limit of defined polynomials: For a function $f \left(x\right)$ having a ratio of two polynomials, if $f \left(x\right) {|}_{x = a}$ is defined, then
$f \left(x\right) {|}_{x = a}$
$\quad \quad = {\lim}_{x \to a -} f \left(x\right)$
$\quad \quad = {\lim}_{x \to a +} f \left(x\right)$

limit of ratio of polynomials defined

To find limits of function $f \left(x\right) = \frac{\textcolor{\mathrm{de} e p s k y b l u e}{3 {x}^{2} - 5 x - 2}}{\textcolor{c \mathmr{and} a l}{{x}^{2} - x - 2}}$ at $x = 2$, first we should check if the function evaluates to $\frac{0}{0}$.

limits of function $f \left(x\right) = \frac{\textcolor{\mathrm{de} e p s k y b l u e}{3 {x}^{2} - 5 x - 2}}{\textcolor{c \mathmr{and} a l}{{x}^{2} - x - 2}}$ at $x = 2$

By Substitution :
$f \left(x\right) {|}_{x = 2}$
$\quad \quad = \frac{\textcolor{\mathrm{de} e p s k y b l u e}{3 {\left(2\right)}^{2} - 5 \times 2 - 2}}{\textcolor{c \mathmr{and} a l}{{\left(2\right)}^{2} - 2 - 2}}$
$\quad \quad = \frac{\textcolor{\mathrm{de} e p s k y b l u e}{12 - 10 - 2}}{\textcolor{c \mathmr{and} a l}{4 - 2 - 2}}$
$\quad \quad = \frac{\textcolor{\mathrm{de} e p s k y b l u e}{0}}{\textcolor{c \mathmr{and} a l}{0}}$

When the function evaluates to $\frac{0}{0}$, examine if the numerator and denominator has a common factor.

factorize the numerator and denominator
$\textcolor{\mathrm{de} e p s k y b l u e}{3 {x}^{2} - 5 x - 2} = \textcolor{\mathrm{de} e p s k y b l u e}{\left(3 x + 1\right) \left(x - 2\right)}$
$\textcolor{c \mathmr{and} a l}{{x}^{2} - x - 2} = \textcolor{c \mathmr{and} a l}{\left(x + 1\right) \left(x - 2\right)}$
Note that the common factor $x - 2$ cannot be canceled at $x = 2$. The function remains indeterminate at $x = 2$.

limits of function $f \left(x\right) = \frac{\textcolor{\mathrm{de} e p s k y b l u e}{3 {x}^{2} - 5 x - 2}}{\textcolor{c \mathmr{and} a l}{{x}^{2} - x - 2}}$ at $x = 2$

Knowing that the numerator and denominator have a common factor, use the function $f \left(x\right) = \frac{\textcolor{\mathrm{de} e p s k y b l u e}{\left(3 x + 1\right) \left(x - 2\right)}}{\textcolor{c \mathmr{and} a l}{\left(x + 1\right) \left(x - 2\right)}}$

left-hand-limit :
${\lim}_{x \to 2 -} f \left(x\right)$
$\quad \quad = \frac{\textcolor{\mathrm{de} e p s k y b l u e}{\left(3 \left(2 - \delta\right) + 1\right) \left(2 - \delta - 2\right)}}{\textcolor{c \mathmr{and} a l}{\left(2 - \delta + 1\right) \left(2 - \delta - 2\right)}}$
$\quad \quad = \frac{\textcolor{\mathrm{de} e p s k y b l u e}{\left(7 - \delta\right) \left(- \delta\right)}}{\textcolor{c \mathmr{and} a l}{\left(3 - \delta\right) \left(- \delta\right)}}$
The $- \delta$ from numerator and denominator can be canceled, as $- \delta$ is not $0$.
$\quad \quad = \frac{\textcolor{\mathrm{de} e p s k y b l u e}{7 - \delta}}{\textcolor{c \mathmr{and} a l}{3 - \delta}}$
In this, compared to $7$ or $3$, the value of $\delta$ is negligible and that is reflected by substituting $\delta = 0$.
$\quad \quad = \frac{\textcolor{\mathrm{de} e p s k y b l u e}{7}}{\textcolor{c \mathmr{and} a l}{3}}$

limits of function $f \left(x\right) = \frac{\textcolor{\mathrm{de} e p s k y b l u e}{3 {x}^{2} - 5 x - 2}}{\textcolor{c \mathmr{and} a l}{{x}^{2} - x - 2}}$ at $x = 2$

Knowing that the numerator and denominator have a common factor, use the function $f \left(x\right) = \frac{\textcolor{\mathrm{de} e p s k y b l u e}{\left(3 x + 1\right) \left(x - 2\right)}}{\textcolor{c \mathmr{and} a l}{\left(x + 1\right) \left(x - 2\right)}}$

right-hand-limit :
${\lim}_{x \to 2 +} f \left(x\right)$
$\quad \quad = \frac{\textcolor{\mathrm{de} e p s k y b l u e}{\left(3 \left(2 + \delta\right) + 1\right) \left(2 + \delta - 2\right)}}{\textcolor{c \mathmr{and} a l}{\left(2 + \delta + 1\right) \left(2 + \delta - 2\right)}}$
$\quad \quad = \frac{\textcolor{\mathrm{de} e p s k y b l u e}{\left(7 + \delta\right) \left(\delta\right)}}{\textcolor{c \mathmr{and} a l}{\left(3 + \delta\right) \left(\delta\right)}}$
The $\delta$ from numerator and denominator can be canceled, as $\delta$ is not $0$.
$\quad \quad = \frac{\textcolor{\mathrm{de} e p s k y b l u e}{7 + \delta}}{\textcolor{c \mathmr{and} a l}{3 + \delta}}$
In this, compared to $7$ or $3$, the value of $\delta$ is negligible and that is reflected by substituting $\delta = 0$.
$\quad \quad = \frac{\textcolor{\mathrm{de} e p s k y b l u e}{7}}{\textcolor{c \mathmr{and} a l}{3}}$

limits of function $f \left(x\right) = \frac{\textcolor{\mathrm{de} e p s k y b l u e}{3 {x}^{2} - 5 x - 2}}{\textcolor{c \mathmr{and} a l}{{x}^{2} - x - 2}}$ at $x = 2$

$f \left(x\right) {|}_{x = 2} = \frac{0}{0}$

${\lim}_{x \to 2 -} f \left(x\right) = \frac{7}{3}$

${\lim}_{x \to 2 +} f \left(x\right) = \frac{7}{3}$

The limits are equal. So the function is defined by limits.

Revising the steps taken to solve the problem...
limits of function $f \left(x\right) = \frac{\textcolor{\mathrm{de} e p s k y b l u e}{3 {x}^{2} - 5 x - 2}}{\textcolor{c \mathmr{and} a l}{{x}^{2} - x - 2}}$ at $x = 2$

We found by Substitution :
$f \left(x\right) {|}_{x = 2}$
$\quad \quad = \frac{0}{0}$

And then factorized the numerator and denominator
$\textcolor{\mathrm{de} e p s k y b l u e}{3 {x}^{2} - 5 x - 2} = \textcolor{\mathrm{de} e p s k y b l u e}{\left(3 x + 1\right) \left(x - 2\right)}$
$\textcolor{c \mathmr{and} a l}{{x}^{2} - x - 2} = \textcolor{c \mathmr{and} a l}{\left(x + 1\right) \left(x - 2\right)}$

Though $\left(x - 2\right)$ cannot be canceled when $x = 2$, we have worked out the left-hand-limit and right-hand-limit results, and know that the factor $x - 2$ will cancel as $\delta$ when limit is calculated.

From this example, a standard result is specified as "if a function evaluate to $\frac{0}{0}$ at an input value, then the expected value of the function can be calculated by canceling out common factors between numerator and denominator."

So without going through the lengthy process of substituting $x = a - \delta$ and $x = a + \delta$, just cancel out the common factor and find the limit for the rest of the function.

If limits of numerator and denominator are $0$, then to find limit
•  the common factors are canceled; or
•  factors are separated to other standard form of limits.
Then find limit using the modified expression.

limit of ratio of polinomials : not defined

To find limits of function $f \left(x\right) = \frac{\textcolor{\mathrm{de} e p s k y b l u e}{3 {x}^{2} - 5 x - 2}}{\textcolor{c \mathmr{and} a l}{{x}^{2} - 4 x + 4}}$ at $x = 2$, let us check if the function evaluates to $\frac{0}{0}$.

limits of function $f \left(x\right) = \frac{\textcolor{\mathrm{de} e p s k y b l u e}{3 {x}^{2} - 5 x - 2}}{\textcolor{c \mathmr{and} a l}{{x}^{2} - 4 x + 4}}$ at $x = 2$

By Substitution :
$f \left(x\right) {|}_{x = 2}$
$\quad \quad = \frac{\textcolor{\mathrm{de} e p s k y b l u e}{3 {\left(2\right)}^{2} - 5 \times 2 - 2}}{\textcolor{c \mathmr{and} a l}{{2}^{2} - 4 \left(2\right) + 4}}$
$\quad \quad = \frac{\textcolor{\mathrm{de} e p s k y b l u e}{12 - 10 - 2}}{\textcolor{c \mathmr{and} a l}{4 - 8 + 4}}$
$\quad \quad = \frac{0}{0}$

When the function evaluates to $\frac{0}{0}$, examine if the numerator and denominator has common factors.
factorize the numerator and denominator
$\textcolor{\mathrm{de} e p s k y b l u e}{3 {x}^{2} - 5 x - 2} = \textcolor{\mathrm{de} e p s k y b l u e}{\left(3 x + 1\right) \left(x - 2\right)}$
$\textcolor{c \mathmr{and} a l}{{x}^{2} - 4 x + 4} = \textcolor{c \mathmr{and} a l}{\left(x - 2\right) \left(x - 2\right)}$

limits of function $f \left(x\right) = \frac{\textcolor{\mathrm{de} e p s k y b l u e}{3 {x}^{2} - 5 x - 2}}{\textcolor{c \mathmr{and} a l}{{x}^{2} - 4 x + 4}}$ at $x = 2$

Knowing that the numerator and denominator have a common factor, use the function $f \left(x\right) = \frac{\textcolor{\mathrm{de} e p s k y b l u e}{\left(3 x + 1\right) \left(x - 2\right)}}{\textcolor{c \mathmr{and} a l}{\left(x - 2\right) \left(x - 2\right)}}$

left-hand-limit :
${\lim}_{x \to 2 -} f \left(x\right)$
$\quad \quad = \frac{\textcolor{\mathrm{de} e p s k y b l u e}{\left(3 \left(2 - \delta\right) + 1\right) \left(2 - \delta - 2\right)}}{\textcolor{c \mathmr{and} a l}{\left(2 - \delta - 2\right) \left(2 - \delta - 2\right)}}$
$\quad \quad = \frac{\textcolor{\mathrm{de} e p s k y b l u e}{\left(7 - \delta\right) \left(- \delta\right)}}{\textcolor{c \mathmr{and} a l}{\left(- \delta\right) \left(- \delta\right)}}$
The $- \delta$ from numerator and denominator can be canceled, as $- \delta$ is not $0$.
$\quad \quad = \frac{\textcolor{\mathrm{de} e p s k y b l u e}{7 - \delta}}{\textcolor{c \mathmr{and} a l}{- \delta}}$
In this, compared to $7$, the value of $\delta$ is negligible and that is reflected by substituting $\delta = 0$.
$\quad \quad = \frac{\textcolor{\mathrm{de} e p s k y b l u e}{- 7}}{\textcolor{c \mathmr{and} a l}{\delta}}$
$\quad \quad = - \infty$

limits of function $f \left(x\right) = \frac{\textcolor{\mathrm{de} e p s k y b l u e}{3 {x}^{2} - 5 x - 2}}{\textcolor{c \mathmr{and} a l}{{x}^{2} - 4 x + 4}}$ at $x = 2$

Knowing that the numerator and denominator have a common factor, use the function $f \left(x\right) = \frac{\textcolor{\mathrm{de} e p s k y b l u e}{\left(3 x + 1\right) \left(x - 2\right)}}{\textcolor{c \mathmr{and} a l}{\left(x - 2\right) \left(x - 2\right)}}$

right-hand-limit :
${\lim}_{x \to 2 -} f \left(x\right)$
$\quad \quad = \frac{\textcolor{\mathrm{de} e p s k y b l u e}{\left(3 \left(2 + \delta\right) + 1\right) \left(2 + \delta - 2\right)}}{\textcolor{c \mathmr{and} a l}{\left(2 + \delta - 2\right) \left(2 + \delta - 2\right)}}$
$\quad \quad = \frac{\textcolor{\mathrm{de} e p s k y b l u e}{\left(7 + \delta\right) \left(\delta\right)}}{\textcolor{c \mathmr{and} a l}{\left(\delta\right) \left(\delta\right)}}$
The $\delta$ from numerator and denominator can be canceled, as $\delta$ is not $0$.
$\quad \quad = \frac{\textcolor{\mathrm{de} e p s k y b l u e}{7 + \delta}}{\textcolor{c \mathmr{and} a l}{\delta}}$
In this, compared to $7$, the value of $\delta$ is negligible and that is reflected by substituting $\delta = 0$.
$\quad \quad = \frac{\textcolor{\mathrm{de} e p s k y b l u e}{7}}{\textcolor{c \mathmr{and} a l}{\delta}}$
$\quad \quad = \infty$

limits of function $f \left(x\right) = \frac{\textcolor{\mathrm{de} e p s k y b l u e}{3 {x}^{2} - 5 x - 2}}{\textcolor{c \mathmr{and} a l}{{x}^{2} - 4 x + 4}}$ at $x = 2$

$f \left(x\right) {|}_{x = 2} = \frac{0}{0}$

${\lim}_{x \to 2 -} f \left(x\right) = - \infty$

${\lim}_{x \to 2 +} f \left(x\right) = \infty$

So the function is not defined.

summary

Limit of polynomials evaluating to $0$ : For a function $f \left(x\right) = \frac{\textcolor{\mathrm{de} e p s k y b l u e}{g \left(x\right)}}{\textcolor{c \mathmr{and} a l}{h \left(x\right)}}$ having a ratio of two polynomials, if $f \left(x\right) {|}_{x = a}$ is $\frac{0}{0}$ then factorize $\textcolor{\mathrm{de} e p s k y b l u e}{g \left(x\right)}$ and $\textcolor{c \mathmr{and} a l}{h \left(x\right)}$ such that
$\textcolor{\mathrm{de} e p s k y b l u e}{g \left(x\right) = {\left(x - a\right)}^{k} {g}_{1} \left(x\right) {g}_{s} \left(x\right)}$
$\textcolor{c \mathmr{and} a l}{h \left(x\right) = {\left(x - a\right)}^{l} {h}_{1} \left(x\right) {h}_{s} \left(x\right)}$
Where the ratio $\frac{\textcolor{\mathrm{de} e p s k y b l u e}{{g}_{s} \left(x\right)}}{\textcolor{c \mathmr{and} a l}{{h}_{s} \left(s\right)}}$ is a standard form for which limit is defined at $x = a$.

Find the limit of the function as
${\lim}_{x \to a} f \left(x\right)$
$\quad \quad = {\lim}_{x \to a} \frac{\textcolor{\mathrm{de} e p s k y b l u e}{{\left(x - a\right)}^{k} {g}_{1} \left(x\right)}}{\textcolor{c \mathmr{and} a l}{{\left(x - a\right)}^{l} {h}_{1} \left(x\right)}}$
$\quad \quad \quad \quad \times {\lim}_{x \to a} \frac{\textcolor{\mathrm{de} e p s k y b l u e}{{g}_{s} \left(x\right)}}{\textcolor{c \mathmr{and} a l}{{h}_{s} \left(x\right)}}$
Cancel out the factors $x - a$ from numerator and denominator.
Then find limit at $x = a$ using the simplified expression.

Outline