 maths > calculus-limits

Limit of functions evaluating to $\infty$

what you'll learn...

Limit of Expressions evaluating to $\infty$

»  Organize the sub-expressions to the following

→  a=0$\frac{a}{\infty} = 0$

→  ±a=$\infty \pm a = \infty$

→  ×a=$\infty \times a = \infty$ when a0$a \ne 0$

→  $\infty \times \infty = \infty$

→  ${\infty}^{n} = \infty$ when $n \ne 0$

→  ${\lim}_{x \to \infty} \frac{x}{x} = 1$

→  ${\lim}_{x \to - \infty} \frac{x}{x} = 1$

→  ${\lim}_{x \to \infty} {a}^{x}$ `=0 text( if ) 0 $\quad \quad \quad = \infty \textrm{\mathmr{if}} a > 1$

The value of function $f \left(x\right) = 3 {x}^{2} + 5 x - 2$ at $x = \infty$ can be found by substituting $x = \infty$

$\left(3 {x}^{2} + 5 x - 2\right)$
$\quad \quad = \left(3 {\left(\infty\right)}^{2} + 5 \infty - 2\right)$
$\quad \quad = \infty$
as ${\infty}^{2} = \infty$; $n \infty = \infty$; and $\infty \pm a = \infty$

The value of function $f \left(x\right) = \frac{1}{3 {x}^{2} + 5 x - 2}$ at $x = \infty$ can be found by substituting $x = \infty$

$\frac{1}{3 {x}^{2} + 5 x - 2}$
$\quad \quad = \frac{1}{3 {\left(\infty\right)}^{2} + 5 \infty - 2}$
$\quad \quad = \frac{1}{\infty}$
$\quad \quad = 0$
as $\frac{1}{\infty} = 0$.

subtraction of infinity

The value of function $f \left(x\right) = 3 {x}^{2} - 5 x - 2$ at $x = \infty$ is first tried with substituting $x = \infty$
$3 {x}^{2} - 5 x - 2$
$\quad \quad = 3 {\infty}^{2} - 5 \infty - 2$
$\quad \quad = \infty - \infty$
$\quad \quad = \frac{0}{0}$

as ${\infty}^{2} = \infty$; $n \infty = \infty$; and $\infty \pm a = \infty$

It is noted that $\infty - \infty$ is neither $\infty$ nor $0$. It is indeterminate value

$\infty - \infty$
$\quad \quad = \left(\frac{1}{0}\right) - \left(\frac{1}{0}\right)$
$\quad \quad = \frac{1 - 1}{0}$
$\quad \quad = \frac{0}{0}$

division of infinity

The value 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 $x = \infty$ is first tried with substituting x equals infinity.

$\frac{\textcolor{\mathrm{de} e p s k y b l u e}{3 {x}^{2} + 5 x - 2}}{\textcolor{c \mathmr{and} a l}{{x}^{2} + x - 2}}$
$\quad \quad = \frac{\textcolor{\mathrm{de} e p s k y b l u e}{3 {\left(\infty\right)}^{2} + 5 \infty - 2}}{\textcolor{c \mathmr{and} a l}{{\infty}^{2} + \infty - 2}}$
$\quad \quad = \frac{\textcolor{\mathrm{de} e p s k y b l u e}{\infty}}{\textcolor{c \mathmr{and} a l}{\infty}}$
$\quad \quad = \frac{0}{0}$

as ${\infty}^{2} = \infty$; $n \infty = \infty$; and $\infty \pm a = \infty$

It is noted that $\infty \div \infty$ is neither $\infty$ nor $0$. It is indeterminate value

$\infty \div \infty$
$\quad \quad = \left(\frac{1}{0}\right) \div \left(\frac{1}{0}\right)$
$\quad \quad = \frac{1}{0} \times \frac{0}{1}$
$\quad \quad = \frac{0}{0}$

watch-out

The forms of expressions evaluate to indeterminate values when computing limit for $\infty$ or $- \infty$ are $\infty \div \infty$ and $\infty - \infty$.

what to do

When we encounter $\infty \div \infty$ or $\infty - \infty$, convert the expression to one of the following forms given on left hand side
${\lim}_{x \to \infty} \frac{x}{x} = 1$
${\lim}_{x \to - \infty} \frac{x}{x} = 1$
$\frac{a}{\infty} = 0$
${\infty}^{n} = \infty$
$n \infty = \infty$
$\infty \pm a = \infty$

examples

Limit of function $f \left(x\right) = \left(3 {x}^{2} - 5 x - 2\right)$ at $x = \infty$
The function evaluates to $\infty - \infty$ at $x = \infty$

The limit of the function is
${\lim}_{x \to \infty} \left(3 {x}^{2} - 5 x - 2\right)$
$\quad \quad = {\lim}_{x \to \infty} {x}^{2} \left(2 - \frac{5}{x} - \frac{2}{x} ^ 2\right)$
$\quad \quad = {\lim}_{x \to \infty} {x}^{2}$
$\quad \quad \quad \quad \times {\lim}_{x \to \infty} \left(2 - \frac{5}{x} - \frac{2}{x} ^ 2\right)$
$\quad \quad = {\infty}^{2} \times \left(2 - 0 - 0\right)$
$\quad \quad = \infty$

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 $x = \infty$
The function evaluates to $\frac{\infty}{\infty}$ at $x = \infty$

The limit of the function is
${\lim}_{x \to \infty} \frac{\textcolor{\mathrm{de} e p s k y b l u e}{3 {x}^{2} + 5 x - 2}}{\textcolor{c \mathmr{and} a l}{{x}^{2} + x - 2}}$
$\quad \quad = {\lim}_{x \to \infty} \frac{\textcolor{\mathrm{de} e p s k y b l u e}{{x}^{2} \left(3 + \frac{5}{x} - \frac{2}{x} ^ 2\right)}}{\textcolor{c \mathmr{and} a l}{{x}^{2} \left(1 + \frac{1}{x} - \frac{2}{x} ^ 2\right)}}$
$\quad \quad = {\lim}_{x \to \infty} \frac{\textcolor{\mathrm{de} e p s k y b l u e}{{x}^{2}}}{\textcolor{c \mathmr{and} a l}{{x}^{2}}}$
$\quad \quad \quad \quad \times {\lim}_{x \to \infty} \frac{\textcolor{\mathrm{de} e p s k y b l u e}{3 + \frac{5}{x} - \frac{2}{x} ^ 2}}{\textcolor{c \mathmr{and} a l}{1 + \frac{1}{x} - \frac{2}{x} ^ 2}}$
$\quad \quad = {\left[{\lim}_{x \to \infty} \frac{\textcolor{\mathrm{de} e p s k y b l u e}{x}}{\textcolor{c \mathmr{and} a l}{x}}\right]}^{2} \times \frac{\textcolor{\mathrm{de} e p s k y b l u e}{3 + 0 - 0}}{\textcolor{c \mathmr{and} a l}{1 + 0 - 0}}$
$\quad \quad = {1}^{2} \times 3$
$\quad \quad = 3$

When evaluating limits to infinity or minus infinity, simplify to known results.

Find the limit of the function ${\lim}_{x \to \infty} \frac{x + 3}{5 x + 4}$

The answer is '$\frac{1}{5}$'.

summary

Evaluating limits to $\infty$ or $- \infty$: Simplify the numerical expressions to one of the following
${\lim}_{x \to \infty} \frac{x}{x} = 1$
${\lim}_{x \to - \infty} \frac{x}{x} = 1$
$\frac{a}{\infty} = 0$
$\infty \pm a = \infty$
$n \infty = \infty$ where $n \ne 0$
$\infty \times \infty = \infty$ or
${\infty}^{n} = \infty$ where $n \ne 0$
And avoid indeterminate values $\frac{\infty}{\infty}$, $\infty - \infty$, $0 \times \infty$, and ${\infty}^{0}$ .

Outline