# Indeterminate Forms

## Indeterminate Forms $$\frac{0}{0}$$

Let $$f\left( x \right)$$ and $$g\left( x \right)$$ be two functions such that

$\lim\limits_{x \to a} f\left( x \right) = 0\;\;\; \text{and}\;\;\lim\limits_{x \to a} g\left( x \right) = 0.$

Then the function $$\frac{{f\left( x \right)}}{{g\left( x \right)}}$$ has the indeterminate form $$\frac{0}{0}$$ at $$x = a.$$ To find the limit at $$x = a$$ when the function $$\frac{{f\left( x \right)}}{{g\left( x \right)}}$$ has the indeterminate form $$\frac{0}{0}$$ at this point, we must factor the numerator and denominator and then reduce the terms that approach zero.

Note: In this topic we do not apply L'Hopital's rule.

## Indeterminate Forms $$\frac{\infty}{\infty}$$

Let $$f\left( x \right)$$ and $$g\left( x \right)$$ be two functions such that

$\lim\limits_{x \to a} f\left( x \right) = \pm \infty\;\;\; \text{and}\;\;\lim\limits_{x \to a} g\left( x \right) = \pm \infty.$

where $$a$$ where a is a real number, or $$+\infty$$ or $$-\infty.$$

It is said that the function $$\frac{{f\left( x \right)}}{{g\left( x \right)}}$$ has the indeterminate form $$\frac{\infty}{\infty}$$ at this point.

To find the limit, we must divide the numerator and denominator by $$x$$ of highest degree.

## Indeterminate Forms $$\infty - \infty,$$ $$0 \cdot \infty,$$ $$\infty^0,$$ $$1^{\infty}$$

Indeterminate forms of these types can usually be treated by putting them into one of the forms $$\frac{0}{0}$$ or $$\frac{\infty}{\infty}.$$

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

Find the limit $\lim\limits_{x \to 1} {\frac{{{x^{20}} - 1}}{{{x^{10}} - 1}}}.$

### Example 2

Calculate $\lim\limits_{y \to - 2} {\frac{{{y^3} + 3{y^2} + 2y}}{{{y^2} - y - 6}}}.$

### Example 3

Calculate $\lim\limits_{x \to \infty } {\frac{{{x^3} + 3x + 5}}{{2{x^3} - 6x + 1}}}.$

### Example 4

Calculate the limit $\lim\limits_{x \to 1} {\frac{{\sqrt[3]{x} - 1}}{{x - 1}}}.$

### Example 1.

Find the limit $\lim\limits_{x \to 1} {\frac{{{x^{20}} - 1}}{{{x^{10}} - 1}}}.$

Solution.

Direct substitution of $$x = 1$$ yields the indeterminate form $$\frac{0}{0}$$ at the point $$x = 1.$$ Therefore, we factor the numerator to get

$\lim\limits_{x \to 1} \frac{{{x^{20}} - 1}}{{{x^{10}} - 1}} = \left[ {\frac{0}{0}} \right] = \lim\limits_{x \to 1} \frac{{{{\left( {{x^{10}}} \right)}^2} - 1}}{{{x^{10}} - 1}} = \lim\limits_{x \to 1} \frac{{\cancel{\left( {{x^{10}} - 1} \right)}\left( {{x^{10}} + 1} \right)}}{\cancel{{x^{10}} - 1}} = \lim\limits_{x \to 1} \left( {{x^{10}} + 1} \right) = {1^{10}} + 1 = 2.$

### Example 2.

Calculate $\lim\limits_{y \to - 2} {\frac{{{y^3} + 3{y^2} + 2y}}{{{y^2} - y - 6}}}.$

Solution.

This is of the form $$\frac{\infty}{\infty}$$ at $$y = -2.$$ We factor the numerator and the denominator:

${y^3} + 3{y^2} + 2y = y\left( {{y^2} + 3y + 2} \right) = y\left( {y + 1} \right)\left( {y + 2} \right).$

Here we used the formula

$a{x^2} + bx + c = a\left( {x - {x_1}} \right)\left( {x - {x_2}} \right),$

where $${x_1},$$ $${x_2}$$ are the solutions of the quadratic equation.

Similarly,

${y^2} - y - 6 = \left( {y - 3} \right)\left( {y + 2} \right).$

Thus, the limit is

$\lim\limits_{y \to - 2} \frac{{{y^3} + 3{y^2} + 2y}}{{{y^2} - y - 6}} = \left[ {\frac{0}{0}} \right] = \lim\limits_{y \to - 2} \frac{{y\left( {y + 1} \right)\cancel{\left( {y + 2} \right)}}}{{\left( {y - 3} \right)\cancel{\left( {y + 2} \right)}}} = \lim\limits_{y \to - 2} \frac{{y\left( {y + 1} \right)}}{{y - 3}} = \frac{{\lim\limits_{y \to - 2} y \cdot \lim\limits_{y \to - 2} \left( {y + 1} \right)}}{{\lim\limits_{y \to - 2} \left( {y - 3} \right)}} = \frac{{ - 2 \cdot \left( { - 1} \right)}}{{ - 5}} = - \frac{2}{5}.$

(by the quotient and product rules for limits).

### Example 3.

Calculate $\lim\limits_{x \to \infty } {\frac{{{x^3} + 3x + 5}}{{2{x^3} - 6x + 1}}}.$

Solution.

Substituting $$x \to \infty$$ shows that this is of the form $$\frac{\infty}{\infty}.$$ Divide the numerator and denominator by $${x^3}$$ (the highest degree in this expression). Thus, we obtain

$\lim\limits_{x \to \infty } \frac{{{x^3} + 3x + 5}}{{2{x^3} - 6x + 1}} = \left[ {\frac{\infty }{\infty }} \right] = \lim\limits_{x \to \infty } \frac{{\frac{{{x^3} + 3x + 5}}{{{x^3}}}}}{{\frac{{2{x^3} - 6x + 1}}{{{x^3}}}}} = \lim\limits_{x \to \infty } \frac{{\frac{{{x^3}}}{{{x^3}}} + \frac{{3x}}{{{x^3}}} + \frac{5}{{{x^3}}}}}{{\frac{{2{x^3}}}{{{x^3}}} - \frac{{6x}}{{{x^3}}} + \frac{1}{{{x^3}}}}} = \lim\limits_{x \to \infty } \frac{{1 + \frac{3}{{{x^2}}} + \frac{5}{{{x^3}}}}}{{2 - \frac{6}{{{x^2}}} + \frac{1}{{{x^3}}}}} = \frac{{\lim\limits_{x \to \infty } \left( {1 + \frac{3}{{{x^2}}} + \frac{5}{{{x^3}}}} \right)}}{{\lim\limits_{x \to \infty } \left( {2 - \frac{6}{{{x^2}}} + \frac{1}{{{x^3}}}} \right)}} = \frac{{\lim\limits_{x \to \infty } 1 + \lim\limits_{x \to \infty } \frac{3}{{{x^2}}} + \lim\limits_{x \to \infty } \frac{5}{{{x^3}}}}}{{\lim\limits_{x \to \infty } 2 - \lim\limits_{x \to \infty } \frac{6}{{{x^2}}} + \lim\limits_{x \to \infty } \frac{1}{{{x^3}}}}} = \frac{{1 + 0 + 0}}{{2 - 0 - 0}} = \frac{1}{2}.$

### Example 4.

Calculate the limit $\lim\limits_{x \to 1} {\frac{{\sqrt[3]{x} - 1}}{{x - 1}}}.$

Solution.

We write the denominator in the form

$x - 1 = {\left( {\sqrt[3]{x}} \right)^3} - {1^3}$

and factor it as difference of cubes:

$x - 1 = {\left( {\sqrt[3]{x}} \right)^3} - {1^3} = \left( {\sqrt[3]{x} - 1} \right)\left( {\sqrt[3]{{{x^2}}} + \sqrt[3]{x} + 1} \right).$

As a result we have

$\lim\limits_{x \to 1} \frac{{\sqrt[3]{x} - 1}}{{x - 1}} = \left[ {\frac{0}{0}} \right] = \lim\limits_{x \to 1} \frac{\cancel{\sqrt[3]{x} - 1}}{{\cancel{\left({\sqrt[3]{x} - 1} \right)}\left( {\sqrt[3]{{{x^2}}} + \sqrt[3]{x} + 1} \right)}} = \lim\limits_{x \to 1} \frac{1}{{\sqrt[3]{{{x^2}}} + \sqrt[3]{x} + 1}} = \frac{1}{{\sqrt[3]{{{1^2}}} + \sqrt[3]{1} + 1}} = \frac{1}{3}.$

See more problems on Page 2.