# Indeterminate Forms

• ### Indeterminate Forms $$\large\frac{0}{0}\normalsize$$

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

${\lim\limits_{x \to a} f\left( x \right) = 0\;\;\;}\kern-0.3pt {\text{and}\;\;\lim\limits_{x \to a} g\left( x \right) = 0.}$

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

Note: In this chapter, we do not apply L’Hopital’s rule.

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

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

${\lim\limits_{x \to a} f\left( x \right) = \pm \infty\;\;\;}\kern-0.3pt {\text{and}\;\;\lim\limits_{x \to a} g\left( x \right) = \pm \infty.}$

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

It is said that the function $$\large\frac{{f\left( x \right)}}{{g\left( x \right)}}\normalsize$$ has the indeterminate form $$\large\frac{\infty}{\infty}\normalsize$$ 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 $$\large\frac{0}{0}\normalsize$$ or $$\large\frac{\infty}{\infty}\normalsize.$$

• ## Solved Problems

Click a problem to see the solution.

### Example 1

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

### Example 2

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

### Example 3

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

### Example 4

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

### Example 5

Find the limit $$\lim\limits_{x \to \pi } {\large\frac{{\cos \frac{x}{2}}}{{\pi – x}}\normalsize}.$$

### Example 6

Calculate $$\lim\limits_{x \to \infty } \left( {\sqrt {{x^2} + 1} – \sqrt {{x^2} – 1} } \right).$$

### Example 7

Find the limit $$\lim\limits_{x \to 4} {\large\frac{{\sqrt {1 + 6x} – 5}}{{\sqrt x – 2}}\normalsize}.$$

### Example 8

Find the limit $$\lim\limits_{x \to \infty } {\large\frac{{{{\left( {2x + 3} \right)}^{10}}{{\left( {3x – 2} \right)}^{20}}}}{{{{\left( {x + 5} \right)}^{30}}}}\normalsize}.$$

### Example 9

Find the limit $$\lim\limits_{x \to e} {\large\frac{{\ln x – 1}}{{x – e}}\normalsize}.$$

### Example 10

Find the limit $$\lim\limits_{t \to + \infty } \left( {\sqrt {t + \sqrt {t + 1} } – \sqrt t } \right).$$

### Example 1.

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

Solution.

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

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

### Example 2.

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

Solution.

This is of the form $$\large\frac{\infty}{\infty}\normalsize$$ 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).

Page 1
Problems 1-2
Page 2
Problems 3-10