Calculus

Limits and Continuity of Functions

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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.

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