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.

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

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

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).