Calculus
Limits and Continuity of Functions
Indeterminate Forms
Problems 1-2
Problems 3-10
Indeterminate Forms \(\large\frac{0}{0}\normalsize\)
Let\(f\left( x \right)\) and \(g\left( x \right)\) be two functions such that
{\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
{\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 on problem description to see 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
{\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\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,
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).
Problems 1-2
Problems 3-10