Contents
We’ll talk in detail about indeterminate forms: what they are, and how to calculate the limit of functions when encountering them. We’ll cover methods to resolve them, along with various examples for each case and their solutions. Let’s begin.
What are Indeterminate Forms?
Indeterminate forms are situations where determining the limit of a function is difficult. We are forced to change the form of the function using many methods or study it more deeply to arrive at the limit. The indeterminate forms are: zero/zero, zero times infinity, ∞/∞, ∞ – ∞.
The following table summarizes indeterminate forms more clearly:
| \[\frac{0}{0}\] | \[\frac{±∞}{±∞}\] |
| \[0 \times ±∞\] | \[+∞ -∞\] |
Methods for Resolving Indeterminate Forms
Indeterminate Form: Zero/Zero
There are 4 methods to resolve this indeterminate form, which we follow according to the function’s shape:
- Factorization (Reduction) Method
- Conjugate Method
- Derivative Definition Method
- Relying on certain well-known, proven limits.
We will provide solved examples for each case.
Example 1: Factorization Method: Find the limit of the following function at x = 1
\[f(x) = \frac{x^2 – 1}{x – 1}\]
By substitution, we reach the indeterminate form 0/0. We resort to the factorization method. We notice that the numerator is a well-known identity (difference of two squares) and can therefore be written in another form:
\[f(x) = \frac{(x+1)(x-1)}{x – 1}\]
\[f(x) =(x+1)\\
\lim_{x \rightarrow 1}(x+1) =2\]
Example 2: Conjugate Method: Find the limit of the following function when x=5
\[f(x)=\frac{ \sqrt{2x-1} -3}{x-5}\]
Solution: We notice the presence of a square root in the numerator. We solve using the second method, which is the conjugate method. That is, we multiply the numerator and denominator by the conjugate of the numerator.
\[f(x)=\frac{(\sqrt{2x-1} -3)(\sqrt{2x-1} +3)}{(x-5)(\sqrt{2x-1} +3)}\]
We notice that the numerator resembles the identity for the difference of two squares. We rewrite the numerator:
\[= \frac{(2x-1) – 3^2}{(x-5)(\sqrt{2x-1} +3)}\]
And the rest of the steps are all simplification and reduction:
\[= \frac{2(x-5)}{(x-5)(\sqrt{2x-1} +3)} \\\
= \frac{2}{(\sqrt{2x-1} +3)} \\\
\lim_{x \rightarrow 5} f(x)=\frac{1}{3}\]
Example 3: Resolving Indeterminacy using the Derivative Definition
Find the limit of the following function as x approaches \(\frac{\pi}{6}\):
\[f(x)=\frac{2\sin x-1}{6x-\pi}\]
In the derivative definition method, the denominator must be in the form x-a and the numerator in the form g(x)-g(a). We modify the denominator to reach the required form.
\[f(x)= \frac{2\sin x-1}{6(x-\frac{\pi}{6})}\\
f(x)= \frac{1}{6} \times \frac{2\sin x-1}{x-\frac{\pi}{6}}\]
We note that \(g(x)= 2\sin x\) and thus \(g(\frac{\pi}{6})= 1\)
The last step is to apply the formula:
\[\lim_{x \rightarrow a} \frac{g(x)-g(a)}{x-a} = g'(a)\]
Therefore, it will be:
\[\lim_{x \rightarrow \frac{\pi}{6}} \frac{2\sin x-1}{6x-\pi}\]
\[=\frac{1}{6}\lim_{x \rightarrow \frac{\pi}{6}} \frac{g(x)-g(\frac{\pi}{6})}{x-\frac{\pi}{6}} =\frac{1}{6} g'(\frac{\pi}{6})\]
We calculate \(g'(x) = 2\cos x\) and thus \(g'(\frac{\pi}{6}) = 2\cos \frac{\pi}{6}=\sqrt{3}\)
Therefore, the limit of the function is:
\[\lim_{x \rightarrow \frac{\pi}{6}} \frac{2\sin x-1}{6x-\pi} =\frac{\sqrt{3}}{6}\]
Resolving the Indeterminate Form: Infinity / Infinity
We distinguish two cases:
- The first case is for a function containing a radical, we use the factorization method.
- The second case is for polynomials.
Example for the first case: Find the limit of the following function:
\[\lim_{x \rightarrow -\infty} \frac{\sqrt{2x^2 + x – 3}}{1-x}\]
Solution: By substitution, we find that the result is: \(\lim_{x \rightarrow -\infty}f(x)= \frac{+\infty}{+\infty}\)
We factor out the term with the highest degree and then simplify.
\[f(x)=\frac{\sqrt{x^2 \left( 2 + \frac{x}{x^2} – \frac{3}{x^2} \right) } }{1-x}\]
\[=\frac{|x|\sqrt{ \left( 2 + \frac{x}{x^2} – \frac{3}{x^2} \right) } }{1-x}: \sqrt{x^2}=|x|\]
Never forget that the square root of x squared = |x|. And there is an important rule that must always be noted:
\[|x| = -x: x \rightarrow -\infty\\
|x| = +x: x \rightarrow +\infty\]
Now we simplify:
\[=\frac{-x\sqrt{ \left( 2 + \frac{x}{x^2} – \frac{3}{x^2} \right) } }{x(\frac{1}{x}-1)}\]
And since:
\[\lim_{x \rightarrow -\infty} \frac{x}{x^2} = \lim_{x \rightarrow -\infty} \frac{3}{x^2} = \lim_{x \rightarrow -\infty} \frac{1}{x} =0\]
Then:
\[\lim_{x \rightarrow -\infty}f(x) = \frac{-\sqrt{2}}{-1} = \sqrt{2}\]
