Subject: Calculus

Indeterminate Form

So far, the limit theorems that we have discussed pertain only to finite forms. We developed techniques and methods in evaluating finite limits. In this article, we'll talk more about indeterminate forms of limits.

Okay, What is an Indeterminate Form!?

Indeterminate forms of limits are limits that deal with the values of a given function when it reaches infinity. We cannot exactly define the values of functions at infinity, can we? Hence the name indeterminate. But, we can predict and approximate some functions at infinity. To begin, let us first consider the function f defined by

f(x)=\frac{x^{2}}{x^{2}+5}

f(x)00.1670.9780.9920.9980.99950.99999
x011525501001000
Table 1: Values of f(x) for increasing values of x.

We can see from Table 1 that as x increases, the function's value f(x) gets closer and closer to 1. In fact, we can make the value of f(x) come very close to 1 as we continue to take lager and larger values of x. In this case, we can come to the correct conclusion that no matter how large x is, the value of the function f(x) will come very close to 1 but will never quite be equal to 1. So, the limit of f is 1 as x goes very large. This can be written as:

\lim_{x \to \infty}\frac{x^{2}}{x^{2}+5}=1

We can verify these by looking at Figure 1 for the graph of f(x). Notice that the curve of f bends away from f(x)=1. It is asymptotic to the limit.


Figure 1: Plot of the function f(x)=\frac{x^{2}}{x^{2}+5}

With indeterminate limits, the symbol "x \to +\infty" denotes that x increases without bound while the symbol "x \to -\infty" means that x decreases without bound.

Theorem 1

If n is a positive integer, then

\lim_{x \to +\infty}\frac{1}{x^{n}}=0

Theorem 2

If n is a positive integer, then

\lim_{x \to -\infty}\frac{1}{x^{n}}=0

The two theorems above seem equal, don't they? The only difference is the region of limit evaluation. The first theorem evaluates from x to positive infinity while the second is from x to negative infinity.

Example #1

Evaluate the following limits.

\lim_{x \to +\infty}\frac{x^{2}+1}{2x^{2}+5}

To solve this, we need to use Theorem 1. But before the theorem is applicable, we still need to divide both the numerators and the denominator by x^{2}. Thus,

\begin{eqnarray} && \lim_{x \to +\infty}\frac{x^{2}+1}{2x^{2}+5} \\ && = \lim_{x \to +\infty}\frac{1+\frac{1}{x^{2}}}{2+\frac{5}{x^{2}}} \\ && = \frac{1+0}{2+0} \\ && = \frac{1}{2} \end{eqnarray}

Thus, with the aid of Theorem 1, the limit of the function is

Example #2

Evaluate the following limit.

\lim_{x \to -\infty}\frac{6x}{2x+5}

Solving directly using the same method above and applying Theorem 2, we get

\begin{eqnarray} && \lim_{x \to -\infty}\frac{6x}{2x+5} \\ && = \lim_{x \to -\infty}\frac{\frac{6x}{x}}{2+\frac{5}{x}} \\ && = \frac{ \lim_{x \to -\infty}\frac{6x}{x}}{ \lim_{x \to -\infty}\left (2+\frac{5}{x}\right )} \\ && = \frac{6+0}{2+0} \\ && = 3 \end{eqnarray}

Therefore, by virtue of Theorem 2, the limit of the function is

\lim_{x \to -\infty}\frac{6x}{2x+5}=3

Example #3

Evaluate the following limits.

\lim_{x \to -\infty}\frac{k\cos^{2}{x}+k\sin^{2}{x}}{x}

Well, to solve for this one, we only have to remember that \cos^{2}{x}+\sin^{2}{x} = 1, thus

\begin{eqnarray} && \lim_{x \to -\infty}\frac{k\cos^{2}{x}+k\sin^{2}{x}}{x} \\ && = k\lim_{x \to -\infty}\frac{\cos^{2}{x}+\sin^{2}{x}}{x} \\ && = k \lim_{x \to -\infty}\frac{1}{x} \\ && = k\cdot 0 \\ && = 0. \end{eqnarray}

Thus, we can say that

\lim_{x \to -\infty}\frac{k\cos^{2}{x}+k\sin^{2}{x}}{x}=0

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