Evaluation of ∞ - ∞ Form

Introduction

Consider the limit

\[ \lim_{x \to a} \bigl[f(x) - g(x)\bigr] \]

under the hypothesis that \(\lim_{x \to a} f(x) = +\infty\) and \(\lim_{x \to a} g(x) = +\infty.\) In this setting, the expression

\[ f(x) - g(x) \]

belongs to the indeterminate form often called \(\infty - \infty\).

One way to see why \(\infty - \infty\) is indeterminate is to transform the difference into a fraction. Observe that:

\[ f(x) - g(x) \;=\; \frac{1}{\,1/f(x)\!} \;-\; \frac{1}{\,1/g(x)\!}. \]

If \(f(x)\) and \(g(x)\) both grow without bound, then \(1/f(x)\) and \(1/g(x)\) both shrink to \(0\). Thus we are looking at

\[ \frac{1}{\,1/f(x)\!} \;-\; \frac{1}{\,1/g(x)\!} \;=\; \frac{\tfrac{1}{g(x)} \;-\; \tfrac{1}{f(x)}}{\tfrac{1}{f(x)} \cdot \tfrac{1}{g(x)}}. \]

Since each of \(\tfrac{1}{f(x)}\) and \(\tfrac{1}{g(x)}\) approaches \(0\), the numerator \(\tfrac{1}{g(x)} - \tfrac{1}{f(x)}\) is of the form \(0 - 0\), and the denominator \(\tfrac{1}{f(x)} \cdot \tfrac{1}{g(x)}\) is also tending to \(0\). Thus we have a \(\tfrac{0}{0}\) indeterminate form, which can be analyzed by l’Hôpital’s Rule or by other techniques suitable for \(0/0\) limits.

Intuitively, we can understand this indeterminate form in this way: As \(x\to a\) \(f\) and \(g\) each become arbitrarily large, but it remains unclear how their rates of growth compare. Depending on which function grows faster, the difference might tend to some finite limit, \(+\infty\), \(-\infty\), or fail to have a limit altogether.

Resolution of ∞ - ∞ Form

A central principle for resolving such expressions is to rewrite or combine \(f\) and \(g\) in such a way that a more familiar indeterminate form—such as \(\tfrac{0}{0}\) or \(\tfrac{\infty}{\infty}\)—emerges. Often, one does this by manipulating the difference into a single fraction, factoring out a dominant term, or using any properties that \(f\) and \(g\) might share. The precise steps will vary according to the structure of the given functions, but the goal is always to convert the problem into something like

\[ \frac{0}{0}, \quad \frac{\infty}{\infty}, \quad 0 \times \infty, \]

One frequent strategy is to introduce a common denominator (effectively taking an “LCM” if \(f\) and \(g\) are fractions) so that \(f(x) - g(x)\) is reorganized into a single quotient. If \(f\) and \(g\) are not initially fractional, one might factor out the main growth behavior from each. For instance, if \(f\) and \(g\) each contain a factor that grows quickly, extracting that factor can reveal whether the difference remains bounded or diverges.

Example

Evaluate

\[ \lim_{x \to \infty} \Bigl[\sqrt{x^2 + 3x} \;-\; \sqrt{x^2 - 1}\Bigr]. \]

Solution. Both \(\sqrt{x^2 + 3x}\) and \(\sqrt{x^2 - 1}\) grow without bound as \(x \to \infty\), so the expression \(\sqrt{x^2 + 3x} - \sqrt{x^2 - 1}\) is an \(\infty - \infty\) form. We resolve it by rationalizing and then factoring out \(x\):

\[ \lim_{x \to \infty} \Bigl[\sqrt{x^2 + 3x} - \sqrt{x^2 - 1}\Bigr] \;=\; \lim_{x \to \infty} \left[ \frac{\bigl(\sqrt{x^2 + 3x} - \sqrt{x^2 - 1}\bigr)\,\bigl(\sqrt{x^2 + 3x} + \sqrt{x^2 - 1}\bigr)} {\sqrt{x^2 + 3x} + \sqrt{x^2 - 1}} \right] \]

\[ =\; \lim_{x \to \infty} \left[ \frac{(x^2 + 3x) - (x^2 - 1)} {\sqrt{x^2 + 3x} + \sqrt{x^2 - 1}} \right] \;=\; \lim_{x \to \infty} \left[ \frac{3x + 1} {\sqrt{x^2 + 3x} + \sqrt{x^2 - 1}} \right]. \]

To make the limit clearer, we divide numerator and denominator by \(x\):

\[ =\; \lim_{x \to \infty} \left[ \frac{\tfrac{3x + 1}{x}} {\tfrac{\sqrt{x^2 + 3x}}{x} + \tfrac{\sqrt{x^2 - 1}}{x}} \right] \;=\; \lim_{x \to \infty} \left[ \frac{3 + \tfrac{1}{x}} {\sqrt{1 + \tfrac{3}{x}} + \sqrt{1 - \tfrac{1}{x^2}}} \right]. \]

As \(x \to \infty\), \(\tfrac{1}{x} \to 0\) and \(\tfrac{1}{x^2} \to 0\). Therefore,

\[ =\; \frac{3 + 0}{\sqrt{1 + 0} + \sqrt{1 - 0}} \;=\; \frac{3}{1 + 1} \;=\; \frac{3}{2}. \]

Hence,

\[ \boxed{\lim_{x \to \infty} \Bigl[\sqrt{x^2 + 3x} \;-\; \sqrt{x^2 - 1}\Bigr] = \frac{3}{2}.} \]

Example

Evaluate

\[ \lim_{x \to \infty} \bigl[\ln(x+1) \;-\; \ln(x)\bigr]. \]

Solution. As \(x \to \infty\), both \(\ln(x+1)\) and \(\ln(x)\) individually grow without bound. Thus the expression \(\ln(x+1) - \ln(x)\) is an \(\infty - \infty\) form. To resolve this, we use a standard logarithm identity to combine the difference into a single logarithm:

\[ \ln(x+1) \;-\; \ln(x) \;=\; \ln\!\Bigl(\frac{x+1}{x}\Bigr). \]

As \(x\) becomes large, \(\tfrac{x+1}{x} = 1 + \tfrac{1}{x}\) approaches \(1\). Hence,

\[ \lim_{x \to \infty} \ln\!\Bigl(\tfrac{x+1}{x}\Bigr) \;=\; \ln\!\Bigl(\lim_{x \to \infty} \bigl(1 + \tfrac{1}{x}\bigr)\Bigr) \;=\; \ln(1) \;=\; 0. \]

Therefore,

\[ \boxed{\lim_{x \to \infty} \bigl[\ln(x+1) - \ln(x)\bigr] = 0.} \]

Example

Evaluate the limit

\[ \lim_{x \to 0}\Bigl(\frac{1}{x^2 + x} \;-\; \frac{1}{\sin x}\Bigr). \]

Solution:

Consider the limit

\[ \lim_{x \to 0}\Bigl(\frac{1}{x^2 + x} \;-\; \frac{1}{\sin x}\Bigr). \]

Because \(x^2 + x \to 0\) and \(\sin x \to 0\) as \(x \to 0\), each term individually tends to infinity, creating the \(\infty - \infty\) form. To resolve this, combine the two fractions into a single quotient inside the limit:

\[ \lim_{x \to 0} \Bigl(\frac{1}{x^2 + x} - \frac{1}{\sin x}\Bigr) = \lim_{x \to 0} \frac{\sin x - (x^2 + x)}{(x^2 + x)\,\sin x}. \]

Next, expand \(\sin x\) near \(0\) using the series \(\sin x = x - \tfrac{x^3}{6} + \text{higher-order terms}.\) Substituting this into the numerator gives

\[ \sin x - (x^2 + x) = \Bigl(x - \tfrac{x^3}{6} + \cdots\Bigr) - (x + x^2) = x - x - x^2 - \tfrac{x^3}{6} + \cdots = -\,x^2 - \tfrac{x^3}{6} + \cdots. \]

In the denominator, \((x^2 + x) \sin x\) expands as \((x^2 + x)\,\bigl(x - \tfrac{x^3}{3!} + \cdots\bigr)\). The leading behavior comes from \(x \cdot x = x^2\), so we write

\[ (x^2 + x)\,\sin x = x^2 \bigl(x - \tfrac{x^3}{6} + \cdots\bigr) + x \bigl(x - \tfrac{x^3}{6} + \cdots\bigr) = x^3 + x^2 + \cdots = x^2 \bigl(1 + x + \cdots\bigr). \]

Thus, inside the limit we have

\[ \frac{-\,x^2 - \tfrac{x^3}{6} + \cdots}{x^2 \bigl(1 + x + \cdots\bigr)}. \]

Factoring out \(x^2\) from numerator and denominator,

\[ \frac{x^2\bigl(-1 - \tfrac{x}{6} + \cdots\bigr)} {x^2\bigl(1 + x + \cdots\bigr)} = \frac{-1 - \tfrac{x}{6} + \cdots}{1 + x + \cdots}. \]

As \(x \to 0\), the higher-order terms vanish, so the ratio goes to \(-1/1 = -1\). Therefore the limit is

\[ \lim_{x \to 0}\Bigl(\frac{1}{x^2 + x} - \frac{1}{\sin x}\Bigr) = -1. \]

Example

Evaluate the limit

\[ \lim_{x \to \infty}\,\biggl(\frac{x^2}{\,x+1\,} \;-\; \frac{x^2 + 3x}{\,x+2\,}\biggr). \]

Solution:

As \(x \to \infty,\) each term \(\tfrac{x^2}{x+1}\) and \(\tfrac{x^2 + 3x}{x+2}\) is a rational function whose numerator has higher degree than its denominator, so each term separately approaches \(\infty.\) This creates an \(\infty - \infty\) form. To resolve this, we combine them into a single fraction inside the limit operator:

\[ \lim_{x \to \infty} \Bigl(\frac{x^2}{\,x+1\,} - \frac{x^2 + 3x}{\,x+2\,}\Bigr) = \lim_{x \to \infty} \frac{x^2(x+2)\;-\;(x^2 + 3x)\,(x+1)}{(x+1)\,(x+2)}. \]

We expand the numerator directly in that limit expression:

\[ = \lim_{x \to \infty} \frac{x^3 + 2x^2 \;-\;\bigl(x^3 + x^2 + 3x^2 + 3x\bigr)}{(x+1)(x+2)} = \lim_{x \to \infty} \frac{x^3 + 2x^2 \;-\;(x^3 + 4x^2 + 3x)}{(x+1)(x+2)} = \lim_{x \to \infty} \frac{x^3 + 2x^2 - x^3 - 4x^2 - 3x}{(x+1)(x+2)} = \lim_{x \to \infty} \frac{-\,2x^2 - 3x}{(x+1)(x+2)}. \]

Next, factor out \(x^2\) from both numerator and denominator inside the limit:

\[ = \lim_{x \to \infty} \frac{x^2\bigl(-2 - \tfrac{3}{x}\bigr)}{(x+1)(x+2)} = \lim_{x \to \infty} \frac{x^2\bigl(-2 - \tfrac{3}{x}\bigr)} {x^2\bigl(\tfrac{x+1}{x}\bigr)\bigl(\tfrac{x+2}{x}\bigr)} = \lim_{x \to \infty} \frac{-2 - \tfrac{3}{x}} {\bigl(1 + \tfrac{1}{x}\bigr)\bigl(1 + \tfrac{2}{x}\bigr)}. \]

As \(x \to \infty,\) the terms \(\tfrac{3}{x}, \tfrac{1}{x},\) and \(\tfrac{2}{x}\) each tend to \(0\). Hence the fraction approaches

\[ \frac{-2}{1 \cdot 1} = -2. \]

Therefore,

\[ \lim_{x \to \infty}\,\biggl(\frac{x^2}{\,x+1\,} \;-\; \frac{x^2 + 3x}{\,x+2\,}\biggr) = -2. \]