Indeterminate Forms

The limit of an expression involving multiple functions can often be evaluated by taking the limits of these functions separately. For instance, if \( \lim\limits_{x\to 2} f(x) = 1 \) and \( \lim\limits_{x\to 2} g(x) = 3 \), then \(\lim\limits_{x\to 2} \big(f(x)+g(x)\big) = 1+3 = 4 \). An indeterminate form is an expression involving two functions whose limit cannot be determined solely from the limits of the individual functions. These forms are common in calculus; indeed, the limit definition of the derivative is the limit of an indeterminate form.

If \( f(x) = \sin(2x),\) then find \( f'(0).\)

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By the limit definition of the derivative,

\[f'(0) = \lim_{h\to 0} \frac{f(0+h)-f(0)}{h} = \lim_{h\to 0} \frac{\sin(2h)-\sin(0)}{h} = \lim_{h \to 0} \frac{\sin(2h)}{h}.\]

The limit of a quotient of functions can often be computed by taking the quotient of the limits, but in this case, the limits of the top and bottom functions are both \(0.\) The expression \( \frac00\) is not meaningful, so computing the limit requires another technique. \((\)In fact the answer is 2—see the wiki on derivatives of trigonometric functions.\()\) \(_\square\)

The most common indeterminate forms stem from evaluating limits of a ratio of functions \(\frac{f(x)}{g(x)}\), namely \(\frac{0}{0}\) and \(\frac{\infty}{\infty}\). The notation is shorthand for a limit of \( \frac{f(x)}{g(x)}\) where the limits of \( f(x) \) and of \( g(x) \) are both \(0 \) or both \( \infty\), respectively.

An indeterminate form \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) can have limit equal to any real number, or the limit may not exist. "Canceling" or other improper manipulations can lead to incorrect answers; see the Common Misconceptions wiki for examples.

For example, these limits are both of the form \( \frac{0}{0} \):

\[ \begin{align} \lim_{x\to 0} \dfrac{ax}{x} &= a \ \ (a\in {\mathbb R})\\\\ \lim_{x\to 0} \dfrac{x\sin\left(\frac1x\right)}{x} &= \text{DNE}. \end{align} \]

Computing these limits, in general, is the fundamental problem of differential calculus since, as noted above, the derivative

\[ f'(x) = \lim_{h\to 0} \dfrac{f(x+h)-f(x)}{h} \]

is the limit of an indeterminate form \( \frac{0}{0} \). If \( f(x) \) and \( g(x)\) are differentiable and \( \frac{f(x)}{g(x)} \) is one of these indeterminate forms, its limit can often be simplified using L'Hopital's rule.

Product: The form \( 0 \cdot \infty \) is indeterminate. (So "0 times anything is 0" does not apply!) It can be converted to the quotient form by changing \( f(x) \) to \(\frac1{\hspace{2mm} \frac{1}{f(x)}\hspace{2mm} } \).

Find \( \lim\limits_{x\to 0^+} x\ln(x). \)

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This is an indeterminate form \( 0 \cdot (-\infty)\). (Some lists classify this as a different form from \( 0 \cdot \infty\), but there is no difference in the techniques used to evaluate the limit.)

Rewrite it as

\[ \begin{align} \lim_{x\to 0^+} x\ln(x) &= \lim_{x\to 0^+} \dfrac{\ln(x)}{\frac1x} &&\left(\text{of the form} \dfrac{-\infty}{\infty}\right) \\ &= \lim_{x\to 0^+} \dfrac{\frac1x}{\hspace{1mm} -\frac1{x^2}\hspace{1mm} } &&\text{(L'Hopital)} \\ &= \lim_{x\to 0^+} (-x) \\&= 0.\ _\square \end{align} \]

Subtraction: The form \( \infty - \infty \) is indeterminate. Again, the general strategy for computing limits of this form is to convert to an indeterminate quotient.

Compute \( \lim\limits_{x\to\infty} \left(\sqrt{x^2+3x+7}-x\right) \).

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Multiply by the "conjugate" to obtain

\[ \begin{align} \lim_{x\to\infty} \left(\sqrt{x^2+3x+7}-x\right)\left( \dfrac{\sqrt{x^2+3x+7}+x}{\sqrt{x^2+3x+7}+x} \right)&= \lim_{x\to\infty} \dfrac{x^2+3x+7-x^2}{\sqrt{x^2+3x+7}+x} \\ &= \lim_{x\to\infty} \dfrac{3x+7}{\sqrt{x^2+3x+7}+x} \\ &= \lim_{x\to\infty} \dfrac{3+\dfrac7x}{\sqrt{1+\dfrac3x+\dfrac{7}{x^2}}+1} \\ &= \dfrac3{\sqrt{1}+1} \\ &= \dfrac32.\ _\square \end{align} \]

Exponential: There are three of these: \( 0^0, \infty^0, 1^\infty \). The first of these is a common misconception, since \( 0^0 = 1 \) in many contexts. But, for instance, \(\lim\limits_{x\to 0} 0^x = 0\), or more exotically, \( \lim\limits_{x\to 0^+} (a^{1/x})^x = a \); this is of the form \( 0^0\) if \( 0 \le a < 1 \).

The strategy for evaluating exponential limits of the above types is to let \( y \) be the function which gives the indeterminate form and then to find the limit of \( \ln(y) \). This turns the problem into the limit of a function with indeterminate product form \( 0 \cdot \infty\).

Find \( \lim\limits_{x\to \infty} \left( 1+\dfrac4{x} \right)^x. \)

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This is an indeterminate form of type \( 1^\infty \). Let \( y = \left( 1+\frac4{x} \right)^x \). Then

\[\begin{align} \lim_{x\to\infty} \ln(y) &= \lim_{x\to\infty} \ln\left( 1+\frac4{x} \right)^x \\ &= \lim_{x\to\infty} x \ln\left( 1+\frac4{x}\right) \\ &= \lim_{x\to\infty} \frac{\ln\left( 1+\frac4{x}\right)}{\frac1x} \\ &= \lim_{x\to\infty} \frac{\frac1{1+\frac4{x}} \left(\frac{-4}{x^2} \right)}{-\frac{1}{x^2}} &&(\text{L'Hopital}) \\ &= \lim_{x\to\infty} \frac4{1+\frac4{x}} \\ &= 4. \end{align}\]

So the original limit is \( e^4 \). \(_\square\)

Other combinations of functions lead to limits that can be determined (possibly with some information about signs—see below) just from the value of the component limits.

Quotient: The fractions \( \frac0{\infty} \) and \( \frac1{\infty} \) are not indeterminate; the limit is \( 0 \).

The fractions \( \frac10\) and \( \frac{\infty}0\) are not indeterminate. If the denominator is positive, the limit is \( \infty \). If the denominator is negative, the limit is \( -\infty \). If the denominator takes both positive and negative values in any neighborhood of the point where the limit is being taken, the limit does not exist.

\[ \begin{align} \lim_{x\to 0^+} \dfrac1{x} &= \infty \\ \lim_{x\to 0^-} \dfrac1{x} &= -\infty \\ \lim_{x\to 0} \dfrac1{x} &= \text{DNE} \end{align} \]

Product: \( \infty \cdot \infty \) is not indeterminate; the limit is \( \infty \).

Exponential: \( 0^\infty \) and \( \infty^\infty \) are not indeterminate; the limits are \( 0 \) and \( \infty \), respectively.

Similarly, \( 0^{-\infty} \) and \( \infty^{-\infty} \) are not indeterminate; the limits are \( \infty \) and \( 0 \), respectively. (In all cases the assumption is that the base of the exponential is a nonnegative function; otherwise the exponential itself is undefined in general.)