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 $\frac{0}{0}$:

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

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

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

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

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

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

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

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.

$$\begin{aligned} \lim_{x\to 0^+} \dfrac1{x} &= \infty \\ \lim_{x\to 0^-} \dfrac1{x} &= -\infty \\ \lim_{x\to 0} \dfrac1{x} &= \text{DNE} \end{aligned}$$

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.)