Indeterminate Forms
You type $\dfrac{x^2-9}{x-3}$ into your graphing calculator and ask for the value at $x=3$. It answers ERROR, because it is trying to compute $\frac00$. So does the limit as $x \to 3$ exist? Is it $0$? Is it undefined? In this page you learn to tell, and to find the answer when plugging in gives nothing useful.
Direct substitution: what can happen
On the AP exam, the first step for any limit is always the same: substitute the value ($x=a$, or think about $x\to\pm\infty$) and see what you get. There are only three kinds of outcome.
| Substitution gives | What it means |
|---|---|
| A number | That number is the limit (the function is continuous there). |
| $\dfrac{k}{0}$ with $k\neq 0$ | Not indeterminate. The function is unbounded, so check the one-sided limits for the sign. |
| $\dfrac00,\ \dfrac{\infty}{\infty},\ \infty-\infty,\ 1^\infty$ | Indeterminate. You cannot conclude anything until you rewrite the expression. |
The $\dfrac{k}{0}$ case and one-sided limits
When the top tends to a nonzero number and the bottom tends to $0$, the function blows up. The sign depends on the side you approach from.
Observe the following graph of the function $f(x)$ and determine, intuitively, the value of the indicated limits. If any limit does not exist or is infinite, specify…
Observe the following graph of the function $f(x)$ and determine, intuitively, the value of the indicated limits. If any limit does not exist or is infinite, specify this clearly.
- a) $\lim_{x \to -2} f(x)$
- b) $\lim_{x \to 0} f(x)$
- c) $\lim_{x \to 3} f(x)$
- d) $\lim_{x \to +\infty} f(x)$
Given the rational function $f(x) = \frac{2}{x-4}$, study its behavior in the vicinity of its vertical asymptote by calculating the one-sided limits: a)…
Given the rational function $f(x) = \frac{2}{x-4}$, study its behavior in the vicinity of its vertical asymptote by calculating the one-sided limits:
- a) $\lim_{x \to 4^-} \frac{2}{x-4}$
- b) $\lim_{x \to 4^+} \frac{2}{x-4}$
Indicate with reasoning whether the overall limit exists at $x = 4$.
$\frac00$ by factoring and common denominators
If direct substitution gives $\frac00$ in a rational expression, the numerator and denominator share a factor $(x-a)$. Factor, cancel it, then substitute again.
Why is canceling legal? A limit only looks at $x$ near $a$, never at $x=a$ itself. For $x\neq a$ the factor $\frac{x-a}{x-a}$ is exactly $1$.
The same idea works when the indeterminate form is $\infty-\infty$ at a finite point: combine the fractions into one, then factor.
Calculate the limit of the function $f(x) = \frac{x^2 - 9}{x - 3}$ when $x$ tends to $3$. Hint: Note that by substituting directly we obtain an indeterminate form of…
Calculate the limit of the function $f(x) = \frac{x^2 - 9}{x - 3}$ when $x$ tends to $3$.
Hint: Note that by substituting directly we obtain an indeterminate form of type $\left[\frac{0}{0}\right]$. Try to factorize the numerator to simplify the expression.
Resolve the following indeterminate form of type $\infty - \infty$. To do this, perform the prior algebraic operation and simplify the resulting expression: $$\lim_{x…
Resolve the following indeterminate form of type $\infty - \infty$. To do this, perform the prior algebraic operation and simplify the resulting expression:
$$\lim_{x \to 1} \left( \frac{6}{x^2 - 1} - \frac{3}{x - 1} \right)$$$\frac00$ with square roots: conjugates
When a square root appears in a $\frac00$ limit, factoring will not work. Multiply the top and bottom by the conjugate (change the sign between the two terms). It uses $(A-B)(A+B)=A^2-B^2$, which makes the root disappear and usually creates the factor $(x-a)$ you need to cancel.
When there are roots in both the numerator and the denominator, multiply by the conjugate of each one (some books call this the double conjugate). Keep the products in factored form so you can see what cancels.
Here is a version with a real-world setting. The pressure in a ventilation duct is $P(v)=\dfrac{v^2-9}{\sqrt{v^2+7}-4}$ pascals, where $v$ is the airflow speed in ft/s. What happens as $v\to3$?
Calculate the following limit by resolving the indeterminate form of type $\frac{0}{0}$ using the conjugate technique: $$\lim_{x \to 4} \frac{x^2 - 16}{\sqrt{x+5} - 3}$$
Calculate the following limit by resolving the indeterminate form of type $\frac{0}{0}$ using the conjugate technique:
$$\lim_{x \to 4} \frac{x^2 - 16}{\sqrt{x+5} - 3}$$Solve the following limit using the double conjugate technique to resolve the indeterminate form $\frac{0}{0}$: $$\lim_{x \to 0} \frac{\sqrt{1+x} -…
Solve the following limit using the double conjugate technique to resolve the indeterminate form $\frac{0}{0}$:
$$\lim_{x \to 0} \frac{\sqrt{1+x} - \sqrt{1-x}}{\sqrt{4+x} - 2}$$A team of aerospace engineers is studying the pressure $P$ (in pascals) inside a ventilation duct. The pressure depends on the airflow speed $v$ according to the…
A team of aerospace engineers is studying the pressure $P$ (in pascals) inside a ventilation duct. The pressure depends on the airflow speed $v$ according to the formula:
$$P(v) = \frac{v^2 - 9}{\sqrt{v^2 + 7} - 4}$$Study the behavior of the pressure when the speed $v$ approaches $3$ ft/s. If the limit is indeterminate, resolve it and indicate whether the pressure tends to a stable value.
Piecewise functions and one-sided limits
Many $\frac00$ problems come with a function defined in pieces, or with absolute values, an exponential like $e^{1/(x-a)}$, or a sine. In all of these, compute the left and right limits separately.
Here is the graph of $f(x)=\begin{cases}\dfrac{x^2-4}{x-2}&\text{if } x<2\\[2mm] ax-1&\text{if } x\ge2\end{cases}$ drawn with $a=1$.
On the left, $\frac{x^2-4}{x-2}=x+2$ (with a hole at $x=2$), which approaches $4$. On the right, with $a=1$, the line $x-1$ starts at height $1$. The sides disagree (a jump), so there is no limit at $x=2$. To make the pieces meet, change $a$ so that the right piece also reaches $4$.
The next example adds a square root and a sine, and asks you to classify the discontinuity.
The next one combines an absolute value with an exponential.
Given the following graph of the function $f(x)$, visually analyze and analytically calculate the one-sided limits at the points of discontinuity to determine if the…
Given the following graph of the function $f(x)$, visually analyze and analytically calculate the one-sided limits at the points of discontinuity to determine if the overall limit exists at $x = 2$:
$$f(x) = \begin{cases} \frac{x^2 - 4}{x - 2} & \text{if } x < 2 \\ ax - 1 & \text{if } x \geq 2 \end{cases}$$Calculate the value of $a$ so that the limit at $x = 2$ is unique and the function does not present a jump.
Consider the piecewise function: $$f(x) = \begin{cases} \frac{x^2 - kx}{x - \sqrt{x}} & \text{if } x > 1 \\ \frac{a \cdot \sin(x-1)}{x^2 - 1} & \text{if } x < 1…
Consider the piecewise function:
$$f(x) = \begin{cases} \frac{x^2 - kx}{x - \sqrt{x}} & \text{if } x > 1 \\ \frac{a \cdot \sin(x-1)}{x^2 - 1} & \text{if } x < 1 \end{cases}$$Find the values of $a$ and $k$ so that the one-sided limits exist at $x=1$ and are equal to $2$. What type of discontinuity does the function have at $x=1$ if we do not define $f(1)$?
Given the function $g(x) = \frac{|x-1|}{x^2-1} \cdot e^{\frac{1}{x-1}}$, analyze the existence of the one-sided limits when $x \to 1$. Identify the indeterminate forms…
Given the function $g(x) = \frac{|x-1|}{x^2-1} \cdot e^{\frac{1}{x-1}}$, analyze the existence of the one-sided limits when $x \to 1$. Identify the indeterminate forms that appear on each side and classify the type of discontinuity at that point.
$\frac{\infty}{\infty}$: polynomials and rational functions
As $x\to\pm\infty$, a polynomial behaves like its leading term. In a quotient, divide the top and bottom by the highest power of $x$ in the denominator: every lower-degree term then goes to $0$.
Applied limits at infinity are the same. A bacteria population $P(t)=\dfrac{20t+50}{t+2}$ (in millions, $t$ in hours) has equal degrees, so $\lim_{t\to+\infty}P(t)=\dfrac{20}{1}=20$ million: the long-run value of the population.
Calculate the following limits of polynomial functions when the variable $x$ tends to infinity. Remember to identify the dominant term in each case: a)…
Calculate the following limits of polynomial functions when the variable $x$ tends to infinity. Remember to identify the dominant term in each case:
- a) $\lim_{x \to +\infty} (-2x^3 + 5x^2 - 1)$
- b) $\lim_{x \to -\infty} (x^4 - 100x^2)$
- c) $\lim_{x \to +\infty} \frac{3x^2 - 5}{10}$
Resolve the indeterminate form of type $\left[\frac{\infty}{\infty}\right]$ in the following limits of rational functions, comparing the degrees of the numerator and the…
Resolve the indeterminate form of type $\left[\frac{\infty}{\infty}\right]$ in the following limits of rational functions, comparing the degrees of the numerator and the denominator:
- a) $\lim_{x \to +\infty} \frac{5x^2 - 3x + 2}{2x^2 + 1}$
- b) $\lim_{x \to +\infty} \frac{x + 5}{x^2 - 3}$
- c) $\lim_{x \to +\infty} \frac{x^3 - 1}{x^2 + 10x}$
A team of scientists is monitoring the growth of a bacteria population in a laboratory. The population (in millions) follows the model given by the function: $$P(t) =…
A team of scientists is monitoring the growth of a bacteria population in a laboratory. The population (in millions) follows the model given by the function:
$$P(t) = \frac{20t + 50}{t + 2}$$where $t$ is the time in hours since the start of the experiment.
What will the bacteria population be if the experiment is prolonged indefinitely? To do this, calculate $\lim_{t \to +\infty} P(t)$.
$\frac{\infty}{\infty}$ with radicals
A root of a polynomial also has a "leading behavior", but be careful with signs:
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Square root: $\sqrt{ax^2+\dots}\approx\sqrt a\,|x|$, because $\sqrt{x^2}=|x|$. For $x\to+\infty$, $|x|=x$. For $x\to-\infty$, $|x|=-x$.
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Odd roots (cube roots): $\sqrt[3]{x^3+\dots}\approx x$ with no absolute value, since $\sqrt[3]{x^3}=x$ for every $x$.
To handle it, factor out the highest power inside the root, or divide everything by $x$ (taking care of $|x|$).
When the top has a root plus an ordinary term and the bottom has a cube root, divide everything by $x$ and treat each piece separately.
Determine the value of the limit at infinity by analysing the degree of the dominant terms in the numerator and the denominator. Pay special attention to the powers…
Determine the value of the limit at infinity by analysing the degree of the dominant terms in the numerator and the denominator. Pay special attention to the powers inside the roots:
$$\lim_{x \to +\infty} \frac{\sqrt{9x^2 + 2x - 1} + 2x}{\sqrt[3]{x^3 + 5x^2}}$$Comparing growth: the hierarchy of infinities
When a limit mixes logarithms, powers and exponentials, there is no need to divide by anything: just find which term in each part grows fastest.
In the study of computer algorithms, it is crucial to know which function grows faster. Based on the hierarchy of infinities, assign the value of the following limits…
In the study of computer algorithms, it is crucial to know which function grows faster. Based on the hierarchy of infinities, assign the value of the following limits ($0$, $+\infty$ or $-\infty$):
- a) $\lim_{x \to +\infty} \frac{\ln(x)}{x}$
- b) $\lim_{x \to +\infty} \frac{e^x}{x^{10}}$
- c) $\lim_{x \to +\infty} \frac{x^2 + 1}{\ln(x)}$
Use the comparison of orders of infinity to solve the following limits without the need for extensive operations. Justify your answer by indicating which function…
Use the comparison of orders of infinity to solve the following limits without the need for extensive operations. Justify your answer by indicating which function "dominates" the other:
- a) $\lim_{x \to +\infty} \frac{e^x + x^5}{100x^6 - \ln(x)}$
- b) $\lim_{x \to +\infty} \frac{\sqrt{x^7 + 1}}{x^3 \cdot \ln(x)}$
Calculate the value of the following limit by analysing the hierarchy of the infinities involved. You must justify which term dominates in the numerator and in the…
Calculate the value of the following limit by analysing the hierarchy of the infinities involved. You must justify which term dominates in the numerator and in the denominator before giving the final result:
$$\lim_{x \to +\infty} \frac{3 \cdot e^{2x} - \ln(x^5 + 1) + x^{10}}{2 \cdot 5^x + \sqrt{x^{20} - 3x^2} - e^x}$$$\infty-\infty$ at infinity
Two huge quantities with the same sign are subtracted. Get rid of the subtraction:
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With square roots: multiply by the conjugate, then divide by the highest power of $x$.
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With cube roots: use $A-B=\dfrac{A^3-B^3}{A^2+AB+B^2}$, which removes the cube root in the numerator.
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With roots of different index: add and subtract an intermediate term (usually $x$) so you can split the limit into two pieces, each with only one kind of root.
Calculate the value of the following expression which presents an indeterminate form of type $\infty - \infty$ with roots of different orders: $$\lim_{x \to +\infty}…
Calculate the value of the following expression which presents an indeterminate form of type $\infty - \infty$ with roots of different orders:
$$\lim_{x \to +\infty} \left[ \sqrt{x^2 + 4x} - \sqrt[3]{x^3 - 6x^2} \right]$$Hint: Add and subtract a suitable intermediate term to separate the limit into two manageable parts.
The number $e$ and $1^\infty$
Here the base $\to1$ and the exponent $\to\pm\infty$. Why is it not just $1$? Try $\left(1+\tfrac1{1000}\right)^{1000}\approx2.7169$: the base is almost $1$, but it is multiplied by itself $1000$ times. The competition between the two tendencies gives the number $e$.
When a sine is involved, as in $\displaystyle\lim_{x\to0}(1+\sin x)^{2/x}$, compute $L=\lim\dfrac2x\cdot\sin x=2\cdot1=2$ using $\frac{\sin x}{x}\to1$. The limit is $e^2$.
Calculate the value of the following limit which results in the indeterminate form of type $[1^\infty]$ using the abbreviated formula for the number $e$: $$\lim_{x \to…
Calculate the value of the following limit which results in the indeterminate form of type $[1^\infty]$ using the abbreviated formula for the number $e$:
$$\lim_{x \to +\infty} \left( 1 + \frac{1}{x} \right)^{3x}$$Calculate the value of the following exponential limit which results in an indeterminate form of type $1^\infty$. Express the result in terms of the number $e$:…
Calculate the value of the following exponential limit which results in an indeterminate form of type $1^\infty$. Express the result in terms of the number $e$:
$$\lim_{x \to 0} (1 + \sin(x))^{\frac{2}{x}}$$$1^\infty$ with rational bases
Problems like user growth or signal strength often hide a $1^\infty$ in a fraction raised to a power of $t$. First check the base tends to $1$ (same degree, same leading coefficient). Then compute $L$.
The base can also have degree $2$. Then $g-1$ is a quotient of polynomials and $L$ is a limit of a rational function.
The graph above is $U(t)=\left(\frac{2t+3}{2t-1}\right)^{t+1}$, a model for users of a new app in millions. Here $g-1=\dfrac{4}{2t-1}$ and $L=\lim\dfrac{4(t+1)}{2t-1}=2$. The curve climbs and flattens out near $e^2\approx7.39$ million users, exactly what $L$ predicts.
In the development of a new social network, the number of active users (in millions) follows the function $U(t) = \left(\frac{2t+3}{2t-1}\right)^{t+1}$, where $t$ is the…
In the development of a new social network, the number of active users (in millions) follows the function $U(t) = \left(\frac{2t+3}{2t-1}\right)^{t+1}$, where $t$ is the time in months since its launch. Calculate the value the number of users tends to when the time increases indefinitely ($t \to +\infty$).
In the design of a new telecommunications antenna, the signal intensity $I$ as a function of the distance $d$ to the repeater is given by the expression: $$I(d) = \left(…
In the design of a new telecommunications antenna, the signal intensity $I$ as a function of the distance $d$ to the repeater is given by the expression:
$$I(d) = \left( \frac{d^2 + 5d}{d^2 - 2d + 1} \right)^{2d + 3}$$Calculate the limiting intensity of the signal when the distance $d$ tends to infinity (indeterminate form of type $1^\infty$).
Finding parameters that make a limit work
Sometimes the unknowns are inside the limit: "Find $a$ and $b$ so that the limit is finite, or equals $4$." The idea is to figure out what must be true for the limit to be finite, and then use the result.
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At a finite point with a denominator that tends to $0$: the numerator must also tend to $0$. Otherwise you would get $\frac{k}{0}$ and the limit would be infinite.
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At infinity with $\infty-\infty$: use the conjugate and see how the leading terms behave.
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"Finite and nonzero": after you find the family of solutions, check that none of them makes the limit equal to $0$.
The graph shows $\sqrt{x^2+ax+1}-\sqrt{x^2+bx}$. Multiply by the conjugate: the numerator is $(a-b)x+1$ and the denominator behaves like $2x$, so the curve levels off at $\dfrac{a-b}{2}$. Move the sliders: for the limit to equal $4$ you need $a-b=8$, as in the starting position $a=8$, $b=0$.
Find the value of the real parameters $a$ and $b$ so that the following limit is finite and equals $4$: $$\lim_{x \to +\infty} \sqrt{x^2 + ax + 1} - \sqrt{x^2 + bx}$$
Find the value of the real parameters $a$ and $b$ so that the following limit is finite and equals $4$:
$$\lim_{x \to +\infty} \sqrt{x^2 + ax + 1} - \sqrt{x^2 + bx}$$Determine the values of the real parameters $a$ and $b$ so that the following limit is a finite real number not equal to zero. Once found, calculate the value of this…
Determine the values of the real parameters $a$ and $b$ so that the following limit is a finite real number not equal to zero. Once found, calculate the value of this limit:
$$\lim_{x \to 2} \frac{\sqrt{x^2 + ax + b} - 2}{x^2 - 5x + 6}$$