Calculating Limits
Type $\dfrac{x^2-25}{x-5}$ into your graphing calculator and evaluate it at $x=5$. You get an error, because the calculator hits $\frac{0}{0}$. Yet the graph is a smooth line heading straight toward the height $10$. How can a function have no value at a point and still be "going" somewhere? In this page you will learn to calculate limits exactly, without tables or guessing.
Direct substitution and infinite limits
Most of the time, finding a limit is easy: you just plug in the number.
Always try substitution first. Look at what you get:
| Substituting $x=a$ gives | What it means |
|---|---|
| A number | That number is the limit. Done. |
| $\dfrac{k}{0}$ with $k\neq 0$ | Not indeterminate. The values blow up to $+\infty$ or $-\infty$. Check the sign on each side. |
| $\dfrac{0}{0}$ (and other forms you'll meet below) | Indeterminate. Substituting tells you nothing yet, so you must rewrite the expression. |
For $\frac{k}{0}$, the sign matters. When $x$ is close to $a$, the denominator is a tiny number that can be positive or negative. Combine its sign with the sign of $k$ to decide between $+\infty$ and $-\infty$ on each side.
The temperature of an electronic component (in degrees Fahrenheit) as a function of time $t$ (in hours) is given by the expression $T(t) = \frac{3t^2 - 5t + 12}{t + 2}$.…
The temperature of an electronic component (in degrees Fahrenheit) as a function of time $t$ (in hours) is given by the expression $T(t) = \frac{3t^2 - 5t + 12}{t + 2}$. Calculate the limit of the temperature as $t$ approaches $3$ hours.
Analyze the one-sided limits of the function $f(x) = \frac{2x + 1}{x - 4}$ as $x$ tends to $4$ and determine whether the overall limit exists at that point.
Analyze the one-sided limits of the function $f(x) = \frac{2x + 1}{x - 4}$ as $x$ tends to $4$ and determine whether the overall limit exists at that point.
The speed of a test vehicle (in ft/s) as a function of time $t$ (in seconds) is defined piecewise as $v(t) = 3t + 2$ if $0 \le t < 4$, and $v(t) = t^2 - 2$ if $t \ge 4$.…
The speed of a test vehicle (in ft/s) as a function of time $t$ (in seconds) is defined piecewise as $v(t) = 3t + 2$ if $0 \le t < 4$, and $v(t) = t^2 - 2$ if $t \ge 4$. Determine using one-sided limits whether the speed function is continuous at time $t = 4$.
One-sided limits and piecewise functions
Two situations force you to work one side at a time: piecewise functions at the point where the formula changes, and absolute values, since $|x|=x$ when $x>0$ and $|x|=-x$ when $x<0$.
For example, $\dfrac{|x|}{x}$ equals $1$ for $x>0$ and $-1$ for $x<0$. Here is $f(x)=\dfrac{|x|}{x}+x^2$ near $0$:
The one-sided limits are $-1$ and $1$. They are different, so $\lim_{x\to0}f(x)$ does not exist.
A one-sided limit can also be infinite, and roots restrict which side makes sense. For $\displaystyle\lim_{x\to 3^-}\dfrac{1}{\sqrt{9-x^2}}$: when $x$ is slightly less than $3$, $9-x^2$ is a tiny positive number, its root is a tiny positive number, and the quotient goes to $+\infty$. From the right, $9-x^2<0$ and the function isn't even defined, so only the left-hand limit exists.
Analyze the existence of the limit at $x = 0$ of the function $f(x) = \frac{|x|}{x} + x^2$ by first calculating its one-sided limits.
Analyze the existence of the limit at $x = 0$ of the function $f(x) = \frac{|x|}{x} + x^2$ by first calculating its one-sided limits.
An electricity company sets its tariff $T(x)$ in dollars as a function of consumption $x$ in kWh. For consumptions less than $100$ kWh, the tariff is $T(x) = 0,15x + 5$.…
An electricity company sets its tariff $T(x)$ in dollars as a function of consumption $x$ in kWh. For consumptions less than $100$ kWh, the tariff is $T(x) = 0,15x + 5$. For consumptions greater than or equal to $100$ kWh, the tariff changes to $T(x) = \frac{ax^2 + 200}{x}$. Find the value of the parameter $a$ so that the limit of the tariff exists as the consumption approaches $100$ kWh.
The relativistic mass of a moving particle is given by $m(v) = \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}}$, where $m_0$ is the rest mass, $v$ is the speed of the particle…
The relativistic mass of a moving particle is given by $m(v) = \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}}$, where $m_0$ is the rest mass, $v$ is the speed of the particle and $c$ is the speed of light. Calculate the limit of the mass as the speed $v$ tends to the speed of light $c$ from the left ($v \to c^-$) and interpret the mathematical result in the physical context.
Indeterminate forms and factoring
Some forms hide the answer. If the pieces compete (one goes to $0$, the other to $\infty$, and so on), the result can be any number, $0$, or infinite.
Let's start with the most common one, $\frac{0}{0}$ with polynomials. If both the numerator and the denominator are $0$ at $x=a$, then by the Remainder Theorem both have a factor $(x-a)$. Factor it, cancel it, and substitute again.
Subtracting fractions. A difference like $\dfrac{1}{x-3}-\dfrac{6}{x^2-9}$ is of the type $\infty-\infty$ at $x=3$. Combine the fractions over a common denominator first. That usually produces a $\frac{0}{0}$ you can factor.
Unknown parameters. Suppose you are told $\displaystyle\lim_{x\to a}\dfrac{P(x)}{x-a}=L$ with $L$ a number. The denominator tends to $0$, so the numerator must also tend to $0$. Otherwise you would get $\frac{k}{0}$ and the limit would be infinite. This gives an equation for the unknowns.
Calculate the following algebraic limit by resolving the indeterminate form through factorization: $\lim_{x \to 5} \frac{x^2 - 25}{x^2 - 7x + 10}$
Calculate the following algebraic limit by resolving the indeterminate form through factorization:
$\lim_{x \to 5} \frac{x^2 - 25}{x^2 - 7x + 10}$
Calculate the value of the limit that presents an indeterminate form of the type subtraction of algebraic fractions:…
Calculate the value of the limit that presents an indeterminate form of the type subtraction of algebraic fractions:
$\lim_{x \to 2} \left( \frac{1}{x - 2} - \frac{4}{x^2 - 4} \right)$
Determine the values of the real parameters $a$ and $b$ so that the following limit condition is satisfied: $\lim_{x \to 1} \frac{x^2 + ax + b}{x^2 - 1} = 3$
Determine the values of the real parameters $a$ and $b$ so that the following limit condition is satisfied:
$\lim_{x \to 1} \frac{x^2 + ax + b}{x^2 - 1} = 3$
Limits with square roots: the conjugate
When a $\frac{0}{0}$ contains a square root, factoring does not help. Instead, multiply the numerator and the denominator by the conjugate. The conjugate of $\sqrt{A}-B$ is $\sqrt{A}+B$. Using $(u-v)(u+v)=u^2-v^2$, the root disappears from that part.
With two radicals, conjugate the numerator the same way.
Sometimes you only need to simplify first, and then check the sign of a one-sided limit. Look at $\displaystyle\lim_{x\to0^+}\dfrac{x\sqrt{25-x^2}}{x^2}$. Cancel one $x$ to get $\dfrac{\sqrt{25-x^2}}{x}$. The numerator tends to $5$ and the denominator tends to $0^+$, so the limit is $+\infty$.
Solve the following limit by multiplying the numerator and the denominator by the corresponding conjugate expression: $\lim_{x \to 3} \frac{\sqrt{x + 6} - 3}{x - 3}$
Solve the following limit by multiplying the numerator and the denominator by the corresponding conjugate expression:
$\lim_{x \to 3} \frac{\sqrt{x + 6} - 3}{x - 3}$
Solve the following limit with indeterminate form $\frac{0}{0}$ containing two radicals in the numerator: $\lim_{x \to 0} \frac{\sqrt{4 + x} - \sqrt{4 - x}}{3x}$
Solve the following limit with indeterminate form $\frac{0}{0}$ containing two radicals in the numerator:
$\lim_{x \to 0} \frac{\sqrt{4 + x} - \sqrt{4 - x}}{3x}$
Consider a rectangle inscribed in a circle of radius $R = 10$ inches. If $x$ is the length of the base of the rectangle, its area is given by the function…
Consider a rectangle inscribed in a circle of radius $R = 10$ inches. If $x$ is the length of the base of the rectangle, its area is given by the function $A(x) = x \sqrt{400 - x^2}$. An architect needs to know how the ratio between the area of the rectangle and the square of its base behaves when the base becomes very small. Calculate $\lim_{x \to 0^+} \frac{A(x)}{x^2}$.
Limits at infinity of rational functions
Now let $x$ get huge. For large $|x|$, in a polynomial the term with the highest power dominates everything else. For example, at $x=1000$, $-2x^3$ is $-2$ billion, while $50x^2$ is only $50$ million. So $\lim_{x\to+\infty}(-2x^3+50x^2+1000)=\lim_{x\to+\infty}(-2x^3)=-\infty$.
For a quotient of polynomials, divide the numerator and the denominator by the highest power of $x$ in the denominator. Every term $\frac{c}{x^n}$ then tends to $0$.
The number of bacteria in a culture follows the growth model $B(x) = -2x^3 + 50x^2 + 1000$, where $x$ is the number of days elapsed. Determine the limit of $B(x)$ as…
The number of bacteria in a culture follows the growth model $B(x) = -2x^3 + 50x^2 + 1000$, where $x$ is the number of days elapsed. Determine the limit of $B(x)$ as $x \to +\infty$ and state the long-term behavior of this culture according to the mathematical model.
The average production cost of $x$ thousand units of an item is given by the function $C(x) = \frac{4x^2 + 150x}{2x^2 + 5}$. Calculate the limit of the average cost as…
The average production cost of $x$ thousand units of an item is given by the function $C(x) = \frac{4x^2 + 150x}{2x^2 + 5}$. Calculate the limit of the average cost as the number of units produced tends to infinity.
Evaluate the limit of the following rational function as the independent variable tends to minus infinity: $\lim_{x \to -\infty} \frac{7x^4 - 3x^2 + 1}{5x^3 + 2x - 8}$
Evaluate the limit of the following rational function as the independent variable tends to minus infinity:
$\lim_{x \to -\infty} \frac{7x^4 - 3x^2 + 1}{5x^3 + 2x - 8}$
Radicals, parameters, and meaning at infinity
A radical behaves like its dominant term. For large positive $x$, $\sqrt{4x^2+x}\approx\sqrt{4x^2}=2x$. To be exact, divide by $x$ and move $x$ inside the root. For $x>0$, $\dfrac{\sqrt{4x^2+x}}{x}=\sqrt{4+\frac1x}$.
Unknown parameters at infinity. If a limit at infinity is given, use the rule for equal degrees and solve for the unknown.
Interpreting limits. A limit at infinity answers "what happens in the long run?" Take the caffeine concentration (in mg/L) in your blood $t$ hours after an energy drink, modeled by $C(t)=\dfrac{40t}{t^2+9}$. The numerator has degree $1$ and the denominator has degree $2$, so $\displaystyle\lim_{t\to+\infty}C(t)=0$. This means that as time passes, the caffeine is eventually cleared from the bloodstream. The concentration approaches, but never reaches, $0$.
Calculate the limit at infinity of the following function containing radicals, by analyzing the degrees of the numerator and the denominator:…
Calculate the limit at infinity of the following function containing radicals, by analyzing the degrees of the numerator and the denominator:
$\lim_{x \to +\infty} \frac{\sqrt{9x^2 + 4x} - 2x}{5x + 1}$
The concentration of a medicine in the bloodstream (in mg/L) after $t$ hours of its administration is given by $C(t) = \frac{25t}{t^2 + 4}$. If the patient does not take…
The concentration of a medicine in the bloodstream (in mg/L) after $t$ hours of its administration is given by $C(t) = \frac{25t}{t^2 + 4}$. If the patient does not take any additional doses, find the limit of the concentration as time tends to infinity and interpret the result.
Calculate the value of the parameter $k$ so that the following equality is satisfied in the limit: $\lim_{x \to +\infty} \frac{kx^3 - 5x + 1}{2x^3 + x^2 - 8} = 4$
Calculate the value of the parameter $k$ so that the following equality is satisfied in the limit:
$\lim_{x \to +\infty} \frac{kx^3 - 5x + 1}{2x^3 + x^2 - 8} = 4$
Exponentials at infinity
For an exponential $b^x$:
| Base | As $x\to+\infty$ | As $x\to-\infty$ |
|---|---|---|
| $b>1$ | $+\infty$ | $0$ |
$0| $0$ |
$+\infty$ |
|
Why? Multiplying again and again by $b>1$ makes things huge. Multiplying by $b<1$ shrinks them toward $0$. Adding a constant just shifts the end behavior: $3\left(\tfrac12\right)^x+2\to 2$ as $x\to+\infty$.
When several exponentials compete, the one with the largest base dominates. So divide the numerator and the denominator by the dominant exponential. Every ratio like $\left(\frac{2}{3}\right)^x$ then tends to $0$.
Calculate the limit of the exponential function $h(x) = 5 \cdot \left(\frac{2}{3}\right)^x + 8$ as $x$ tends to $+\infty$.
Calculate the limit of the exponential function $h(x) = 5 \cdot \left(\frac{2}{3}\right)^x + 8$ as $x$ tends to $+\infty$.
Calculate the following limit at infinity involving exponential functions with different bases: $\lim_{x \to +\infty} \frac{3^{x+1} - 5^x}{2^x + 5^{x-1}}$
Calculate the following limit at infinity involving exponential functions with different bases:
$\lim_{x \to +\infty} \frac{3^{x+1} - 5^x}{2^x + 5^{x-1}}$
The form ∞ − ∞ with square roots
Something like $\sqrt{x^2+8x}-x$ is $\infty-\infty$: both terms grow without bound. You can't tell the result by looking. The fix is again the conjugate, then divide by $x$.
Calculate the following limit which presents an indeterminate form of the type $\infty - \infty$ involving square roots:…
Calculate the following limit which presents an indeterminate form of the type $\infty - \infty$ involving square roots:
$\lim_{x \to +\infty} \left( \sqrt{x^2 + 6x} - x \right)$
Oblique asymptotes
If $f(x)$ gets closer and closer to a slanted line as $x\to\pm\infty$, that line is an oblique (slant) asymptote. This happens, for instance, when the numerator's degree is exactly one more than the denominator's.
Let the rational function be $f(x) = \frac{x^3 - 8}{x^2 - 4}$. Find the equation of the oblique asymptote line $y = mx + n$ of this function by rigorously calculating…
Let the rational function be $f(x) = \frac{x^3 - 8}{x^2 - 4}$. Find the equation of the oblique asymptote line $y = mx + n$ of this function by rigorously calculating the limits $m = \lim_{x \to \infty} \frac{f(x)}{x}$ and $n = \lim_{x \to \infty} (f(x) - mx)$.
The form 1^∞ and the number e
The number $e\approx2.718$ is defined by $\displaystyle\lim_{u\to\infty}\left(1+\frac1u\right)^u=e$. The base is close to $1$, but the huge exponent keeps pushing the result up. The two effects balance at $e$, not at $1$.
The same rule works at a finite point: $\displaystyle\lim_{x\to0}(1+3x)^{1/x}$ has $L=\lim\dfrac1x\cdot3x=3$, so the limit is $e^3$.
Determine the value of the following limit by resolving the indeterminate form associated with the number $e$:…
Determine the value of the following limit by resolving the indeterminate form associated with the number $e$:
$\lim_{x \to +\infty} \left( \frac{x + 5}{x - 2} \right)^{3x}$
Calculate the parametric limit $\lim_{x \to +\infty} \left( \frac{2x^2 + 3}{2x^2 - x} \right)^{ax+1}$ and determine the value of $a$ so that the exact result is $e^2$.
Calculate the parametric limit $\lim_{x \to +\infty} \left( \frac{2x^2 + 3}{2x^2 - x} \right)^{ax+1}$ and determine the value of $a$ so that the exact result is $e^2$.
Back to the start: at $x=5$ the function $\dfrac{x^2-25}{x-5}$ gives $\frac{0}{0}$. Factoring, $\dfrac{(x-5)(x+5)}{x-5}=x+5$ for every $x\ne5$, so the graph is the line $y=x+5$ with a hole at $x=5$, and the limit is $5+5=10$.