Continuity of a Function at a Point
A parking lot charges $\$40$ if you stay up to one hour, and as soon as you go over the hour the charge "jumps" to $\$60$. If you graph cost against time, you have to lift your pen right at the one-hour mark. How can you detect that jump without drawing? And how do you decide whether a formula with letters ($a$, $b$, $k$) can be adjusted so that there is no jump? That is what this page is about.
The three conditions
A function is continuous at a point if its graph can be traced there without lifting your pen. To check it with numbers, we use the limit.
Why three? Because the graph can break in three ways: there is no point (condition 1 fails), the two sides arrive at different heights (condition 2 fails), or the point sits at a height different from the one the two sides "expect" (condition 3 fails).
Check whether the function $f(x) = e^{x-1}$ is continuous at the point $x = 1$. For a complete answer, you must verify whether the value of the function at that point…
Check whether the function $f(x) = e^{x-1}$ is continuous at the point $x = 1$. For a complete answer, you must verify whether the value of the function at that point coincides with the limit of the function as $x$ approaches 1.
Observe the following graph of the function $f(x)$ and analyse its behavior at the point with x-coordinate $x = 2$. State whether the function is continuous at that…
Observe the following graph of the function $f(x)$ and analyse its behavior at the point with x-coordinate $x = 2$. State whether the function is continuous at that point by checking the three conditions for continuity: existence of $f(2)$, existence of $\lim_{x \to 2} f(x)$ and the equality of both values.
Types of discontinuity
Depending on which condition fails and how the one-sided limits behave, the discontinuity is classified as follows:
| Type | What happens |
|---|---|
| Removable | The limit exists and is finite, but $f(a)$ does not exist or is different from the limit. |
| Finite jump | Both one-sided limits exist and are finite, but they are different. |
| Infinite jump | At least one of the one-sided limits is $+\infty$ or $-\infty$ (there is a vertical asymptote). |
Given the function $f(x) = \frac{1}{x + 4}$, justify why it is not continuous at the point $x = -4$. To do so, calculate the one-sided limits $\lim_{x \to -4^-} f(x)$…
Given the function $f(x) = \frac{1}{x + 4}$, justify why it is not continuous at the point $x = -4$. To do so, calculate the one-sided limits $\lim_{x \to -4^-} f(x)$ and $\lim_{x \to -4^+} f(x)$ and describe the type of discontinuity it has.
Study the continuity of the function $f(x) = \frac{x^2 - 1}{x - 1}$ at the point $x = 1$. If it has a discontinuity, state what type it is (removable, jump, or infinite)…
Study the continuity of the function $f(x) = \frac{x^2 - 1}{x - 1}$ at the point $x = 1$. If it has a discontinuity, state what type it is (removable, jump, or infinite) and explain whether there is any way to redefine the function so that it is continuous at that point.
Classify all the discontinuities of the function $g(x)$ and determine its vertical asymptotes by analysing the behavior of the function near the values where it is not…
Classify all the discontinuities of the function $g(x)$ and determine its vertical asymptotes by analysing the behavior of the function near the values where it is not defined:
$$g(x) = \frac{x^3 - 4x}{x^2 - 2x}$$How would you define the function at the points of removable discontinuity so that it becomes continuous at them?
Piecewise functions: one-sided limits
In a piecewise function, the point where the expression changes is the suspect. There you calculate the two one-sided limits, each with its own formula, and the value $f(a)$ with the formula that includes the equality.
Given the following piecewise function, calculate the left-hand and right-hand limits at the boundary point $x = 3$ and determine whether the function is continuous at…
Given the following piecewise function, calculate the left-hand and right-hand limits at the boundary point $x = 3$ and determine whether the function is continuous at that point:
$f(x) = \begin{cases} 2x - 4 & \text{if } x < 3 \\ \sqrt{x + 1} & \text{if } x \ge 3 \end{cases}$
Visually analyse the continuity of the function represented in the following graph. Identify the value of $x$ where the function has a discontinuity and classify it…
Visually analyse the continuity of the function represented in the following graph. Identify the value of $x$ where the function has a discontinuity and classify it based on the behavior of its left-hand and right-hand branches.
Discontinuities with absolute value
Absolute value works like a piecewise function: it changes expression where its argument is zero.
Given the function: $$f(x) = \frac{x^2 - 1}{|x| - 1}$$ Study its continuity at the points $x = 1$ and $x = -1$. If there are discontinuities, state what type they are…
Given the function:
$$f(x) = \frac{x^2 - 1}{|x| - 1}$$Study its continuity at the points $x = 1$ and $x = -1$. If there are discontinuities, state what type they are and whether there is any way to redefine the function to remove them.
Study the continuity of the following function across its domain, classifying any points of discontinuity you find: $$f(x) = \frac{|x - 2|}{x^2 - x - 2}$$ Accompany your…
Study the continuity of the following function across its domain, classifying any points of discontinuity you find:
$$f(x) = \frac{|x - 2|}{x^2 - x - 2}$$Accompany your analytical study with a diagram of the one-sided limits at the critical points.
Study the continuity of the following function over its entire domain, classifying in detail each of its points of discontinuity:…
Study the continuity of the following function over its entire domain, classifying in detail each of its points of discontinuity:
$f(x) = \dfrac{|x^2 - 1|}{x^2 - 3x + 2}$
Afterwards, if there is any point where the discontinuity is removable, propose a redefinition of the function at that point to turn it into a function that is continuous there. If there is none, justify it.
Exponentials and logarithms: finite jump or infinite jump?
With exponentials like $2^{1/x}$, the two sides can behave very differently, even though the function "looks calm." Do not assume that a point outside the domain always gives an asymptote: calculate the one-sided limits.
Study the continuity of the function $f(x) = \frac{1}{1 + 2^{1/x}}$ at the point $x = 0$. To do so, calculate the one-sided limits $\lim_{x \to 0^-} f(x)$ and…
Study the continuity of the function $f(x) = \frac{1}{1 + 2^{1/x}}$ at the point $x = 0$. To do so, calculate the one-sided limits $\lim_{x \to 0^-} f(x)$ and $\lim_{x \to 0^+} f(x)$. What type of discontinuity does the function have at that point?
Finding parameters that make a function continuous
This is the most common problem: a piecewise function with letters ($a$, $b$, $k$) and you have to choose them so that there are no jumps.
Determine the value of the parameter $k$ so that the function $g(x)$ is continuous on the entire real line $\mathbb{R}$:…
Determine the value of the parameter $k$ so that the function $g(x)$ is continuous on the entire real line $\mathbb{R}$:
$g(x) = \begin{cases} x + k & \text{if } x \le 2 \\ x^2 - 1 & \text{if } x > 2 \end{cases}$
A surveillance drone follows an altitude trajectory defined by the function $h(t)$ (in meters) with respect to time $t$ (in seconds). Due to a change in the control…
A surveillance drone follows an altitude trajectory defined by the function $h(t)$ (in meters) with respect to time $t$ (in seconds). Due to a change in the control system, the function changes at $t = 5$:
$h(t) = \begin{cases} t^2 - 10 & \text{if } 0 \le t < 5 \\ 3t & \text{if } t \ge 5 \end{cases}$
Check whether the transition in altitude is continuous at time $t = 5$ to prevent the drone from undergoing a sudden movement.
Given the function dependent on the parameter $k \in \mathbb{R}$: $$f(x) = \begin{cases} \frac{x^2 - kx}{x - 3} & \text{if } x < 3 \\ 2x - 5 & \text{if } x \geq 3…
Given the function dependent on the parameter $k \in \mathbb{R}$:
$$f(x) = \begin{cases} \frac{x^2 - kx}{x - 3} & \text{if } x < 3 \\ 2x - 5 & \text{if } x \geq 3 \end{cases}$$Calculate the value of $k$ so that the function has a jump discontinuity (finite jump) at $x = 3$. Is there any value of $k$ that makes it continuous at that point? Justify your answer by computing the one-sided limits.
When the limit is $0/0$
If at the junction point a piece gives $0/0$, you have to simplify before setting things equal. There are two tools: factoring (if it is rational) and rationalizing, that is, multiplying numerator and denominator by the conjugate (if there are roots).
If the piece is rational and must be continuous at a point where the denominator is zero, the numerator has to be zero as well (otherwise, there is an asymptote).
Determine the values of $a$ and $b$ so that the function $f(x)$ is continuous everywhere in $\mathbb{R}$: $$f(x) = \begin{cases} \frac{x+2}{x-1} & \text{if } x < -2 \\ a…
Determine the values of $a$ and $b$ so that the function $f(x)$ is continuous everywhere in $\mathbb{R}$:
$$f(x) = \begin{cases} \frac{x+2}{x-1} & \text{if } x < -2 \\ a \times x + b & \text{if } -2 \leq x \leq 3 \\ \sqrt{x^2 + 7} & \text{if } x > 3 \end{cases}$$Explain in detail the process followed, checking that the left-hand and right-hand limits are equal at each boundary point.
Consider the function $f(x)$ defined by: $$f(x) = \begin{cases} \ln(x+1) & \text{if } -1 < x < 0 \\ x^2 - a & \text{if } 0 \leq x < 2 \\ b \times 2^x & \text{if } x \geq…
Consider the function $f(x)$ defined by:
$$f(x) = \begin{cases} \ln(x+1) & \text{if } -1 < x < 0 \\ x^2 - a & \text{if } 0 \leq x < 2 \\ b \times 2^x & \text{if } x \geq 2 \end{cases}$$Find the values of $a$ and $b$ so that the function is continuous on its domain. Once found, state whether the function is continuous at $x = -1$.
An industrial cooling system regulates its power $P$ (in kilowatts) according to the outside temperature $T$ (in degrees Celsius) by means of the following function:…
An industrial cooling system regulates its power $P$ (in kilowatts) according to the outside temperature $T$ (in degrees Celsius) by means of the following function:
$$P(T) = \begin{cases} \dfrac{a \times e^T}{e^T + 1} & \text{if } T \leq 0 \\[2mm] \dfrac{\sqrt{T+4} - 2}{T} & \text{if } T > 0 \end{cases}$$Determine the value of the constant $a$ so that the power transition is continuous when the temperature passes through $0\,^\circ C$.
A new engine prototype is being tested whose acceleration $a$ (in $m/s^2$) depends on time $t$ (in seconds) according to the expression: $$a(t) = \begin{cases} t^2 + m &…
A new engine prototype is being tested whose acceleration $a$ (in $m/s^2$) depends on time $t$ (in seconds) according to the expression:
$$a(t) = \begin{cases} t^2 + m & \text{if } 0 \leq t < 2 \\ \frac{t-2}{\sqrt{t+7} - 3} & \text{if } t > 2 \end{cases}$$Calculate the value of $m$ so that the acceleration does not undergo sudden jumps (is continuous) at time $t = 2$ s.
A team of robotics engineers has programmed the motion of a mechanical arm. The speed of the end of the arm (in m/s) follows the path:…
A team of robotics engineers has programmed the motion of a mechanical arm. The speed of the end of the arm (in m/s) follows the path:
$v(t) = \begin{cases} \frac{\sqrt{t+5}-3}{t-4} & \text{if } 0 \leq t < 4 \\ \frac{a}{12} & \text{if } t = 4 \\ \frac{t^2 - 16}{t^2 - 3t - 4} \times b & \text{if } t > 4 \end{cases}$
Determine the values of $a$ and $b$ so that the robot's motion is continuous (without sudden jerks) at time $t = 4$ seconds.
Given the function depending on the parameter $k$: $f(x) = \frac{x^3 - kx^2 + 3x}{x^2 - x}$ a) Find the value of $k$ so that the function has a removable discontinuity…
Given the function depending on the parameter $k$:
$f(x) = \frac{x^3 - kx^2 + 3x}{x^2 - x}$
- a) Find the value of $k$ so that the function has a removable discontinuity at $x = 1$.
- b) For that value of $k$, study the continuity at $x = 0$ and determine whether any vertical asymptote exists.
- c) Does any value of $k$ exist that makes the function continuous across all of $\mathbb{R}$? Justify your answer.
Extension: notable limits and limits at infinity
Limit as $x$ grows without bound
With quotients of polynomials, you solve it by dividing numerator and denominator by the highest power of $x$: the terms with $\frac{1}{x}$, $\frac{1}{x^2}$… tend to $0$.
Three notable limits
Given the function $f(x) = \left( \frac{2x^2 + 3}{2x^2 - 1} \right)^{ax^2}$: a) Find the value of the parameter $a \in \mathbb{R}$ such that…
Given the function $f(x) = \left( \frac{2x^2 + 3}{2x^2 - 1} \right)^{ax^2}$:
- a) Find the value of the parameter $a \in \mathbb{R}$ such that $\lim_{x \to \infty} f(x) = e^2$.
- b) With the value of $a$ obtained in the previous section, define a new function $g(x)$ such that:
$g(x) = \begin{cases} f(x) & \text{if } x \ge 1 \\ \frac{x^2 + bx + c}{x - 1} & \text{if } x < 1 \end{cases}$
Determine $b$ and $c$ so that $g(x)$ is continuous at $x = 1$.
A telecommunications company is adjusting the power of a 5G antenna. The signal loss $L$ (in decibels) depends on the distance $d$ (in km) according to the expression:…
A telecommunications company is adjusting the power of a 5G antenna. The signal loss $L$ (in decibels) depends on the distance $d$ (in km) according to the expression:
$L(d) = \begin{cases} \frac{k \times \ln(d+1)}{d} & \text{if } 0 < d < 1 \\ \sqrt{d^2 + 3} - a & \text{if } d \geq 1 \end{cases}$
The power transition is required to be smooth and continuous so as not to interrupt calls at the 1 km boundary. Calculate the values of $k$ and $a$ knowing that as the distance tends to 0 km, the signal loss tends to 2 dB.
In the design of a new component for a particle accelerator, the magnetic field strength as a function of position $x$ (in millimeters) is defined by the following…
In the design of a new component for a particle accelerator, the magnetic field strength as a function of position $x$ (in millimeters) is defined by the following function:
$f(x) = \begin{cases} \frac{e^{ax} - 1}{x} & \text{if } x < 0 \\ b & \text{if } x = 0 \\ \frac{x^2 - 4}{x - 2} \times \cos(\pi x) & \text{if } x > 0 \end{cases}$
Determine the values of the real parameters $a$ and $b$ so that the magnetic field strength is continuous at the point $x = 0$. Once found, would the function be continuous at the point $x = 2$? Reason your answer analytically.
A consequence: Bolzano's theorem
To show that two graphs $f$ and $g$ intersect, apply Bolzano to the difference $h=f-g$.
Apply Bolzano's Theorem to show that the graphs of the functions $f(x) = \ln(x)$ and $g(x) = e^{-x}$ intersect at least once within the interval $(1, e)$. Justify with…
Apply Bolzano's Theorem to show that the graphs of the functions $f(x) = \ln(x)$ and $g(x) = e^{-x}$ intersect at least once within the interval $(1, e)$. Justify with reasoning whether this point of intersection is unique in that interval based on the growth of the functions.
Back to the beginning: the parking lot charge has a finite jump at the one-hour mark (the one-sided limits are $\$40$ and $\$60$), and that is why it cannot be drawn without lifting your pen.