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Limits and continuity · Limits

Continuity of a Function at a Point

Before you start, make sure you know: Continuity of functions (High School Year 2 (semesters 3 and 4))
Before you start, make sure you know: Absolute value function (High School Year 2 (semesters 3 and 4))

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.

Continuous graph at x = a: a smooth curve passing through the filled point (a, f(a)) with no breaks x = a f(a) the limit equals f(a): no break

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).

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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.

Solve
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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.

x→2⁻ x→2⁺ f(x)→2 f(x)→3 1 2 3 4 5 1 2 3 4 x y f(x)
Solve

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).
Three graphs: a removable discontinuity with a hollow point on the curve and the function value at a different height, a finite jump with two different heights, and an infinite jump with a vertical asymptote Removable Finite jump Infinite jump
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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.

Solve
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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.

Solve
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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?

Solve

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.

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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:

$f(x) = \begin{cases} 2x - 4 & \text{if } x < 3 \\ \sqrt{x + 1} & \text{if } x \ge 3 \end{cases}$

Solve
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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.

x = −2 f(x) x y −4 −3 −1 1 2 3 4 1 −1 0
Solve

Discontinuities with absolute value

Absolute value works like a piecewise function: it changes expression where its argument is zero.

Graph with two separate branches at x = 2: the left one lies below the axis and the right one above it, a finite jump x = 2 left: −1/(x+1) right: 1/(x+1)
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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.

Solve
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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.

Solve
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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.

Solve

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.

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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?

Solve

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.

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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 $\mathbb{R}$:

$g(x) = \begin{cases} x + k & \text{if } x \le 2 \\ x^2 - 1 & \text{if } x > 2 \end{cases}$

Solve
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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.

Solve
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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.

Solve

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).

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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.

Solve
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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 $x = -1$.

Solve
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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 $0\,^\circ C$.

Solve
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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.

1 2 3 4 5 6 2 4 6 8 10 t (s) a (m/s²) a(t) = t² + m a(t) = (t - 2) / (√(t + 7) - 3) Limit = 6
Solve
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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.

Solve
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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 $k$:

$f(x) = \frac{x^3 - kx^2 + 3x}{x^2 - x}$

  1. a) Find the value of $k$ so that the function has a removable discontinuity at $x = 1$.
  2. b) For that value of $k$, study the continuity at $x = 0$ and determine whether any vertical asymptote exists.
  3. c) Does any value of $k$ exist that makes the function continuous across all of $\mathbb{R}$? Justify your answer.
Solve

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

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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 $f(x) = \left( \frac{2x^2 + 3}{2x^2 - 1} \right)^{ax^2}$:

  1. a) Find the value of the parameter $a \in \mathbb{R}$ such that $\lim_{x \to \infty} f(x) = e^2$.
  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 $x = 1$.

Solve
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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.

Solve
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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.

Solve

A consequence: Bolzano's theorem

To show that two graphs $f$ and $g$ intersect, apply Bolzano to the difference $h=f-g$.

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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.

x y 0 1 e 1 c f(x) = ln(x) g(x) = e⁻ˣ f(1) < g(1) f(e) > g(e)
Solve

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.

Summary

See all the exercises in this section