Continuity

What Nour says

1. 1. Continuity at a real number · Continuity of some common functions

   Today, the whole chapter: "Continuity", Year 3, Mathematics section.

   We will build this step by step, with a drawing and a worked example in each part.

   By the end, you will know how to check continuity and how to solve equations using it.

   Let's start with continuity at a real number.

   A function f is continuous at a real number a when its curve has no break near a.

   That is, you can draw the curve near a without lifting your pen.

   The exact definition uses a limit, written on the board now.

   This means the limit of f(x) as x gets close to a equals the value f(a) itself.

   Look at this curve: it jumps at x equals 2.

   The limit from the left is not the same as the limit from the right.

   So this function is not continuous at x equals 2.

   Now compare with a smooth curve, with no break at all.

   Here the limit as x approaches any point a equals the value of the function there.

   This curve is continuous everywhere, so it is continuous at every real number.

   Some functions are already known to be continuous on their whole domain.

   That includes polynomial functions, the square root function, sine and cosine.

   We call this continuity of some common functions: results we accept without proof.

   Now let's work through a full example, step by step.

   Consider the function f defined by f(x) equals x squared plus 1 for x less than 1.

   And f(x) equals 3x minus 1 for x greater than or equal to 1.

   We want to know if f is continuous at a equals 1.

   First, we compute the limit coming from the left, with x less than 1.

   As x approaches 1 from the left, x squared plus 1 approaches 2.

   Next, we compute the limit coming from the right, with x greater than 1.

   As x approaches 1 from the right, 3x minus 1 approaches 2.

   Both limits equal 2, and this is also the value of f at 1, since 3 times 1 minus 1 is 2.

   Since both sides agree with f of 1, the function is continuous at a equals 1.

   We write the final answer clearly, in a frame.

   Notice the method: find the left limit, the right limit, and compare both to f(a).

   If all three values match, the function is continuous at that point.

   If even one differs, the function has a break there, and is not continuous.

   This idea is the base for everything else in this chapter.

   Common functions like polynomials, square roots, sine, and cosine are continuous everywhere they are defined.

   Keep this definition in mind: it will come back in every section today.

   What does it mean for a function to be continuous at a point a?

2. 2. Continuity of |f| · Algebraic operations on continuous functions

   Let's move to the continuity of |f| and operations on continuous functions.

   If f is continuous at a, then the absolute value of f is also continuous at a.

   That is, taking the absolute value never breaks a curve that was already connected.

   Here is f, a simple line, continuous everywhere.

   Now we fold the negative part up, to get the absolute value of f.

   The curve now has a corner at x equals 1, but no break, no jump.

   A corner is fine for continuity, only a jump or a hole is a problem.

   Next, continuous functions behave well under the usual operations.

   We call this algebraic operations on continuous functions: that is, sums, products and quotients of continuous functions stay continuous.

   A quotient needs one extra condition: the denominator must not vanish at a.

   This lets us build new continuous functions from simple known ones, without checking limits again.

   Now let's apply this to a full worked example.

   Let f be the function defined by f(x) equals x squared minus 4, on all of ℝ.

   We want to study the continuity of g, the absolute value of f, at a equals 2.

   First, f is a polynomial, so f is continuous everywhere, including at a equals 2.

   We compute f of 2: 2 squared minus 4 gives 0.

   Since f is continuous at a equals 2, and the absolute value keeps continuity, g is continuous there too.

   We see g of 2 also equals 0, since the absolute value of 0 is 0.

   The curve of g touches zero but keeps going without any break.

   So g, the absolute value of f, is continuous at a equals 2.

   Notice the reasoning: start from a function already known to be continuous.

   Then apply the rule that taking the absolute value preserves continuity.

   The same logic works for sums, products, and quotients with a nonzero denominator.

   You never need to recompute limits from scratch for these combined functions.

   This saves a lot of work compared to checking continuity by definition every time.

   Keep this toolbox in mind: it lets you build continuity from simple pieces.

   Why is g, the absolute value of f, continuous at a point where f is continuous?

3. 3. Continuity of √f · Right continuity. Left continuity

   Now let's look at the continuity of √f, and right and left continuity.

   If f is continuous at a, and f stays nonnegative near a, then √f is continuous at a.

   That is, taking a square root of a continuous positive function keeps it continuous.

   Here f of x equals x plus 1, a continuous line, positive on this interval.

   Now we draw √f, the square root of f.

   This curve is still smooth, with no break anywhere on this interval.

   Next, let's talk about right continuity and left continuity.

   A function is right continuous at a when the limit from the right equals f of a.

   That is, approaching a from values greater than a gives the correct value.

   Left continuity is the same idea, but approaching a from values smaller than a.

   A function is continuous at a exactly when it is both left and right continuous there.

   Let's see a full worked example now.

   Let f be defined by f(x) equals x for x less than 0, and f(x) equals √x for x greater than or equal to 0.

   We check the continuity of f at a equals 0.

   First, the left limit: as x approaches 0 from values less than 0, f of x approaches 0.

   So f is left continuous at 0, since the left limit equals f of 0, which is 0.

   Now the right limit: as x approaches 0 from values greater than 0, √x approaches 0.

   So f is also right continuous at 0, since this limit also equals f of 0.

   Since both one-sided limits equal f of 0, f is continuous at a equals 0.

   Remember this method: check the left limit, then the right limit, then compare both to f of a.

   Square roots of continuous positive functions, and one-sided limits, both build on the same core idea.

   What must be true for a function to be continuous at a, in terms of one-sided limits?

4. 4. Continuity on an interval

   Let's move on to continuity on an interval.

   A function is continuous on an interval when it is continuous at every point of that interval.

   That is, the curve has no break at all, for any point inside the interval.

   This curve is continuous on the whole interval from 0 to plus infinity.

   You can draw this entire curve without ever lifting your pen.

   Now look at a function with a break inside an interval.

   There is a jump at x equals 2, so this function is not continuous on an interval containing 2.

   But on an interval that avoids the point 2, this same function is continuous.

   So continuity on an interval always depends on checking every point, one by one in theory.

   In practice, we rely on known continuous functions and the operations we studied before.

   Polynomials, square roots, sine and cosine are continuous on their whole domain of definition.

   Let's apply this idea to a full example now.

   Let f be defined by f(x) equals the quotient of 1 by x minus 2.

   We want to study the continuity of f on the interval from 3 to plus infinity.

   First, f is a quotient of two polynomials, both continuous everywhere on ℝ.

   A quotient of continuous functions is continuous wherever the denominator does not vanish.

   Here the denominator x minus 2 is zero only at x equals 2.

   On the interval from 3 to plus infinity, x minus 2 is always positive and never zero.

   So f, this quotient function, is continuous on the whole interval from 3 to plus infinity.

   The key step was checking the denominator never vanishes on that interval.

   This same method works for any quotient: find where the denominator is zero, then exclude it.

   What must you check to know if a quotient function is continuous on an interval?

5. 5. Image of an interval by a continuous function

   Now let's study the image of an interval by a continuous function.

   This is a key theorem: a continuous function sends an interval to another interval.

   That is, there is no gap in the set of values f takes on that interval.

   Take f of x equals x squared, on the interval from 0 to 2.

   And the other endpoint, at x equals 2.

   f of 0 equals 0, and f of 2 equals 4.

   Since f is continuous, every value between 0 and 4 is reached by f somewhere on this interval.

   So the image of the interval from 0 to 2 is exactly the interval from 0 to 4.

   This result is called the image of an interval by a continuous function: that is, continuity guarantees the image is also one single interval.

   If f were not continuous, the image could have a gap, missing some values in between.

   This broken function skips values between its two pieces, so its image has a hole.

   When f is also strictly monotonic, we get a very useful special case.

   The image interval's endpoints are simply f of a and f of b, in the same order.

   If f is strictly decreasing instead, the endpoints just swap places.

   Let's see this in a full worked example.

   Let f be defined by f(x) equals x cubed minus 3x, on the interval from 0 to 2.

   Suppose we know f is continuous and strictly increasing on the interval from 1 to 2.

   We want to find the image of the interval from 1 to 2 by f.

   First, we compute f of 1: 1 cubed minus 3 times 1 gives -2.

   Next, we compute f of 2: 2 cubed minus 3 times 2 gives 2.

   Since f is continuous and strictly increasing here, the image is the interval from f of 1 to f of 2.

   So the image of the interval from 1 to 2 is the interval from minus 2 to 2.

   The method: check continuity, check strict monotonicity, then compute f at both endpoints.

   The order of the endpoints in the image depends on whether f increases or decreases.

   This theorem is very useful before solving equations, which is our final topic today.

   If f is continuous and strictly increasing on [a;b], what is the image f([a;b])?

6. 6. Solving equations of the form f(x) = k

   Now, our last topic: solving equations of the form f(x) equals k.

   This uses the image of an interval theorem we just saw, to prove a solution exists.

   If f is continuous and strictly monotonic on [a;b], and k lies between f(a) and f(b), there is exactly one solution.

   Here f of x equals x squared minus 1, continuous and strictly increasing on the interval from 0 to 3.

   f of 0 equals -1, and f of 3 equals 8.

   Take k equals 2, which lies between -1 and 8.

   The horizontal line at height k must cross the curve exactly once, since f is strictly increasing.

   This equation f of x equals k has exactly one solution, called x zero, in the interval.

   To find x zero more precisely, we use a method called dichotomy.

   That is, we cut the interval in half repeatedly, keeping the half where the solution must lie.

   At each step, we check the sign of f at the midpoint, to pick the correct half.

   This gives smaller and smaller intervals, closing in on x zero step by step.

   Let's apply all of this to a full worked example.

   Let f be defined by f(x) equals x cubed plus x minus 1, on the interval from 0 to 1.

   f is continuous and strictly increasing on this interval, as a sum of increasing functions.

   We compute f of 0: 0 cubed plus 0 minus 1 gives -1.

   And f of 1: 1 cubed plus 1 minus 1 gives 1.

   Since 0 lies between -1 and 1, the equation f of x equals 0 has a unique solution here.

   We call this solution alpha, somewhere between 0 and 1.

   Now we apply dichotomy: we test the midpoint, 0.5.

   f of 0.5 equals 0.5 cubed plus 0.5 minus 1, which gives -0.375.

   This value is negative, so alpha lies between 0.5 and 1, not between 0 and 0.5.

   We repeat this process, halving the interval again and again, to get closer to alpha.

   Each repetition divides the interval length in half, giving better precision on alpha.

   This method never gives the exact value, but it approaches it as closely as we want.

   That closes our chapter: continuity at a point, on an interval, and solving equations with it.

   In the dichotomy method, what do we test at each step to choose the next half-interval?

   That completes our chapter on continuity, from a single point to whole intervals and equations.

   You now have the tools to check continuity and to prove a solution exists and locate it.

   Practice these two exercises, and review the method steps whenever you get stuck.