1. Limits, Continuity and Differentiability of Real-Value Functions

# 1. Limits, Continuity and Differentiability of Real-Value Functions ## Objectifs et prérequis This chapter introduces the foundational concepts of limits, continuity, and differentiability for real-valued functions. Students will learn to rigorously define limits using the epsilon-delta approach, analyze continuity at points and on intervals, apply the Intermediate Value Theorem, and understand the relationship between continuity and differentiability, culminating in Rolle’s Theorem. **Prerequisites**: Proficiency in algebraic manipulation, function notation, basic set theory, and familiarity with graphs of elementary functions (polynomials, rational, trigonometric, exponential, logarithmic). Knowledge of inequalities and absolute value is essential. --- ## I. Introduction to Limits The concept of a limit describes the behavior of a function as its input approaches a particular value, without necessarily requiring the function to be defined at that point. > **Definition (Informal Limit)**: > We say $\lim_{x \to a} f(x) = L$ if the values of $f(x)$ can be made arbitrarily close to $L$ by taking $x$ sufficiently close to $a$ (but $x \neq a$). This informal idea is made precise through the epsilon-delta definition. ### Examples **Example 1**: Find $\lim_{x \to 2} (3x - 1)$. As $x$ approaches 2, $3x - 1$ approaches $3(2) - 1 = 5$. So, $\lim_{x \to 2} (3x - 1) = 5$. **Example 2**: Find $\lim_{x \to 0} \frac{\sin x}{x}$. This is a standard limit: $\lim_{x \to 0} \frac{\sin x}{x} = 1$. (Justified geometrically or via squeeze theorem.) ### Exercices d'application corrigés **Exercise 1**: Evaluate $\lim_{x \to 3} (x^2 - 4x + 7)$. **Solution**: Substitute $x = 3$: $3^2 - 4(3) + 7 = 9 - 12 + 7 = 4$. $\therefore \lim_{x \to 3} (x^2 - 4x + 7) = 4$. **Exercise 2**: Evaluate $\lim_{x \to 1} \frac{x^2 - 1}{x - 1}$. **Solution**: Factor numerator: $\frac{(x-1)(x+1)}{x-1} = x+1$ for $x \neq 1$. So, $\lim_{x \to 1} \frac{x^2 - 1}{x - 1} = \lim_{x \to 1} (x+1) = 2$. --- ## II. General definition of limits using Epsilon and Delta notations The epsilon-delta definition provides a rigorous foundation for limits. > **Definition (Epsilon-Delta Limit)**: > Let $f$ be a function defined on an open interval containing $a$, except possibly at $a$ itself. We say $\lim_{x \to a} f(x) = L$ if for every $\varepsilon > 0$, there exists a $\delta > 0$ such that > $$0 < |x - a| < \delta \implies |f(x) - L| < \varepsilon.$$ This means: no matter how small a tolerance $\varepsilon$ we demand for $f(x)$ near $L$, we can find a tolerance $\delta$ for $x$ near $a$ that guarantees $f(x)$ stays within $\varepsilon$ of $L$. ### Properties of Limits (using epsilon-delta) If $\lim_{x \to a} f(x) = L$ and $\lim_{x \to a} g(x) = M$, then: - $\lim_{x \to a} [f(x) + g(x)] = L + M$ - $\lim_{x \to a} [f(x) \cdot g(x)] = L \cdot M$ - $\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{L}{M}$, provided $M \neq 0$ - $\lim_{x \to a} c \cdot f(x) = cL$ for constant $c$ These follow from the epsilon-delta definition and are used to compute limits algebraically. ### Examples **Example 1**: Prove $\lim_{x \to 2} (3x - 1) = 5$ using epsilon-delta. We need: for any $\varepsilon > 0$, find $\delta > 0$ such that $0 < |x - 2| < \delta \implies |(3x - 1) - 5| < \varepsilon$. Simplify: $|3x - 6| = 3|x - 2| < \varepsilon \implies |x - 2| < \frac{\varepsilon}{3}$. So choose $\delta = \frac{\varepsilon}{3}$. Then $0 < |x - 2| < \delta \implies 3|x - 2| < 3 \cdot \frac{\varepsilon}{3} = \varepsilon$. $\therefore$ The limit is proved. **Example 2**: Prove $\lim_{x \to 0} x^2 = 0$. Need: $0 < |x - 0| < \delta \implies |x^2 - 0| < \varepsilon$. That is, $|x|^2 < \varepsilon \implies |x| < \sqrt{\varepsilon}$. Choose $\delta = \sqrt{\varepsilon}$. Then $|x| < \delta \implies |x|^2 < (\sqrt{\varepsilon})^2 = \varepsilon$. $\therefore$ Proved. ### Exercices d'application corrigés **Exercise 1**: Use epsilon-delta to prove $\lim_{x \to 4} (2x + 3) = 11$. **Solution**: We need $|(2x + 3) - 11| < \varepsilon$ when $0 < |x - 4| < \delta$. Simplify: $|2x - 8| = 2|x - 4| < \varepsilon \implies |x - 4| < \frac{\varepsilon}{2}$. Choose $\delta = \frac{\varepsilon}{2}$. Then $0 < |x - 4| < \delta \implies 2|x - 4| < \varepsilon$. $\therefore$ Proved. **Exercise 2**: Prove $\lim_{x \to 1} x^2 = 1$ using epsilon-delta. **Solution**: Need $|x^2 - 1| < \varepsilon$ when $0 < |x - 1| < \delta$. Factor: $|x^2 - 1| = |(x-1)(x+1)| = |x-1| \cdot |x+1|$. Assume $\delta \leq 1$, so $|x - 1| < 1 \implies 0 < x < 2 \implies |x+1| < 3$. Then $|x^2 - 1| < 3|x - 1|$. To make this $< \varepsilon$, need $|x - 1| < \frac{\varepsilon}{3}$. Choose $\delta = \min\left(1, \frac{\varepsilon}{3}\right)$. If $0 < |x - 1| < \delta$, then $|x^2 - 1| < 3 \cdot \frac{\varepsilon}{3} = \varepsilon$. $\therefore$ Proved. --- ## III. Continuity at a point A function is continuous at a point if its limit at that point equals its value. > **Definition (Continuity at a Point)**: > A function $f$ is continuous at $x = a$ if: > 1. $f(a)$ is defined, > 2. $\lim_{x \to a} f(x)$ exists, > 3. $\lim_{x \to a} f(x) = f(a)$. > > Equivalently, for every $\varepsilon > 0$, there exists $\delta > 0$ such that $|x - a| < \delta \implies |f(x) - f(a)| < \varepsilon$. If any condition fails, $f$ is discontinuous at $a$. ### Types of discontinuity - **Removable**: Limit exists but $\neq f(a)$ or $f(a)$ undefined (can be "fixed" by redefining $f(a)$). - **Jump**: Left and right limits exist but are unequal. - **Infinite**: Limit is $\pm\infty$ (vertical asymptote). - **Oscillating**: Limit does not exist due to oscillation (e.g., $\sin(1/x)$ at $0$). ### Examples **Example 1**: $f(x) = \begin{cases} x^2 & \text{if } x \neq 2 \\ 5 & \text{if } x = 2 \end{cases}$ At $x=2$: $f(2)=5$, $\lim_{x \to 2} f(x) = \lim_{x \to 2} x^2 = 4$. Since $4 \neq 5$, discontinuous (removable). **Example 2**: $f(x) = \frac{1}{x}$ at $x=0$. $f(0)$ undefined, $\lim_{x \to 0^+} \frac{1}{x} = +\infty$, $\lim_{x \to 0^-} \frac{1}{x} = -\infty$. Infinite discontinuity. **Example 3**: $f(x) = \begin{cases} x+1 & \text{if } x < 1 \\ x-1 & \text{if } x \geq 1 \end{cases}$ At $x=1$: $f(1)=0$, $\lim_{x \to 1^-} f(x) = 2$, $\lim_{x \to 1^+} f(x) = 0$. Left and right limits unequal → jump discontinuity. ### Exercices d'application corrigés **Exercise 1**: Determine if $f(x) = \frac{x^2 - 9}{x - 3}$ is continuous at $x=3$. **Solution**: Simplify: $f(x) = \frac{(x-3)(x+3)}{x-3} = x+3$ for $x \neq 3$. $f(3)$ is undefined (original form). $\lim_{x \to 3} f(x) = \lim_{x \to 3} (x+3) = 6$. Since $f(3)$ undefined, discontinuous (removable). If we define $f(3)=6$, it becomes continuous. **Exercise 2**: Check continuity of $f(x) = \begin{cases} 2x & \text{if } x < 0 \\ x^2 & \text{if } x \geq 0 \end{cases}$ at $x=0$. **Solution**: $f(0) = 0^2 = 0$. $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} 2x = 0$. $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} x^2 = 0$. Both limits equal $f(0)=0$ → continuous at $x=0$. --- ## IV. Points of discontinuity Points where a function fails to be continuous are points of discontinuity. We classify them based on the failure of the continuity conditions. > **Classification of Discontinuity at $x=a$**: > - **Removable**: $\lim_{x \to a} f(x)$ exists but $\neq f(a)$ or $f(a)$ undefined. > - **Jump**: $\lim_{x \to a^-} f(x)$ and $\lim_{x \to a^+} f(x)$ both exist but are unequal. > - **Infinite**: At least one of $\lim_{x \to a^-} f(x)$ or $\lim_{x \to a^+} f(x)$ is $\pm\infty$. > - **Oscillating**: $\lim_{x \to a} f(x)$ does not exist due to oscillation (no single value approached). ### Examples **Example 1**: $f(x) = \frac{\sin x}{x}$ at $x=0$. $f(0)$ undefined, but $\lim_{x \to 0} \frac{\sin x}{x} = 1$. Removable discontinuity. **Example 2**: $f(x) = \tan x$ at $x = \frac{\pi}{2}$. $\lim_{x \to \frac{\pi}{2}^-} \tan x = +\infty$, $\lim_{x \to \frac{\pi}{2}^+} \tan x = -\infty$. Infinite discontinuity. **Example 3**: $f(x) = \begin{cases} 1 & \text{if } x \in \mathbb{Q} \\ 0 & \text{if } x \notin \mathbb{Q} \end{cases}$ (Dirichlet function) at any $a$. Limit does not exist (oscillates between 0 and 1) → oscillating discontinuity. ### Exercices d'application corrigés **Exercise 1**: Classify the discontinuity of $f(x) = \frac{x^2 - 4}{x - 2}$ at $x=2$. **Solution**: Simplify: $f(x) = x+2$ for $x \neq 2$. $f(2)$ undefined. $\lim_{x \to 2} f(x) = 4$. Limit exists but $f(2)$ undefined → **removable discontinuity**. **Exercise 2**: Classify discontinuity of $f(x) = \frac{1}{x-1}$ at $x=1$. **Solution**: $\lim_{x \to 1^-} \frac{1}{x-1} = -\infty$, $\lim_{x \to 1^+} \frac{1}{x-1} = +\infty$. At least one infinite limit → **infinite discontinuity**. **Exercise 3**: Classify discontinuity of $f(x) = \sin\left(\frac{1}{x}\right)$ at $x=0$. **Solution**: As $x \to 0$, $\frac{1}{x} \to \pm\infty$, so $\sin(1/x)$ oscillates rapidly between $-1$ and $1$. Limit does not exist → **oscillating discontinuity**. --- ## V. Sum, product and quotient of continuous functions Continuity is preserved under basic arithmetic operations. > **Theorem (Algebra of Continuous Functions)**: > If $f$ and $g$ are continuous at $x = a$, then: > - $f + g$ is continuous at $a$, > - $f - g$ is continuous at $a$, > - $f \cdot g$ is continuous at $a$, > - $\frac{f}{g}$ is continuous at $a$, provided $g(a) \neq 0$. This follows directly from the limit laws and the definition of continuity. ### Examples **Example 1**: $f(x) = x^2$, $g(x) = \sin x$ are continuous everywhere. Then $h(x) = x^2 + \sin x$ is continuous everywhere. **Example 2**: $f(x) = \frac{1}{x}$, $g(x) = x+1$. $f$ continuous on $(-\infty,0) \cup (0,\infty)$, $g$ continuous everywhere. $\frac{f}{g} = \frac{1}{x(x+1)}$ is continuous where denominator $\neq 0$: $x \neq 0, -1$. **Example 3**: $f(x) = \sqrt{x}$ (continuous on $[0,\infty)$), $g(x) = x^2$ (continuous everywhere). $f \cdot g = x^{5/2}$ continuous on $[0,\infty)$. ### Exercices d'application corrigés **Exercise 1**: Where is $h(x) = \frac{x^2 + 1}{x - 3}$ continuous? **Solution**: Numerator $x^2+1$ continuous everywhere (polynomial). Denominator $x-3$ continuous everywhere, zero at $x=3$. Quotient continuous where denominator $\neq 0$ → $(-\infty, 3) \cup (3, \infty)$. **Exercise 2**: Let $f(x) = \begin{cases} x & \text{if } x < 1 \\ 2 & \text{if } x \geq 1 \end{cases}$, $g(x) = x^2$. Is $f \cdot g$ continuous at $x=1$? **Solution**: $f$ discontinuous at $x=1$ (jump: $\lim_{x \to 1^-} f = 1$, $\lim_{x \to 1^+} f = 2$, $f(1)=2$). $g$ continuous everywhere. Product: $(f \cdot g)(x) = \begin{cases} x \cdot x^2 = x^3 & x < 1 \\ 2 \cdot x^2 & x \geq 1 \end{cases}$. At $x=1$: $(f \cdot g)(1) = 2 \cdot 1 = 2$. $\lim_{x \to 1^-} (f \cdot g)(x) = \lim_{x \to 1^-} x^3 = 1$. $\lim_{x \to 1^+} (f \cdot g)(x) = \lim_{x \to 1^+} 2x^2 = 2$. Left limit $=1 \neq$ right limit $=2$ → jump discontinuity. $\therefore f \cdot g$ is **not** continuous at $x=1$, even though $g$ is continuous. *(Note: Theorem requires both $f$ and $g$ continuous at $a$; here $f$ is not.)* --- ## VI. Continuity of the composite of two functions The composition of continuous functions is continuous. > **Theorem (Continuity of Composite Functions)**: > If $g$ is continuous at $x = a$, and $f$ is continuous at $x = g(a)$, then the composite function $f \circ g$ defined by $(f \circ g)(x) = f(g(x))$ is continuous at $x = a$. This follows from: $\lim_{x \to a} f(g(x)) = f\left( \lim_{x \to a} g(x) \right) = f(g(a))$, since $g$ continuous at $a$ implies $\lim_{x \to a} g(x) = g(a)$, and $f$ continuous at $g(a)$. ### Examples **Example 1**: $g(x) = x^2$ (continuous everywhere), $f(x) = \sin x$ (continuous everywhere). Then $f(g(x)) = \sin(x^2)$ is continuous everywhere. **Example 2**: $g(x) = \sqrt{x}$ (continuous on $[0,\infty)$), $f(x) = \frac{1}{x}$ (continuous on $(-\infty,0) \cup (0,\infty)$). $f \circ g(x) = \frac{1}{\sqrt{x}}$. $g$ continuous at $a>0$, $g(a)=\sqrt{a}>0$, and $f$ continuous at $\sqrt{a}$ (since $\sqrt{a} \neq 0$). So $f \circ g$ continuous on $(0,\infty)$. At $x=0$: $g(0)=0$, but $f$ not continuous at $0$ → $f \circ g$ not continuous at $0$. **Example 3**: $g(x) = |x|$ (continuous everywhere), $f(x) = x^2$ (continuous everywhere). $f(g(x)) = (|x|)^2 = x^2$ → continuous everywhere. ### Exercices d'application corrigés **Exercise 1**: Let $g(x) = x + 1$, $f(x) = \frac{1}{x}$. Is $f \circ g$ continuous at $x=0$? **Solution**: $g(0) = 0+1 = 1$. $g$ continuous at $0$ (polynomial). $f$ continuous at $x=1$? Yes, since $f(x)=1/x$ continuous for $x \neq 0$, and $1 \neq 0$. Thus $f \circ g$ continuous at $x=0$. $(f \circ g)(x) = f(g(x)) = \frac{1}{x+1}$, which is indeed continuous at $x=0$ (value $=1$). **Exercise 2**: Let $g(x) = \sin x$, $f(x) = \sqrt{x}$. Is $f \circ g$ continuous at $x=0$? **Solution**: $g(0) = \sin 0 = 0$. $g$ continuous at $0$ (yes). $f$ continuous at $x=0$? $f(x)=\sqrt{x}$ continuous on $[0,\infty)$, so yes at $0$ (right-continuous, but defined and limit matches). Thus $f \circ g$ continuous at $x=0$. $(f \circ g)(x) = \sqrt{\sin x}$. At $x=0$: value $=0$. $\lim_{x \to 0} \sqrt{\sin x} = \sqrt{\lim_{x \to 0} \sin x} = \sqrt{0} = 0$ → continuous. **Exercise 3**: Let $g(x) = \frac{1}{x}$, $f(x) = x^2$. Is $f \circ g$ continuous at $x=0$? **Solution**: $g$ not defined at $x=0$ → $g$ not continuous at $0$. Theorem requires $g$ continuous at $a$; condition fails. Indeed, $(f \circ g)(x) = \left(\frac{1}{x}\right)^2 = \frac{1}{x^2}$, undefined at $x=0$, limit $+\infty$ → not continuous. --- ## VII. Continuity on an interval I. A function is continuous on an interval if it is continuous at every point in that interval. > **Definition (Continuity on an Interval)**: > A function $f$ is continuous on an open interval $(a,b)$ if it is continuous at every $x \in (a,b)$. > It is continuous on a closed interval $[a,b]$ if: > - It is continuous on $(a,b)$, > - $\lim_{x \to a^+} f(x) = f(a)$ (right-continuous at $a$), > - $\lim_{x \to b^-} f(x) = f(b)$ (left-continuous at $b$). For half-open intervals, adjust accordingly. ### Examples **Example 1**: $f(x) = x^2$ is continuous on $(-\infty, \infty)$ (everywhere). **Example 2**: $f(x) = \frac{1}{x}$ is continuous on $(-\infty, 0)$ and $(0, \infty)$, but not on $[-1,1]$ because discontinuous at $0$. **Example 3**: $f(x) = \sqrt{x}$ is continuous on $[0, \infty)$: - Continuous on $(0,\infty)$ (as composition of continuous functions), - Right-continuous at $0$: $\lim_{x \to 0^+} \sqrt{x} = 0 = f(0)$. **Example 4**: $f(x) = \begin{cases} x & \text{if } 0 \leq x < 1 \\ 2 & \text{if } x = 1 \end{cases}$ on $[0,1]$. Continuous on $(0,1)$? Yes (identity). At $x=0$: right-continuous? $\lim_{x \to 0^+} f(x) = 0 = f(0)$ → yes. At $x=1$: left-continuous? $\lim_{x \to 1^-} f(x) = 1$, but $f(1)=2$ → not left-continuous. So not continuous on $[0,1]$. ### Exercices d'application corrigés **Exercise 1**: Determine where $f(x) = \frac{x+2}{x^2 - 4}$ is continuous. **Solution**: Factor denominator: $x^2 - 4 = (x-2)(x+2)$. So $f(x) = \frac{x+2}{(x-2)(x+2)} = \frac{1}{x-2}$ for $x \neq -2$. Discontinuous where denominator zero: $x=2$ and $x=-2$. At $x=-2$: original form undefined, but simplified form $\frac{1}{x-2}$ has limit $\frac{1}{-4} = -0.25$. So removable discontinuity at $x=-2$, infinite at $x=2$. Thus continuous on $(-\infty, -2) \cup (-2, 2) \cup (2, \infty)$. **Exercise 2**: Is $f(x) = \begin{cases} x^2 & \text{if } x < 1 \\ 2x & \text{if } x \geq 1 \end{cases}$ continuous on $[0,2]$? **Solution**: Check continuity at $x=1$ (only potential issue). $f(1) = 2(1) = 2$. $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} x^2 = 1$. $\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} 2x = 2$. Left limit $=1 \neq$ right limit $=2$ → jump discontinuity at $x=1$. Thus not continuous on $[0,2]$ (fails at interior point). --- ## VIII. The Intermediate Value theorem for continuous functions The Intermediate Value Theorem (IVT) guarantees that a continuous function takes on all values between its endpoints. > **Theorem (Intermediate Value Theorem)**: > Let $f$ be continuous on the closed interval $[a,b]$, and let $N$ be any number between $f(a)$ and $f(b)$. > Then there exists at least one $c \in (a,b)$ such that $f(c) = N$. This relies on the completeness of the real numbers and the definition of continuity. It does not hold for discontinuous functions. ### Key Points - IVT only applies if $f$ is continuous on $[a,b]$. - It guarantees existence of $c$, not uniqueness or how to find it. - Common application: proving roots exist (set $N=0$). ### Examples **Example 1**: $f(x) = x^3 - x - 2$ on $[1,2]$. $f(1) = 1 - 1 - 2 = -2$, $f(2) = 8 - 2 - 2 = 4$. Since $-2 < 0 < 4$ and $f$ continuous (polynomial), IVT guarantees $\exists c \in (1,2)$ s.t. $f(c)=0$. Indeed, $f(1.5)=3.375 - 1.5 - 2 = -0.125$, $f(1.6)=4.096 - 1.6 - 2 = 0.496$ → root in $(1.5,1.6)$. **Example 2**: $f(x) = \cos x$ on $[0, \pi]$. $f(0)=1$, $f(\pi)=-1$. For any $N \in [-1,1]$, $\exists c \in (0,\pi)$ s.t. $\cos c = N$. E.g., $N=0$ → $c=\pi/2$. **Example 3**: $f(x) = \frac{1}{x}$ on $[-1,1]$. Not continuous on $[-1,1]$ (discontinuous at $0$), so IVT does not apply. Indeed, $f(-1)=-1$, $f(1)=1$, but $f(x)$ never takes value $0$ (no $c$ s.t. $1/c=0$). ### Exercices d'application corrigés **Exercise 1**: Use IVT to show that $f(x) = x^5 - x + 1$ has a root in $[-2,-1]$. **Solution**: $f$ is polynomial → continuous everywhere. $f(-2) = (-32) - (-2) + 1 = -32 + 2 + 1 = -29$. $f(-1) = (-1) - (-1) + 1 = -1 + 1 + 1 = 1$. Since $f(-2) = -29 < 0 < 1 = f(-1)$, by IVT, $\exists c \in (-2,-1)$ such that $f(c)=0$. **Exercise 2**: Show that the equation $\sin x = x - 1$ has a solution in $[0, \pi/2]$. **Solution**: Rewrite as $f(x) = \sin x - x + 1 = 0$. $f$ continuous (sum of continuous functions). $f(0) = \sin 0 - 0 + 1 = 0 - 0 + 1 = 1$. $f(\pi/2) = \sin(\pi/2) - \pi/2 + 1 = 1 - \pi/2 + 1 = 2 - \pi/2 \approx 2 - 1.57 = 0.43 > 0$. Both positive? Wait, need sign change. Try $x=2$: but $[0,\pi/2] \approx [0,1.57]$, $f(1.5) = \sin(1.5) - 1.5 + 1 \approx 0.997 - 0.5 = 0.497 > 0$. $f(0)=1>0$, $f(\pi/2)>0$ → no sign change. Mistake: equation is $\sin x = x - 1$ → $f(x) = \sin x - (x - 1) = \sin x - x + 1$. At $x=0$: $0 - 0 + 1 = 1 > 0$. At $x=\pi/2 \approx 1.57$: $1 - 1.57 + 1 = 0.43 > 0$. Still positive. Try $x=2$ (outside interval): $\sin 2 \approx 0.909$, $2-1=1$ → $0.909 < 1$ → $f(2)<0$. But $[0,\pi/2]$ ends at ~1.57. Check $x=1.8$: $\sin 1.8 \approx 0.974$, $1.8-1=0.8$ → $0.974 > 0.8$ → $f>0$. Actually, $f(x) = \sin x - x + 1$, derivative $f'(x)=\cos x - 1 \leq 0$, so $f$ decreasing on $[0,\pi/2]$. $f(0)=1$, $f(\pi/2)=2 - \pi/2 \approx 0.43 > 0$ → always positive → no root in $[0,\pi/2]$. **Correction**: Use interval $[\pi/2, \pi]$. $f(\pi/2) \approx 0.43 > 0$, $f(\pi) = \sin \pi - \pi + 1 = 0 - \pi + 1 \approx -2.14 < 0$. So root in $[\pi/2, \pi]$. But for $[0,\pi/2]$, no root. **Revised Exercise**: Show $\sin x = x - 1$ has a solution in $[\pi/2, \pi]$. **Solution**: $f(x) = \sin x - x + 1$ continuous. $f(\pi/2) = 1 - \pi/2 + 1 = 2 - \pi/2 > 0$ (since $\pi<4$). $f(\pi) = 0 - \pi + 1 = 1 - \pi < 0$. Since $f(\pi/2)>0$ and $f(\pi)<0$, by IVT, $\exists c \in (\pi/2, \pi)$ s.t. $f(c)=0$ → $\sin c = c - 1$. **Exercise 3**: Does IVT apply to $f(x) = \frac{1}{x}$ on $[-1,1]$ to conclude there is a $c$ with $f(c)=0$? **Solution**: No, because $f$ is not continuous on $[-1,1]$ (discontinuous at $x=0$). IVT requires continuity on the closed interval; condition fails. Indeed, no such $c$ exists since $1/c=0$ has no solution. --- ## IX. Application of the Intermediate Value Theorem (I.V.T.) IVT is widely used to prove existence of solutions, roots, and fixed points. ### Key Applications 1. **Root location**: If $f(a)$ and $f(b)$ have opposite signs and $f$ continuous on $[a,b]$, then $\exists c \in (a,b)$ with $f(c)=0$. 2. **Fixed points**: If $f: [a,b] \to [a,b]$ is continuous, then $\exists c \in [a,b]$ s.t. $f(c)=c$ (consider $g(x)=f(x)-x$). 3. **Equilibrium**: In physics/chemistry, IVT shows equilibrium states exist. 4. **Range coverage**: Continuous functions on intervals attain all intermediate values. ### Examples **Example 1**: Prove $\sqrt{2}$ exists using IVT. Let $f(x)=x^2-2$ on $[1,2]$. $f(1)=-1<0$, $f(2)=4-2=2>0$, $f$ continuous → $\exists c \in (1,2)$ s.t. $c^2-2=0$ → $c=\sqrt{2}$. **Example 2**: Show that any continuous function $f: [0,1] \to [0,1]$ has a fixed point. Define $g(x)=f(x)-x$. $g$ continuous (difference of continuous). $g(0)=f(0)-0 \geq 0$ (since $f(0)\in[0,1]$). $g(1)=f(1)-1 \leq 0$ (since $f(1)\in[0,1]$). If $g(0)=0$ or $g(1)=0$, fixed point at endpoint. Else, $g(0)>0$ and $g(1)<0$ → by IVT, $\exists c \in (0,1)$ s.t. $g(c)=0$ → $f(c)=c$. **Example 3**: A thermometer in a room reads $15^\circ C$ at 8am and $25^\circ C$ at 12pm. Show it must have read $20^\circ C$ at some time. Let $T(t)$ be temperature at time $t$ (hours after 8am). $T$ continuous (temperature changes continuously). $T(0)=15$, $T(4)=25$. Since $15 < 20 < 25$, by IVT, $\exists c \in (0,4)$ s.t. $T(c)=20$. ### Exercices d'application corrigés **Exercise 1**: Use IVT to show that the equation $e^x = 3x$ has a solution in $(0,1)$. **Solution**: Define $f(x)=e^x - 3x$. $f$ continuous (exponential and polynomial). $f(0)=e^0 - 0 = 1 - 0 = 1 > 0$. $f(1)=e^1 - 3 \approx 2.718 - 3 = -0.282 < 0$. Since $f(0)>0$ and $f(1)<0$, by IVT, $\exists c \in (0,1)$ s.t. $f(c)=0$ → $e^c = 3c$. **Exercise 2**: Prove that the function $f(x)=x^3 - 2x + 2$ has a root in $[-2,-1]$. **Solution**: $f$ polynomial → continuous. $f(-2)=(-8) - 2(-2) + 2 = -8 + 4 + 2 = -2$. $f(-1)=(-1) - 2(-1) + 2 = -1 + 2 + 2 = 3$. $f(-2)=-2<0<3=f(-1)$ → by IVT, $\exists c \in (-2,-1)$ s.t. $f(c)=0$. **Exercise 3**: A monk climbs a mountain path starting at 6am and reaches the top at 6pm. He spends the night at the top. Next day, he descends the same path, starting at 6am and reaching the bottom at 6pm. Prove there is a time when he is at the same point on the path at the same time on both days. **Solution**: Let $f(t)$ = position on path (distance from bottom) at time $t$ hours after 6am on day 1 (ascent). Let $g(t)$ = position on path at time $t$ hours after 6am on day 2 (descent). Both $f$ and $g$ continuous (motion is continuous). $f(0)=0$ (bottom at 6am day1), $f(12)=L$ (top at 6pm day1, $L$=path length). $g(0)=L$ (top at 6am day2), $g(12)=0$ (bottom at 6pm day2). Define $h(t)=f(t)-g(t)$. $h$ continuous. $h(0)=0 - L = -L < 0$. $h(12)=L - 0 = L > 0$. Since $h(0)<0< h(12)$, by IVT, $\exists c \in (0,12)$ s.t. $h(c)=0$ → $f(c)=g(c)$. Thus at time $c$ hours after 6am, he is at same point on both days. ---