The Hook: The Speed Camera Paradox
John Abraham rides a superbike on a highway. Two speed cameras are 10 km apart.
Camera 1 (2:00 PM): Photo taken Camera 2 (2:05 PM): Photo taken
Time difference = 5 minutes = 1/12 hour
Average speed = 10 km ÷ (1/12 hour) = 120 km/h
Speed limit = 80 km/h
Cop: “You were speeding!” John: “But both cameras showed I was at 79 km/h at those exact moments!”
Question: Can John escape the ticket?
Answer: NO! Rolle’s theorem proves that at some point between the cameras, his instantaneous speed must have been 120 km/h!
This is the power of Mean Value Theorems — they guarantee certain things happen, even if you don’t see them directly.
Interactive: Visualize Mean Value Theorems
See how the tangent line parallels the secant line somewhere in between!
Rolle’s Theorem
Statement
Rolle’s Theorem:
Let $f(x)$ be a function such that:
- $f$ is continuous on $[a, b]$
- $f$ is differentiable on $(a, b)$
- $f(a) = f(b)$ (same value at endpoints)
Then there exists at least one point $c \in (a, b)$ such that:
$$\boxed{f'(c) = 0}$$In words: “If you return to the same height, you must have reached a turning point somewhere!”
Geometric Interpretation
Imagine climbing a mountain and returning to the same elevation — you must have reached a peak or valley somewhere (where slope = 0).
Graph: If endpoints have same $y$-value and curve is smooth, there’s at least one horizontal tangent in between.
Example 1
Verify Rolle’s theorem for $f(x) = x^2 - 5x + 6$ on $[2, 3]$.
Check conditions:
- Polynomial → continuous on $[2, 3]$ ✓
- Polynomial → differentiable on $(2, 3)$ ✓
- $f(2) = 4 - 10 + 6 = 0$, $f(3) = 9 - 15 + 6 = 0$ → $f(2) = f(3)$ ✓
Find $c$:
$$f'(x) = 2x - 5 = 0 \Rightarrow x = \frac{5}{2} = 2.5$$Since $2 < 2.5 < 3$, $c = 2.5 \in (2, 3)$ ✓
Rolle’s theorem is verified!
Example 2: Application
Show that $x^3 + x - 1 = 0$ has exactly one real root.
Solution:
Let $f(x) = x^3 + x - 1$
Existence of root:
- $f(0) = -1 < 0$
- $f(1) = 1 > 0$
By Intermediate Value Theorem, there exists $c \in (0, 1)$ where $f(c) = 0$ → At least one root exists.
Uniqueness (only one root):
Assume there are two roots $a$ and $b$ with $a < b$, so $f(a) = f(b) = 0$.
By Rolle’s theorem, there exists $c \in (a, b)$ where $f'(c) = 0$.
But $f'(x) = 3x^2 + 1 > 0$ for all $x$ (always positive!)
Contradiction! So there cannot be two roots.
Conclusion: Exactly one real root exists. ∎
Lagrange’s Mean Value Theorem (LMVT)
Statement
Lagrange’s Mean Value Theorem (LMVT):
Let $f(x)$ be a function such that:
- $f$ is continuous on $[a, b]$
- $f$ is differentiable on $(a, b)$
Then there exists at least one point $c \in (a, b)$ such that:
$$\boxed{f'(c) = \frac{f(b) - f(a)}{b - a}}$$In words: “Instantaneous rate at some point = Average rate over the interval”
Geometric Interpretation
The slope of the tangent at some point $c$ equals the slope of the chord joining $(a, f(a))$ and $(b, f(b))$.
Physical meaning (Speed Camera paradox):
Instantaneous velocity at some moment = Average velocity over the journey
Connection to Rolle’s Theorem
LMVT is a generalization of Rolle’s theorem!
If $f(a) = f(b)$, then $\frac{f(b) - f(a)}{b - a} = 0$, so LMVT says $f'(c) = 0$ (which is Rolle’s theorem).
Example 1
Verify LMVT for $f(x) = x^2$ on $[1, 3]$.
Check conditions:
- Polynomial → continuous on $[1, 3]$ ✓
- Polynomial → differentiable on $(1, 3)$ ✓
Find $c$:
$$\frac{f(3) - f(1)}{3 - 1} = \frac{9 - 1}{2} = 4$$ $$f'(x) = 2x$$Set $f'(c) = 4$:
$$2c = 4 \Rightarrow c = 2$$Since $1 < 2 < 3$, $c = 2 \in (1, 3)$ ✓
LMVT is verified!
Example 2: Application
If $f'(x) = 0$ for all $x$ in an interval, prove that $f(x)$ is constant.
Proof:
Let $a, b$ be any two points in the interval with $a < b$.
By LMVT, there exists $c \in (a, b)$ such that:
$$f'(c) = \frac{f(b) - f(a)}{b - a}$$Given $f'(x) = 0$ for all $x$, so $f'(c) = 0$:
$$0 = \frac{f(b) - f(a)}{b - a} \Rightarrow f(b) = f(a)$$Since $a$ and $b$ are arbitrary, $f$ is constant. ∎
Cauchy’s Mean Value Theorem
Statement
Cauchy’s Mean Value Theorem:
Let $f(x)$ and $g(x)$ be functions such that:
- Both are continuous on $[a, b]$
- Both are differentiable on $(a, b)$
- $g'(x) \neq 0$ for any $x \in (a, b)$
Then there exists at least one point $c \in (a, b)$ such that:
$$\boxed{\frac{f'(c)}{g'(c)} = \frac{f(b) - f(a)}{g(b) - g(a)}}$$Connection to LMVT
If we take $g(x) = x$, then $g'(x) = 1$ and:
$$\frac{f'(c)}{1} = \frac{f(b) - f(a)}{b - a}$$This is exactly Lagrange’s MVT! So LMVT is a special case of Cauchy’s MVT.
Parametric Form Interpretation
If $x = g(t)$ and $y = f(t)$ (parametric), then:
$$\frac{dy/dt}{dx/dt}\bigg|_{t=c} = \frac{y(b) - y(a)}{x(b) - x(a)}$$The slope of the tangent at some $t = c$ equals the slope of the chord.
L’Hospital’s Rule
The Indeterminate Form Problem
Limits like $\lim_{x \to 0} \frac{\sin x}{x}$ give $\frac{0}{0}$ form (indeterminate).
L’Hospital’s Rule is a powerful technique to evaluate such limits!
Statement
L’Hospital’s Rule:
If $\lim_{x \to a} f(x) = 0$ and $\lim_{x \to a} g(x) = 0$ (or both $\to \infty$), and if $\lim_{x \to a} \frac{f'(x)}{g'(x)}$ exists, then:
$$\boxed{\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}}$$Applicable for indeterminate forms:
- $\frac{0}{0}$
- $\frac{\infty}{\infty}$
Other forms can be converted: $0 \cdot \infty$, $\infty - \infty$, $0^0$, $1^\infty$, $\infty^0$
Example 1: Basic Application
Evaluate $\lim_{x \to 0} \frac{\sin x}{x}$
Direct substitution: $\frac{0}{0}$ (indeterminate)
Apply L’Hospital:
$$\lim_{x \to 0} \frac{\sin x}{x} = \lim_{x \to 0} \frac{(\sin x)'}{(x)'} = \lim_{x \to 0} \frac{\cos x}{1} = 1$$Example 2: Multiple Applications
Evaluate $\lim_{x \to 0} \frac{e^x - 1 - x}{x^2}$
Form: $\frac{0}{0}$
First application:
$$\lim_{x \to 0} \frac{e^x - 1 - x}{x^2} = \lim_{x \to 0} \frac{e^x - 1}{2x}$$Still $\frac{0}{0}$!
Second application:
$$= \lim_{x \to 0} \frac{e^x}{2} = \frac{1}{2}$$Example 3: $\frac{\infty}{\infty}$ Form
Evaluate $\lim_{x \to \infty} \frac{\ln x}{x}$
Form: $\frac{\infty}{\infty}$
Apply L’Hospital:
$$\lim_{x \to \infty} \frac{\ln x}{x} = \lim_{x \to \infty} \frac{1/x}{1} = \lim_{x \to \infty} \frac{1}{x} = 0$$Example 4: Converting $0 \cdot \infty$ Form
Evaluate $\lim_{x \to 0^+} x \ln x$
Form: $0 \cdot (-\infty)$ (indeterminate)
Convert to $\frac{0}{0}$ or $\frac{\infty}{\infty}$:
$$x \ln x = \frac{\ln x}{1/x}$$(now $\frac{-\infty}{\infty}$ form)
Apply L’Hospital:
$$\lim_{x \to 0^+} \frac{\ln x}{1/x} = \lim_{x \to 0^+} \frac{1/x}{-1/x^2} = \lim_{x \to 0^+} \frac{x^2}{-x} = \lim_{x \to 0^+} (-x) = 0$$Example 5: $1^\infty$ Form
Evaluate $\lim_{x \to 0^+} (1 + x)^{1/x}$
Form: $1^\infty$ (indeterminate)
Method: Let $y = (1 + x)^{1/x}$, take $\ln$:
$$\ln y = \frac{1}{x} \ln(1 + x) = \frac{\ln(1 + x)}{x}$$Form: $\frac{0}{0}$
Apply L’Hospital:
$$\lim_{x \to 0^+} \frac{\ln(1 + x)}{x} = \lim_{x \to 0^+} \frac{1/(1+x)}{1} = 1$$So $\lim \ln y = 1 \Rightarrow \lim y = e^1 = e$
$$\boxed{\lim_{x \to 0^+} (1 + x)^{1/x} = e}$$Important Results and Applications
Result 1: Increasing/Decreasing Functions
Using MVT to test monotonicity:
Let $f$ be continuous on $[a, b]$ and differentiable on $(a, b)$.
- If $f'(x) > 0$ for all $x \in (a, b)$, then $f$ is strictly increasing on $[a, b]$
- If $f'(x) < 0$ for all $x \in (a, b)$, then $f$ is strictly decreasing on $[a, b]$
Proof idea: Use LMVT on any two points to show $f(b) > f(a)$ when $f' > 0$.
Result 2: Bounding Functions
If $|f'(x)| \leq M$ on $[a, b]$, then by LMVT:
$$|f(b) - f(a)| = |f'(c)| \cdot |b - a| \leq M|b - a|$$This bounds the change in $f$.
Common Indeterminate Forms and Techniques
| Form | Example | Technique |
|---|---|---|
| $\frac{0}{0}$ | $\frac{\sin x}{x}$ | Direct L’Hospital |
| $\frac{\infty}{\infty}$ | $\frac{x^2}{e^x}$ | Direct L’Hospital |
| $0 \cdot \infty$ | $x \ln x$ | Rewrite as $\frac{\ln x}{1/x}$ |
| $\infty - \infty$ | $x - \ln x$ | Combine into single fraction |
| $1^\infty$ | $(1+x)^{1/x}$ | Take $\ln$, apply L’Hospital, exponentiate |
| $0^0$ | $x^x$ | Take $\ln$, apply L’Hospital, exponentiate |
| $\infty^0$ | $x^{1/x}$ | Take $\ln$, apply L’Hospital, exponentiate |
Memory Tricks
“Return to the same height → Must have turned around somewhere”
If $f(a) = f(b)$, then $f'(c) = 0$ for some $c$.
Think: Throwing a ball up — it returns to hand height, so velocity was zero at the peak!
“At some moment, your speed equals your average speed”
$$f'(c) = \frac{f(b) - f(a)}{b - a}$$Speed camera paradox!
“When in doubt, differentiate separately!”
$$\lim \frac{f(x)}{g(x)} = \lim \frac{f'(x)}{g'(x)}$$(for $\frac{0}{0}$ or $\frac{\infty}{\infty}$)
NOT $\left(\frac{f}{g}\right)'$ — differentiate numerator and denominator separately!
Take Logarithm for $1^\infty$, $0^0$, $\infty^0$ forms!
Then apply L’Hospital, then exponentiate: $\lim y = e^{\lim \ln y}$
Practice Problems
Level 1: Foundation (NCERT)
Question: Verify Rolle’s theorem for $f(x) = x(x-1)$ on $[0, 1]$.
Solution:
$f(0) = 0$, $f(1) = 0$ → $f(0) = f(1)$ ✓
Continuous and differentiable (polynomial) ✓
$$f'(x) = 2x - 1 = 0 \Rightarrow x = \frac{1}{2} \in (0, 1)$$✓
Question: Evaluate $\lim_{x \to 0} \frac{e^x - 1}{x}$ using L’Hospital’s rule.
Solution:
Form: $\frac{0}{0}$
$$\lim_{x \to 0} \frac{e^x - 1}{x} = \lim_{x \to 0} \frac{e^x}{1} = 1$$Level 2: JEE Main
Question: Verify LMVT for $f(x) = x^3 - 3x$ on $[0, 2]$ and find the value of $c$.
Solution:
$$\frac{f(2) - f(0)}{2 - 0} = \frac{(8-6) - 0}{2} = 1$$ $$f'(x) = 3x^2 - 3$$Set $f'(c) = 1$:
$$3c^2 - 3 = 1 \Rightarrow c^2 = \frac{4}{3} \Rightarrow c = \frac{2}{\sqrt{3}} \in (0, 2)$$✓
Question: Evaluate $\lim_{x \to 0} \frac{\tan x - x}{x^3}$
Solution:
Form: $\frac{0}{0}$
First L’Hospital:
$$\lim_{x \to 0} \frac{\sec^2 x - 1}{3x^2}$$Still $\frac{0}{0}$!
Second L’Hospital:
$$\lim_{x \to 0} \frac{2\sec^2 x \tan x}{6x}$$Still $\frac{0}{0}$!
Third L’Hospital:
$$\lim_{x \to 0} \frac{2[\sec^2 x \sec^2 x + \tan x \cdot 2\sec^2 x \tan x]}{6}$$ $$= \frac{2[1 \cdot 1 + 0]}{6} = \frac{1}{3}$$Question: Evaluate $\lim_{x \to \infty} \frac{x^2}{e^x}$
Solution:
Form: $\frac{\infty}{\infty}$
First L’Hospital:
$$\lim_{x \to \infty} \frac{2x}{e^x}$$Still $\frac{\infty}{\infty}$!
Second L’Hospital:
$$\lim_{x \to \infty} \frac{2}{e^x} = 0$$Conclusion: Exponential grows faster than polynomial!
Level 3: JEE Advanced
Question: Show that the equation $x^5 + x - 1 = 0$ has exactly one real root in $(0, 1)$.
Solution:
Let $f(x) = x^5 + x - 1$
Existence: $f(0) = -1 < 0$, $f(1) = 1 > 0$
By IVT, there exists $c \in (0, 1)$ where $f(c) = 0$. ✓
Uniqueness: Assume two roots $a, b$ with $0 < a < b < 1$ and $f(a) = f(b) = 0$.
By Rolle’s theorem, $\exists c \in (a, b)$ where $f'(c) = 0$.
But $f'(x) = 5x^4 + 1 > 0$ for all $x$ (always positive!).
Contradiction! So exactly one root exists. ∎
Question: Evaluate $\lim_{x \to 0} \frac{x - \sin x}{x^3}$ without L’Hospital’s rule, then verify using L’Hospital.
Solution (Series Method):
$$\sin x = x - \frac{x^3}{6} + \frac{x^5}{120} - \ldots$$ $$x - \sin x = \frac{x^3}{6} - \frac{x^5}{120} + \ldots$$ $$\frac{x - \sin x}{x^3} = \frac{1}{6} - \frac{x^2}{120} + \ldots \to \frac{1}{6}$$Using L’Hospital (3 times):
$$\lim_{x \to 0} \frac{x - \sin x}{x^3} = \lim_{x \to 0} \frac{1 - \cos x}{3x^2}$$ $$= \lim_{x \to 0} \frac{\sin x}{6x} = \lim_{x \to 0} \frac{\cos x}{6} = \frac{1}{6}$$✓
Question: Evaluate $\lim_{x \to 0} \left(\frac{\sin x}{x}\right)^{1/x^2}$
Solution:
Form: $1^\infty$
Let $y = \left(\frac{\sin x}{x}\right)^{1/x^2}$
$$\ln y = \frac{1}{x^2} \ln\left(\frac{\sin x}{x}\right) = \frac{\ln(\sin x) - \ln x}{x^2}$$Form: $\frac{-\infty}{\infty} = \frac{\infty}{\infty}$ after adjusting signs
Actually, as $x \to 0$: $\sin x \approx x - \frac{x^3}{6}$
$$\frac{\sin x}{x} \approx 1 - \frac{x^2}{6} \to 1$$So $\ln\left(\frac{\sin x}{x}\right) \approx \ln\left(1 - \frac{x^2}{6}\right) \approx -\frac{x^2}{6}$
$$\lim \ln y = \lim_{x \to 0} \frac{-x^2/6}{x^2} = -\frac{1}{6}$$ $$\lim y = e^{-1/6}$$Common Mistakes to Avoid
Wrong: $\lim \frac{f(x)}{g(x)} = \lim \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}$ ✗
Correct: $\lim \frac{f(x)}{g(x)} = \lim \frac{f'(x)}{g'(x)}$ ✓
Differentiate numerator and denominator separately!
Wrong: $\lim_{x \to 0} \frac{x^2 + 1}{x + 2} = \lim_{x \to 0} \frac{2x}{1}$ ✗
The original limit is $\frac{1}{2}$ (not indeterminate), so don’t use L’Hospital!
Only use L’Hospital for $\frac{0}{0}$ or $\frac{\infty}{\infty}$ forms!
Before applying MVT/Rolle’s:
- Check continuity on $[a, b]$
- Check differentiability on $(a, b)$
- For Rolle’s, check $f(a) = f(b)$
Don’t just mechanically solve $f'(x) = 0$!
Quick Revision Box
| Theorem | Conditions | Conclusion |
|---|---|---|
| Rolle’s | Continuous on $[a,b]$, differentiable on $(a,b)$, $f(a)=f(b)$ | $\exists c: f'(c) = 0$ |
| Lagrange MVT | Continuous on $[a,b]$, differentiable on $(a,b)$ | $\exists c: f'(c) = \frac{f(b)-f(a)}{b-a}$ |
| Cauchy MVT | Both continuous on $[a,b]$, differentiable on $(a,b)$, $g' \neq 0$ | $\exists c: \frac{f'(c)}{g'(c)} = \frac{f(b)-f(a)}{g(b)-g(a)}$ |
| L’Hospital | $\frac{0}{0}$ or $\frac{\infty}{\infty}$ form | $\lim \frac{f}{g} = \lim \frac{f'}{g'}$ |
JEE Strategy Tips
Weightage: MVT appears in 2-3 JEE questions, L’Hospital in 3-4 questions (very common!).
Time-Saver: For “exactly one root” problems, combine IVT (existence) + Rolle’s (uniqueness via contradiction).
Common Pattern: JEE loves:
- $\frac{0}{0}$ forms with trig/exponential
- $1^\infty$ forms (use logarithm!)
- Proving monotonicity using $f' > 0$
Trap Alert: L’Hospital can be applied repeatedly, but check the form each time! Don’t blindly keep differentiating.
Advanced Tip: Sometimes series expansion is faster than multiple L’Hospital applications.
Exam Hack: If L’Hospital gets too messy after 2-3 attempts, try another method (series, algebraic manipulation).
Teacher’s Summary
Rolle’s theorem: Equal endpoints → horizontal tangent somewhere ($f'(c) = 0$)
Lagrange MVT: Instantaneous rate at some point = Average rate over interval
L’Hospital’s rule: For $\frac{0}{0}$ or $\frac{\infty}{\infty}$, differentiate top and bottom separately
Applications: Proving uniqueness of roots, testing monotonicity, evaluating tricky limits
Converting indeterminate forms: Use algebra, logarithms, or rewrite to apply L’Hospital
“Mean Value Theorems: The guarantee that somewhere in between, something special happens!”
Related Topics
Within Limits, Continuity & Differentiability
- Continuity — MVT requires continuity
- Differentiability — MVT requires differentiability
- Limits — L’Hospital evaluates limits
- Standard Limits — Many can be proved using L’Hospital
- Maxima-Minima — Rolle’s theorem connects to extrema
Applications
- Curve Sketching — Using monotonicity from MVT
- Differential Equations — Existence and uniqueness theorems
Mathematical Foundations
- Functions — Understanding behavior on intervals