Limits, Continuity & Differentiability
KEAM – Mathematics
Study Notes – Clear, Concise & Complete
Purpose: Quickly refresh the core ideas, remember the most common pitfalls, and practise a few typical problems that appear on the KEAM exam.
1. Simple Explanation
| Concept | What it Means (in plain words) |
|---|---|
| Limit | Describes the value a function approaches as the input gets arbitrarily close to a certain point. |
| Continuity | A function is continuous at a point if the limit at that point equals the function’s value there. |
| Differentiability | A function is differentiable at a point if the slope of the tangent line (the derivative) exists there. Differentiability implies continuity, but not vice‑versa. |
2. Key Points
2.1 Limits
| Property | Symbolic Form | Remark |
|---|---|---|
| Definition | (\displaystyle \lim_{x\to a} f(x)=L) | For every (\varepsilon>0) there exists (\delta>0) such that ( |
| Two‑sided limit | (\lim_{x\to a^-} f(x)=\lim_{x\to a^+} f(x)=L) | If both one‑sided limits exist and are equal. |
| Limit Laws | <ul><li>Additivity: (\displaystyle \lim(f+g)=\lim f+\lim g)</li><li>Multiplication: (\displaystyle \lim(fg)=\lim f\cdot\lim g)</li><li>Quotient (denominator ≠ 0): (\displaystyle \lim\frac{f}{g}=\frac{\lim f}{\lim g})</li></ul> | Apply repeatedly. |
| Indeterminate Forms | (0/0,\ \infty/\infty,\ 0\cdot\infty,\ \infty-\infty,\ 0^0,\ \infty^0,\ 1^\infty) | Requires special handling (algebra, expansions, L’Hospital). |
| L’Hospital’s Rule | If (\displaystyle \lim_{x\to a}\frac{f(x)}{g(x)}) gives (0/0) or (\infty/\infty), then (\displaystyle \lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{f'(x)}{g'(x)}) (provided the new limit exists). | Do once unless another indeterminate form appears. |
2.2 Continuity
| Criterion | Formal statement | Practical check |
|---|---|---|
| Pointwise continuity | (\displaystyle \lim_{x\to a} f(x)=f(a)) | Verify limit exists and equals (f(a)). |
| Types of discontinuity | <ul><li>Removable: limit exists but ≠ value.</li><li>Jump: left & right limits exist but differ.</li><li>Infinite (essential): limit is ±∞.</li></ul> | Look at graphs or algebraic form. |
| Continuity on an interval | Continuous at every point in the interval. | Check endpoints for one‑sided continuity. |
| Common continuous functions | Polynomials, rational (denominator ≠ 0), trigonometric, exponential, logarithmic, roots (odd roots). | These are continuous wherever they are defined. |
2.3 Differentiability
| Concept | Details |
|---|---|
| Derivative definition | (\displaystyle f'(a)=\lim_{h\to0}\frac{f(a+h)-f(a)}{h}) |
| Differentiability ⇒ Continuity | If (f) is differentiable at (a), then (f) is continuous at (a). |
| Necessary condition | The graph must have a well‑defined tangent; no corners, cusps, or vertical tangents. |
| Rules | <ul><li>Sum/Difference, Product, Quotient, Chain, Power, Exponential, Logarithmic, Trigonometric.</li></ul> |
| Derivative of (\sqrt[n]{x}) | For (n) odd: (f'(x)=\dfrac{1}{n}x^{\frac{1}{n}-1}); for (n) even: defined only for (x>0). |
| Differentiability fails | At points where the function is not continuous or where the slope tends to infinity (vertical tangent). |
3. Worked Example
Problem
Find
- (\displaystyle \lim_{x\to 1}\frac{x^2-1}{x-1}).
- The derivative of (f(x)=\sqrt{x}) at (x=0).
- (\displaystyle \lim_{x\to 0}\frac{\sin x}{x}).
Solution
1. Limit (\displaystyle \frac{x^2-1}{x-1}) as (x\to1)
- Notice that substituting (x=1) gives (0/0) (indeterminate).
- Factorise numerator: (x^2-1=(x-1)(x+1)).
- Cancel the common factor:
[ \frac{x^2-1}{x-1}=x+1 \quad (x\neq1) ] - Now take the