1. Existence of a limit. The limit of a function f(x) as x → a is said to exist and be finite when the left-hand and right-hand limits are equal and finite. In symbolic form:

2. Fundamental theorems on limits. Let g(x) → m and f(x) → l as x → a. If l and m exist (finite), then the following hold:






For example:

3. Standard limits.









4. Squeeze Theorem. If f(x) ≤ g(x) ≤ h(x) for all x near a (except possibly at a) and

with



and hence limx→a g(x) = l.
5. Indeterminate forms.

Note.
6. Useful methods to evaluate limits.
The common expansions (as referenced) are:









1. Definition of continuity at a point. A function f(x) is said to be continuous at x = c if

that is, the left-hand limit and the right-hand limit at x = c exist, are equal and equal to f(c):


In words: LHL at x = c = RHL at x = c = f(c).
Continuity at x = a is meaningful only if the function is defined in an immediate neighbourhood of a (it need not be defined at a for the limit to exist).
2. Reasons for discontinuity.


Geometrically, such points appear as breaks or jumps in the graph. For example, a typical graph may be discontinuous at x = 1, 2 and 3.
3. Type-1: Removable discontinuity. If limx→c f(x) exists but is not equal to f(c) (or f(c) is not defined), the discontinuity is removable. We can redefine f(c) suitably to make f continuous at c.
Removable discontinuities include:




Type-2: Non-removable discontinuity. If limx→c f(x) does not exist then the discontinuity cannot be removed by redefining f(c). These include:
In all cases of non-removable discontinuity, limx→a f(x) does not exist.

Note. In the adjacent example graph:
4. Jump of discontinuity and piecewise continuity. The absolute finite difference between the RHL and LHL at x = c is called the jump of discontinuity. A function with finitely many jumps in an interval I is called piecewise continuous or sectionally continuous on that interval.
5. Continuity of common functions. All polynomials, trigonometric, exponential and logarithmic functions are continuous on their domains.
6. Arithmetic of continuous functions. If f and g are continuous at x = c, then:

7. Intermediate value theorem. If f is continuous on a closed interval I and a, b ∈ I, then for any y0 between f(a) and f(b) there exists c ∈ (a, b) such that f(c) = y0. Consequently:
Important remarks about products and continuity.





7. Continuity in an interval.


Functions continuous on [a, b] satisfy the properties given above under the Intermediate Value Theorem.
8. Single point continuity. Functions that are continuous at only one point are said to have single point continuity. Example: f(x) =

1. Right-hand and left-hand derivatives. By definition, the derivative at x = a (if it exists) is

(i) The right-hand derivative f′(a+) is defined by:

provided the limit exists and is finite.
(ii) The left-hand derivative f′(a-) is defined by:

provided the limit exists and is finite.
We also write f′(a+) = f′+(a) and f′(a-) = f′-(a).
Geometrically, the existence of a finite derivative at x = a means a unique tangent of finite slope can be drawn at that point.
(iii) Relation between differentiability and continuity.
If f′(a) exists then f is differentiable at x = a ⇒ f is continuous at x = a.
Sketch of reasoning (informal):
For f′(x) =


Note: If f is differentiable at every point of its domain, then it is continuous on that domain.
The converse is not true: continuity does not imply differentiability. For example, functions like f(x) =

Important observations:
3. Derivability on an interval. A function f is said to be differentiable over an interval if it is differentiable at each point of the interval. For the closed interval [a, b], f is differentiable on [a, b] if:
Notes and algebraic properties of derivatives.

6. A useful result (product with zero value). Suppose f and g are defined on an interval containing x0. If f is differentiable at x0 with f(x0) = 0 and g is continuous at x0, then F(x) = f(x)·g(x) is differentiable at x0. Example: F(x) = sin x · x2/3 is differentiable at x = 0.
Example 1. Evaluate limx→0 (sin x)/x.
Sol.
Using the standard trigonometric limit:

Therefore limx→0 (sin x)/x = 1.
Example 2. If f(x) = x·sin(1/x) for x ≠ 0 and f(0) = 0, show f is continuous at x = 0 but not differentiable there.
Sol.
For continuity:
For x ≠ 0, |x·sin(1/x)| ≤ |x|.
As x → 0, |x| → 0, so x·sin(1/x) → 0 = f(0). Hence f is continuous at 0.
For differentiability:
Consider the difference quotient:
[f(h) - f(0)]/h = [h·sin(1/h) - 0]/h = sin(1/h).
As h → 0 this oscillates between -1 and 1 and does not have a limit. Hence f′(0) does not exist.
Example 3. Show that if f′(a) exists then f is continuous at a.
Sol.
By definition f′(a) = limh→0 [f(a + h) - f(a)]/h exists (finite).
Then limh→0 [f(a + h) - f(a)] = limh→0 h·[f(a + h) - f(a)]/h = 0.
Therefore limx→a f(x) = f(a). Hence f is continuous at a.







































































173 videos|510 docs|154 tests |
173 videos|510 docs|154 tests |
![]() | Explore Courses for JEE exam |