Limit, Continuity and Differentiability

These come up constantly. Memorize them, recognize them, use them — most JEE limit problems collapse to one of these.

1. lim(x→0) [sin x / x] = 1

2. lim(x→0) [(1 − cos x) / x²] = 1/2

3. lim(x→0) [tan x / x] = 1

4. lim(x→0) [(eˣ − 1) / x] = 1

5. lim(x→0) [(aˣ − 1) / x] = ln a

6. lim(x→0) [ln(1 + x) / x] = 1

7. lim(x→0) [(1 + x)^(1/x)] = e

8. lim(x→a) [(xⁿ − aⁿ) / (x − a)] = n · a^(n−1)

Indeterminate forms that often appear:

• 0/0
• ∞/∞
• ∞ − ∞
• 0 · ∞
• 1^∞
• 0^0
• ∞^0

For each, transform to a form where you can apply L'Hôpital, factor, rationalize, or use a standard limit.

L'Hôpital's rule (for 0/0 or ∞/∞ only):
lim(x→a) f(x)/g(x) = lim(x→a) f'(x)/g'(x)

provided the latter exists. Useful but easy to overuse — often the standard limits or factoring are faster.

Worked example. lim(x→0) [sin 5x / sin 3x] = lim(x→0) [(sin 5x / 5x) · (3x / sin 3x) · (5/3)] = 1 · 1 · 5/3 = 5/3.

A function f is continuous at x = c if all three conditions hold:

1. f(c) is defined.
2. lim_{x→c} f(x) exists (LHL = RHL).
3. lim_{x→c} f(x) = f(c).

Geometric interpretation: you can draw the graph through x = c without lifting your pen.

TYPES OF DISCONTINUITY

1. Removable (point) discontinuity. Limit exists but f(c) ≠ limit (or f(c) undefined).
Example: f(x) = (x² − 4)/(x − 2). At x = 2, simplifies to f(x) = x + 2, so limit is 4. But f(2) is undefined. Filling in f(2) = 4 makes it continuous.

2. Jump discontinuity. LHL and RHL exist but are different.
Example: signum function sgn(x): −1 for x<0, 0 for x=0, 1 for x>0. Jump at 0.

3. Infinite discontinuity. Limit is ±∞.
Example: f(x) = 1/x at x = 0.

4. Oscillatory. Limit doesn't exist due to wild oscillation.
Example: f(x) = sin(1/x) at x = 0.

Continuity on intervals:

• Continuous on [a, b]: continuous at every point inside, and right-continuous at a, left-continuous at b.
• A polynomial is continuous everywhere.
• Rational functions are continuous except where denominator = 0.
• |x| is continuous everywhere (but not differentiable at x=0).
• e^x, sin x, cos x are continuous everywhere.

Intermediate Value Theorem (IVT): if f is continuous on [a, b] and N is between f(a) and f(b), then there exists c ∈ (a, b) with f(c) = N.

Application: prove existence of a root. If f(a) and f(b) have opposite signs and f is continuous, ∃ c in (a,b) with f(c) = 0.

DIFFERENTIABILITY

A function f is differentiable at x = c if:

f'(c) = lim_{h→0} [f(c + h) − f(c)] / h exists (and is finite).

Equivalently: LHD = RHD (left-hand and right-hand derivatives are equal).

Theorem: differentiable ⇒ continuous. (But not conversely — |x| is continuous at 0 but not differentiable.)

Non-differentiable points (even when continuous):

1. Corners (sharp bends): |x| at 0 — LHD = −1, RHD = +1.
2. Cusps: tangent line is vertical. f(x) = x^(2/3) at 0.
3. Vertical tangent: f(x) = x^(1/3) at 0 — derivative → ∞.
4. Discontinuity — automatic non-differentiability.

DIFFERENTIATION RULES

• (c)' = 0.
• (xⁿ)' = n x^(n−1).
• (sin x)' = cos x.
• (cos x)' = −sin x.
• (tan x)' = sec²x.
• (eˣ)' = eˣ.
• (ln x)' = 1/x.
• (aˣ)' = aˣ ln a.

Sum: (f + g)' = f' + g'.
Product: (fg)' = f'g + fg'.
Quotient: (f/g)' = (f'g − fg')/g².
Chain rule: (f(g(x)))' = f'(g(x)) · g'(x).

Implicit differentiation: for relations like x² + y² = 25, differentiate both sides w.r.t. x, treating y as a function. 2x + 2y(dy/dx) = 0 → dy/dx = −x/y.

Parametric differentiation: if x = f(t), y = g(t):
dy/dx = (dy/dt) / (dx/dt).

Worked example. Check continuity of f(x) = (x² − 1)/(x − 1) at x = 1.

f(1) = 0/0 — undefined. But limit:
lim_{x→1} (x² − 1)/(x − 1) = lim (x − 1)(x + 1)/(x − 1) = lim (x + 1) = 2.

Since f(1) ≠ limit (f(1) undefined), there's a removable discontinuity at x = 1.