Limits, Continuity & Differentiability

The backbone of calculus — master limit evaluation techniques, continuity analysis, differentiation rules, and applications of derivatives for JEE Mains & Advanced.

Limits Continuity Differentiability Differentiation Rules Applications of Derivatives

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Limits

Limit evaluation is the foundation of calculus. JEE tests algebraic manipulation, standard limits, L'Hôpital's rule, and squeeze theorem extensively.

◆ Standard Limits

Must-Know Standard Limits

\(\displaystyle\lim_{x \to 0}\frac{\sin x}{x} = 1\), \(\displaystyle\lim_{x \to 0}\frac{\tan x}{x} = 1\), \(\displaystyle\lim_{x \to 0}\frac{1 - \cos x}{x^2} = \frac{1}{2}\)
\(\displaystyle\lim_{x \to 0}\frac{e^x - 1}{x} = 1\), \(\displaystyle\lim_{x \to 0}\frac{a^x - 1}{x} = \ln a\)
\(\displaystyle\lim_{x \to 0}\frac{\ln(1+x)}{x} = 1\), \(\displaystyle\lim_{x \to 0}\frac{(1+x)^n - 1}{x} = n\)
\(\displaystyle\lim_{x \to 0}(1+x)^{1/x} = e\), \(\displaystyle\lim_{n \to \infty}\left(1+\frac{1}{n}\right)^n = e\)

★ Example

Evaluate \(\displaystyle\lim_{x \to 0}\frac{\sin 5x}{\tan 3x}\).

\(\displaystyle\frac{\sin 5x}{\tan 3x} = \frac{\sin 5x}{5x} \cdot \frac{3x}{\tan 3x} \cdot \frac{5}{3} \to 1 \cdot 1 \cdot \frac{5}{3} = \frac{5}{3}\).

◆ L'Hôpital's Rule & Squeeze Theorem

L'Hôpital's Rule If \(\lim_{x \to a}\frac{f(x)}{g(x)}\) is of the form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), then: \[\lim_{x \to a}\frac{f(x)}{g(x)} = \lim_{x \to a}\frac{f'(x)}{g'(x)}\] provided the right-hand limit exists.

Squeeze (Sandwich) Theorem If \(g(x) \leq f(x) \leq h(x)\) near \(x = a\) and \(\lim_{x \to a}g(x) = \lim_{x \to a}h(x) = L\), then \(\lim_{x \to a}f(x) = L\).

★ Example

Evaluate \(\displaystyle\lim_{x \to 0}\frac{e^x - 1 - x}{x^2}\).

This is \(\frac{0}{0}\). Applying L'Hôpital's rule: \(\frac{e^x - 1}{2x}\), still \(\frac{0}{0}\).
Apply again: \(\frac{e^x}{2} \to \frac{1}{2}\).
JEE Shortcut: Using Taylor expansion, \(e^x = 1 + x + \frac{x^2}{2} + \cdots\), so numerator \(\approx \frac{x^2}{2}\). Answer = \(\frac{1}{2}\).

◆ \(1^\infty\) Form

Key Formula for \(1^\infty\) Form If \(\lim_{x \to a}f(x) = 1\) and \(\lim_{x \to a}g(x) = \infty\), then: \[\lim_{x \to a}[f(x)]^{g(x)} = e^{\lim_{x \to a}(f(x)-1) \cdot g(x)}\]

★ Example

Evaluate \(\displaystyle\lim_{x \to 0}\left(\frac{\sin x}{x}\right)^{1/x^2}\).

This is \(1^\infty\) form. The exponent is \(\frac{1}{x^2}\left(\frac{\sin x}{x} - 1\right)\).
Using Taylor: \(\frac{\sin x}{x} = 1 - \frac{x^2}{6} + \cdots\), so \(\frac{\sin x}{x} - 1 \approx -\frac{x^2}{6}\).
Limit of exponent \(= \frac{1}{x^2} \cdot \left(-\frac{x^2}{6}\right) = -\frac{1}{6}\).
Answer: \(e^{-1/6}\).

Continuity

Continuity connects limits to function values. JEE frequently tests piecewise functions and the Intermediate Value Theorem.

◆ Definition & Types of Discontinuity

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, and (3) \(\lim_{x \to a}f(x) = f(a)\).

Types of Discontinuity

Removable: \(\lim_{x \to a}f(x)\) exists but \(\neq f(a)\) (or \(f(a)\) undefined). Can be "fixed" by redefining \(f(a)\).
Jump (first kind): Both one-sided limits exist but \(\lim_{x \to a^-}f(x) \neq \lim_{x \to a^+}f(x)\). Jump \(= |L^+ - L^-|\).
Infinite/Essential (second kind): At least one one-sided limit is \(\pm\infty\) or does not exist.

★ Example

Determine the type of discontinuity of \(f(x) = \frac{|x|}{x}\) at \(x = 0\).

\(\lim_{x \to 0^+}\frac{|x|}{x} = 1\) and \(\lim_{x \to 0^-}\frac{|x|}{x} = -1\).
Since the one-sided limits exist but are unequal, this is a jump discontinuity with jump \(= 2\).

◆ Intermediate Value Theorem (IVT)

Intermediate Value Theorem If \(f\) is continuous on \([a,b]\) and \(k\) is any value between \(f(a)\) and \(f(b)\), then there exists at least one \(c \in (a,b)\) such that \(f(c) = k\).

JEE Application: To prove a polynomial equation has a root in \((a,b)\), show \(f(a)\) and \(f(b)\) have opposite signs.

Differentiability

Differentiability is a stronger condition than continuity. JEE Advanced regularly tests the subtle distinction between the two.

◆ Definition & Relation with Continuity

Definition — Differentiability \(f\) is differentiable at \(x = a\) if: \[f'(a) = \lim_{h \to 0}\frac{f(a+h) - f(a)}{h}\] exists (as a finite limit). Equivalently, the left-hand derivative (LHD) equals the right-hand derivative (RHD).

Key Relationship

Differentiable \(\Rightarrow\) Continuous (always true)
Continuous \(\not\Rightarrow\) Differentiable (e.g., \(|x|\) at \(x=0\))

★ Example

Check differentiability of \(f(x) = x|x|\) at \(x = 0\).

\(f(x) = \begin{cases}x^2 & x \geq 0 \\ -x^2 & x < 0\end{cases}\).
RHD \(= \lim_{h \to 0^+}\frac{h^2 - 0}{h} = 0\). LHD \(= \lim_{h \to 0^-}\frac{-h^2 - 0}{h} = 0\).
LHD = RHD = 0. So \(f\) is differentiable at \(x = 0\) with \(f'(0) = 0\).

Differentiation Rules

Master the chain rule, product rule, implicit differentiation, parametric differentiation, and logarithmic differentiation for JEE speed.

◆ Core Rules

Differentiation Rules

Chain Rule: \(\frac{d}{dx}f(g(x)) = f'(g(x)) \cdot g'(x)\)
Product Rule: \((uv)' = u'v + uv'\)
Quotient Rule: \(\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}\)

◆ Implicit, Parametric & Logarithmic Differentiation

Special Techniques

Implicit: Differentiate both sides w.r.t. \(x\), treating \(y\) as a function of \(x\). Collect \(\frac{dy}{dx}\) terms.
Parametric: If \(x = f(t)\), \(y = g(t)\), then \(\frac{dy}{dx} = \frac{g'(t)}{f'(t)}\), and \(\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}(dy/dx)}{f'(t)}\).
Logarithmic: Take \(\ln\) of both sides when \(y = [f(x)]^{g(x)}\). Then \(\ln y = g(x)\ln f(x)\) and differentiate.

★ Example

Find \(\frac{dy}{dx}\) if \(y = x^x\).

Taking \(\ln\): \(\ln y = x \ln x\).
Differentiating: \(\frac{1}{y}\frac{dy}{dx} = \ln x + 1\).
\(\frac{dy}{dx} = x^x(\ln x + 1)\).

Applications of Derivatives

Tangent/normal, monotonicity, maxima/minima, Rolle's theorem, and MVT — the most application-heavy unit in JEE calculus.

◆ Tangent, Normal & Rate of Change

Tangent & Normal At the point \((x_0, y_0)\) on \(y = f(x)\):

Tangent: \(y - y_0 = f'(x_0)(x - x_0)\)
Normal: \(y - y_0 = -\frac{1}{f'(x_0)}(x - x_0)\) (provided \(f'(x_0) \neq 0\))
Length of tangent: \(\frac{|y_0|}{|\sin\theta|}\), Length of normal: \(|y_0 \sec\theta|\) where \(\tan\theta = f'(x_0)\)

◆ Monotonicity & Maxima/Minima

First Derivative Test

\(f'(x) > 0\) on \((a,b)\) \(\Rightarrow f\) is strictly increasing on \((a,b)\)
\(f'(x) < 0\) on \((a,b)\) \(\Rightarrow f\) is strictly decreasing on \((a,b)\)
If \(f'\) changes sign from \(+\) to \(-\) at \(c\), then \(f(c)\) is a local maximum
If \(f'\) changes sign from \(-\) to \(+\) at \(c\), then \(f(c)\) is a local minimum

Second Derivative Test At a critical point \(c\) (where \(f'(c) = 0\)):

\(f''(c) < 0 \Rightarrow\) local maximum
\(f''(c) > 0 \Rightarrow\) local minimum
\(f''(c) = 0 \Rightarrow\) test is inconclusive (use first derivative test or higher order)

★ Example

Find the maximum value of \(f(x) = x^3 - 3x\) on \([-2, 2]\).

\(f'(x) = 3x^2 - 3 = 3(x-1)(x+1) = 0 \Rightarrow x = \pm 1\).
\(f(-2) = -8 + 6 = -2\), \(f(-1) = -1 + 3 = 2\), \(f(1) = 1 - 3 = -2\), \(f(2) = 8 - 6 = 2\).
Maximum value = 2 at \(x = -1\) and \(x = 2\).

◆ Rolle's Theorem & Mean Value Theorem

Rolle's Theorem If \(f\) is continuous on \([a,b]\), differentiable on \((a,b)\), and \(f(a) = f(b)\), then there exists \(c \in (a,b)\) such that \(f'(c) = 0\).

Lagrange's Mean Value Theorem (LMVT) If \(f\) is continuous on \([a,b]\) and differentiable on \((a,b)\), then there exists \(c \in (a,b)\) such that: \[f'(c) = \frac{f(b) - f(a)}{b - a}\]

★ Example

Verify LMVT for \(f(x) = x^2\) on \([1, 3]\) and find \(c\).

\(f\) is a polynomial, hence continuous and differentiable everywhere.
\(\frac{f(3)-f(1)}{3-1} = \frac{9-1}{2} = 4\). We need \(f'(c) = 2c = 4\), so \(c = 2 \in (1,3)\). ✓

★ Key Takeaways

For \(1^\infty\) forms, use \(e^{\lim(f-1) \cdot g}\) — never attempt direct substitution.
Differentiability implies continuity, but not vice versa. Always check LHD = RHD at suspect points.
Logarithmic differentiation handles \([f(x)]^{g(x)}\) forms that no other rule can directly tackle.
For global extrema on \([a,b]\), evaluate \(f\) at all critical points AND endpoints — compare all values.
LMVT is Rolle's theorem for unequal endpoints — geometrically, there exists a point where the tangent is parallel to the secant.

📝 Practice Problems

Problem 1

Evaluate \(\displaystyle\lim_{x \to 0}\frac{x - \sin x}{x^3}\).

Show Solution ▼

Using Taylor: \(\sin x = x - \frac{x^3}{6} + \cdots\). So \(\frac{x - \sin x}{x^3} = \frac{x^3/6 + \cdots}{x^3} \to \frac{1}{6}\). Alternatively, apply L'Hôpital three times: \(\frac{1-\cos x}{3x^2} \to \frac{\sin x}{6x} \to \frac{1}{6}\).

Problem 2

Let \(f(x) = \begin{cases}\frac{\sin(p+1)x + \sin x}{x} & x < 0 \\ q & x = 0 \\ \frac{\sqrt{x+x^2}-\sqrt{x}}{x^{3/2}} & x > 0\end{cases}\). Find \(p\) and \(q\) so that \(f\) is continuous at \(x = 0\).

Show Solution ▼

LHL: \(\lim_{x \to 0^-}\frac{\sin(p+1)x + \sin x}{x} = (p+1) + 1 = p + 2\). RHL: Rationalise \(\frac{\sqrt{x(1+x)}-\sqrt{x}}{x^{3/2}} = \frac{\sqrt{1+x}-1}{x} = \frac{x/2 + \cdots}{x} \to \frac{1}{2}\). For continuity: \(p + 2 = q = \frac{1}{2}\). So \(p = -\frac{3}{2}\), \(q = \frac{1}{2}\).

Problem 3

Find the number of points where \(f(x) = |x^2 - 1| + |x^2 - 4|\) is non-differentiable.

Show Solution ▼

Critical points where inner expressions change sign: \(x^2 - 1 = 0 \Rightarrow x = \pm 1\) and \(x^2 - 4 = 0 \Rightarrow x = \pm 2\). Check each: at \(x = \pm 1\), the function has a corner (non-differentiable). At \(x = \pm 2\), also corners. 4 points of non-differentiability: \(x = -2, -1, 1, 2\).

Problem 4

If \(y = (\sin x)^{\tan x}\), find \(\frac{dy}{dx}\) at \(x = \frac{\pi}{4}\).

Show Solution ▼

Taking \(\ln y = \tan x \cdot \ln(\sin x)\). Differentiating: \(\frac{1}{y}\frac{dy}{dx} = \sec^2 x \cdot \ln(\sin x) + \tan x \cdot \frac{\cos x}{\sin x} = \sec^2 x \cdot \ln(\sin x) + 1\). At \(x = \pi/4\): \(y = (\frac{1}{\sqrt{2}})^1 = \frac{1}{\sqrt{2}}\), \(\sec^2(\pi/4) = 2\), \(\ln(\frac{1}{\sqrt{2}}) = -\frac{1}{2}\ln 2\). So \(\frac{dy}{dx} = \frac{1}{\sqrt{2}}(2 \cdot (-\frac{\ln 2}{2}) + 1) = \frac{1}{\sqrt{2}}(1 - \ln 2)\).

Problem 5

Find the interval(s) where \(f(x) = x^3 - 6x^2 + 9x + 1\) is strictly increasing.

Show Solution ▼

\(f'(x) = 3x^2 - 12x + 9 = 3(x^2 - 4x + 3) = 3(x-1)(x-3)\). \(f'(x) > 0\) when \(x < 1\) or \(x > 3\). Strictly increasing on \((-\infty, 1) \cup (3, \infty)\).

🎯 Interactive Quiz

1. \(\displaystyle\lim_{x \to 0}\frac{\tan x - \sin x}{x^3}\) equals:

A 0

B \(\frac{1}{2}\)

C 1

D \(\infty\)

2. The function \(f(x) = x^{2/3}\) at \(x = 0\) is:

A Differentiable but not continuous

B Continuous but not differentiable

C Neither continuous nor differentiable

D Both continuous and differentiable

3. If \(f(x) = e^x \sin x\), then \(f'(\pi)\) equals:

A \(e^\pi\)

B \(-e^\pi\)

C 0

D \(-2e^\pi\)

4. The number of real roots of \(e^x = x^2\) is:

A 0

B 1

C 2

D 3

5. The minimum value of \(f(x) = e^x + e^{-x}\) is:

A 1

B 2

C \(e\)

D \(\frac{1}{e}\)