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Limits, Continuity & Differentiability Formulas For JEE 2027
Limits, Continuity and Differentiability forms the foundation of calculus in JEE Mathematics. This chapter covers standard limits, one-sided limits, continuity conditions and differentiability, helping students understand how functions behave near a point. A clear understanding of these concepts also makes it easier to study derivatives and other calculus topics.
This blog brings together important formulas, key concepts and worked examples for JEE 2027 revision. Use a JEE Mains Formula sheet to revise essential results, then take a Free JEE Mains mock test to assess your understanding and identify areas that need more practice. Combining formula revision with problem-solving can help you improve accuracy and build confidence.
Limits: Definition and One-Sided Limits
Limit: We write $$\displaystyle\lim_{x \to a} f(x) = L$$ to mean: as $$x$$ gets arbitrarily close to $$a$$ (but $$x \neq a$$), the value $$f(x)$$ gets arbitrarily close to $$L$$.
Existence of a Limit
The limit $$\displaystyle\lim_{x \to a} f(x) = L$$ exists if and only if:
$$$\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = L$$$
Both one-sided limits must exist and be equal.
Worked Example
Evaluate $$\displaystyle\lim_{x \to 1} \frac{x^2 - 1}{x - 1}$$.
Factor: $$x^2 - 1 = (x-1)(x+1)$$.
$$\displaystyle\lim_{x \to 1} \frac{(x-1)(x+1)}{x-1} = \lim_{x \to 1} (x+1) = 1 + 1 =$$ 2
Algebra of Limits
Let $$\lim_{x \to a} f(x) = L$$ and $$\lim_{x \to a} g(x) = M$$. Then:
- $$\lim_{x \to a} [f(x) \pm g(x)] = L \pm M$$
- $$\lim_{x \to a} [f(x) \cdot g(x)] = L \cdot M$$
- $$\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{L}{M}$$, provided $$M \neq 0$$
- $$\lim_{x \to a} [f(x)]^n = L^n$$
Indeterminate Forms
Seven Indeterminate Forms
| $$\dfrac{0}{0}$$ | $$\dfrac{\infty}{\infty}$$ | $$0 \cdot \infty$$ | $$\infty - \infty$$ | $$0^0$$ | $$\infty^0$$ | $$1^\infty$$ |
These forms require algebraic manipulation, standard limits, or L'Hopital's rule to resolve.
Standard Limits Formula
Algebraic Standard Limit
$$$\lim_{x \to a} \frac{x^n - a^n}{x - a} = n \, a^{n-1}$$$
where $$n$$ is any rational number and $$a \neq 0$$.
Worked Example
Evaluate $$\displaystyle\lim_{x \to 2} \frac{x^5 - 32}{x - 2}$$.
Here $$a = 2$$, $$n = 5$$: $$5 \cdot 2^{4} = 5 \cdot 16 =$$ 80
Trigonometric Standard Limits
| Limit | Value |
|---|---|
| $$\displaystyle\lim_{x \to 0} \frac{\sin x}{x}$$ | $$1$$ ($$x$$ in radians) |
| $$\displaystyle\lim_{x \to 0} \frac{\tan x}{x}$$ | $$1$$ |
| $$\displaystyle\lim_{x \to 0} \frac{1 - \cos x}{x^2}$$ | $$\dfrac{1}{2}$$ |
| $$\displaystyle\lim_{x \to 0} \frac{\sin^{-1} x}{x}$$ | $$1$$ |
| $$\displaystyle\lim_{x \to 0} \frac{\tan^{-1} x}{x}$$ | $$1$$ |
Worked Example
Evaluate $$\displaystyle\lim_{x \to 0} \frac{\sin 3x}{\sin 7x}$$.
$$= \displaystyle\lim_{x \to 0} \frac{\sin 3x}{3x} \cdot \frac{7x}{\sin 7x} \cdot \frac{3x}{7x} = 1 \cdot 1 \cdot \frac{3}{7} =$$ $$\dfrac{3}{7}$$
Tip: Quick shortcut: $$\displaystyle\lim_{x \to 0} \frac{\sin ax}{\sin bx} = \frac{a}{b}$$ and $$\displaystyle\lim_{x \to 0} \frac{\tan ax}{\tan bx} = \frac{a}{b}$$.
Exponential and Logarithmic Standard Limits
| Limit | Value |
|---|---|
| $$\displaystyle\lim_{x \to 0} \frac{e^x - 1}{x}$$ | $$1$$ |
| $$\displaystyle\lim_{x \to 0} \frac{a^x - 1}{x}$$ | $$\ln a$$ |
| $$\displaystyle\lim_{x \to 0} \frac{\ln(1 + x)}{x}$$ | $$1$$ |
| $$\displaystyle\lim_{x \to 0} (1 + x)^{1/x}$$ | $$e$$ |
| $$\displaystyle\lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n$$ | $$e$$ |
Worked Example
Evaluate $$\displaystyle\lim_{x \to 0} \frac{e^{3x} - 1}{x}$$.
$$= \displaystyle\lim_{x \to 0} \frac{e^{3x} - 1}{3x} \cdot 3 = 1 \cdot 3 =$$ 3
The $$1^\infty$$ Form
$$1^\infty$$ Indeterminate Form
If $$\displaystyle\lim_{x \to a} f(x) = 1$$ and $$\displaystyle\lim_{x \to a} g(x) = \infty$$, then:
$$$\lim_{x \to a} [f(x)]^{g(x)} = e^{\displaystyle\lim_{x \to a}\, g(x)\,[f(x) - 1]}$$$
Worked Example
Evaluate $$\displaystyle\lim_{x \to 0} (1 + 3x)^{2/x}$$.
This is $$1^\infty$$ form: $$= e^{\lim_{x \to 0} \frac{2}{x} \cdot 3x} = e^{6} =$$ $$e^{6}$$
Tip: For $$1^\infty$$ form, always identify $$f(x)$$ and $$g(x)$$, compute $$g(x) \cdot [f(x) - 1]$$, and raise $$e$$ to that limit.
Squeeze Theorem (Sandwich Theorem)
If $$g(x) \leq f(x) \leq h(x)$$ for all $$x$$ near $$a$$, and
$$$\lim_{x \to a} g(x) = \lim_{x \to a} h(x) = L$$$
then $$\displaystyle\lim_{x \to a} f(x) = L$$.
Worked Example
Evaluate $$\displaystyle\lim_{x \to 0} x^2 \sin\!\left(\frac{1}{x}\right)$$.
$$-x^2 \leq x^2 \sin\!\left(\frac{1}{x}\right) \leq x^2$$
Both bounds $$\to 0$$ as $$x \to 0$$. By Squeeze Theorem, the limit is 0.
L'Hopital's Rule Formula
If $$\displaystyle\lim_{x \to a} f(x) = 0$$ and $$\displaystyle\lim_{x \to a} g(x) = 0$$ (or both $$\to \pm\infty$$), 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. The rule can be applied repeatedly if needed.
Note: L'Hopital's rule only applies to $$\frac{0}{0}$$ or $$\frac{\infty}{\infty}$$ forms. Always verify the form before applying. You differentiate numerator and denominator separately — this is not the quotient rule.
Worked Example
Evaluate $$\displaystyle\lim_{x \to 0} \frac{x - \sin x}{x^3}$$.
Form: $$\frac{0}{0}$$. Apply L'Hopital repeatedly:
$$\displaystyle\lim_{x \to 0} \frac{1 - \cos x}{3x^2} \;\xrightarrow{\;0/0\;}\; \lim_{x \to 0} \frac{\sin x}{6x} = \frac{1}{6} \cdot 1 =$$ $$\dfrac{1}{6}$$
Continuity Formula
Continuity at a Point: A function $$f(x)$$ is continuous at $$x = a$$ if: (1) $$f(a)$$ is defined, (2) $$\displaystyle\lim_{x \to a} f(x)$$ exists, (3) $$\displaystyle\lim_{x \to a} f(x) = f(a)$$.
$$f$$ is continuous at $$x = a$$ $$\iff$$ $$\displaystyle\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = f(a)$$
Types of Discontinuity
- Removable discontinuity: The limit exists but $$f(a)$$ is either undefined or $$f(a) \neq \lim_{x \to a} f(x)$$. Can be "fixed" by redefining $$f(a)$$.
- Jump discontinuity: Both one-sided limits exist but are not equal. $$\lim_{x \to a^-} f(x) \neq \lim_{x \to a^+} f(x)$$.
- Infinite discontinuity: At least one of the one-sided limits is $$\pm \infty$$.
Worked Example
Find the value of $$k$$ so that $$f(x) = \begin{cases} \frac{\sin 2x}{x}, & x \neq 0 \\ k, & x = 0 \end{cases}$$ is continuous at $$x = 0$$.
$$\displaystyle\lim_{x \to 0} \frac{\sin 2x}{x} = \lim_{x \to 0} \frac{\sin 2x}{2x} \cdot 2 = 2$$
So $$k =$$ 2.
Important Continuous Functions
- Polynomial functions are continuous everywhere on $$\mathbb{R}$$.
- $$\sin x$$, $$\cos x$$ are continuous everywhere.
- $$e^x$$, $$\ln x$$ (for $$x > 0$$) are continuous on their domains.
- $$|x|$$ is continuous everywhere.
Tip: In JEE problems on continuity, always check the "boundary points" where the function definition changes (for piecewise functions). Those are the only points where continuity can fail.
Differentiability Formula
Derivative at a Point: The derivative of $$f$$ at $$x = a$$ is: $$f'(a) = \displaystyle\lim_{h \to 0} \frac{f(a + h) - f(a)}{h}$$. If this limit exists and is finite, $$f$$ is differentiable at $$x = a$$.
Left-Hand and Right-Hand Derivatives
Left-hand derivative (LHD): $$f'(a^-) = \displaystyle\lim_{h \to 0^+} \frac{f(a) - f(a - h)}{h}$$
Right-hand derivative (RHD): $$f'(a^+) = \displaystyle\lim_{h \to 0^+} \frac{f(a + h) - f(a)}{h}$$
$$f$$ is differentiable at $$x = a$$ $$\iff$$ $$f'(a^-) = f'(a^+)$$ (and both are finite).
Worked Example
Is $$f(x) = |x|$$ differentiable at $$x = 0$$?
LHD $$= -1$$, RHD $$= 1$$
Since LHD $$\neq$$ RHD, $$f(x) = |x|$$ is not differentiable at $$x = 0$$ (sharp V-corner). Note: it is continuous at 0.
Continuity vs. Differentiability
Key Relationship
- Differentiable $$\Rightarrow$$ Continuous: If $$f$$ is differentiable at $$a$$, then $$f$$ is continuous at $$a$$.
- Continuous $$\not\Rightarrow$$ Differentiable: A function can be continuous at a point without being differentiable there (e.g., $$|x|$$ at 0).
Worked Example
Find $$a$$ and $$b$$ so that $$f(x) = \begin{cases} x^2, & x \leq 1 \\ ax + b, & x > 1 \end{cases}$$ is differentiable at $$x = 1$$.
Continuity: $$a + b = 1$$ … (i)
Differentiability: LHD $$= 2(1) = 2$$, RHD $$= a$$, so $$a = 2$$ … (ii)
From (i): $$b = 1 - 2 = -1$$.
$$a = 2, b = -1$$
Tip: In JEE, differentiability problems often ask you to find unknown constants in piecewise functions. Always set up two equations: one from continuity and one from matching derivatives.
Quick Reference
Standard Limits Summary
| Limit | Result |
|---|---|
| $$\displaystyle\lim_{x \to a} \frac{x^n - a^n}{x - a}$$ | $$n \cdot a^{n-1}$$ |
| $$\displaystyle\lim_{x \to 0} \frac{\sin x}{x} = \lim_{x \to 0} \frac{\tan x}{x}$$ | $$1$$ |
| $$\displaystyle\lim_{x \to 0} \frac{1 - \cos x}{x^2}$$ | $$\frac{1}{2}$$ |
| $$\displaystyle\lim_{x \to 0} \frac{e^x - 1}{x}$$ | $$1$$ |
| $$\displaystyle\lim_{x \to 0} \frac{a^x - 1}{x}$$ | $$\ln a$$ |
| $$\displaystyle\lim_{x \to 0} \frac{\ln(1+x)}{x}$$ | $$1$$ |
| $$\displaystyle\lim_{x \to 0} (1+x)^{1/x}$$ | $$e$$ |
Limits, Continuity & Differentiability Formulas For JEE 2027: Conclusion
Limits, Continuity and Differentiability forms the foundation of calculus and deserves focused attention in your JEE 2027 preparation. Understanding standard limits, continuity conditions and one-sided derivatives can help you approach questions with greater clarity. Use a JEE Mains Maths Formula sheet to revise essential results, and practise different question types to understand when each formula applies.
As you revise JEE Mains Maths Important chapters, connect this topic with differentiation and other calculus concepts to strengthen your understanding. Review mistakes, revisit challenging examples and solve previous years’ questions regularly. Combining formula revision with consistent practice can help you improve accuracy and approach calculus problems with confidence.
