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Limits, Continuity & Differentiability Formulas For JEE 2026
Limits, Continuity and Differentiability is a key chapter in JEE Mathematics because it builds the foundation of calculus. This topic usually carries 2–3 questions in the exam, making it quite important. It includes concepts like standard limits, continuity, and differentiability, which help students understand how different functions behave and change.
Students also come across useful techniques such as L’Hôpital’s rule, the Squeeze theorem, and different types of discontinuity, which are commonly used in solving problems. When these concepts are clear, it becomes easier to handle advanced calculus topics. Regular practice and revising formulas play a big role in scoring well. For quick revision, students can also use a well-organized JEE Mains Maths Formula PDF to go through important formulas and concepts easily.
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 2026, PDF: Conclusion
Limits, Continuity, and Differentiability form the backbone of calculus and play a crucial role in JEE preparation. A clear understanding of limits, standard results, and concepts like continuity and differentiability helps students solve complex problems with confidence. Regular practice of different question types, along with a strong grasp of formulas, makes this chapter much easier to handle.
To score well, students should focus on revising key concepts frequently and applying them in problem-solving. Topics like L’Hôpital’s rule, standard limits, and continuity conditions are frequently tested, so mastering them is essential. With consistent practice and smart revision strategies, this chapter can become a high-scoring area in the exam.
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