Limits And Derivatives — Complete Guide with Solved Examples

What Limits and Derivatives Actually Mean

Most students see limits as a chapter of algebra tricks. That framing will cost you marks. A limit answers one honest question: what value does a function approach as its input gets close to some number? Not what it equals there — what it approaches.

Think of it this way. You’re walking toward a wall. The limit is where you’re headed, not whether you actually hit the wall. The function might not even be defined at that point — and the limit can still exist. That distinction is what the entire chapter is built on.

Derivatives come naturally from limits. The derivative of a function at a point tells us its instantaneous rate of change — how fast the output is changing at that exact input. We build this by taking a limit of the average rate of change over a shrinking interval. This is not just theory; it’s the engine behind every optimization problem in JEE Main and every application-based question in CBSE Class 12.

Together, limits and derivatives form the gateway to calculus. Class 11 introduces them intuitively; Class 12 takes them further with rules and applications. JEE Main expects you to use them fluently under time pressure. We’ll cover all of it here.

Key Terms and Definitions

Limit of a function: We say $\lim_{x \to a} f(x) = L$ if $f(x)$ gets arbitrarily close to $L$ as $x$ approaches $a$ from either side. The function need not be defined at $x = a$.

Left-hand limit (LHL): $\lim_{x \to a^-} f(x)$ — the value $f(x)$ approaches as $x$ comes from the left.

Right-hand limit (RHL): $\lim_{x \to a^+} f(x)$ — the value $f(x)$ approaches as $x$ comes from the right.

Existence condition: A limit exists if and only if LHL = RHL. If they differ, the limit does not exist (DNE).

Continuity: $f$ is continuous at $x = a$ if three things hold simultaneously:

1. $f(a)$ is defined
2. $\lim_{x \to a} f(x)$ exists
3. $\lim_{x \to a} f(x) = f(a)$

Derivative: The derivative of $f$ at $x = a$ is:

This limit, when it exists, gives the slope of the tangent to the curve at $x = a$.

Differentiability: $f$ is differentiable at $x = a$ if this limit exists. Every differentiable function is continuous, but not every continuous function is differentiable (classic example: $|x|$ at $x = 0$).

Standard Limit Results You Must Know

Before solving anything, memorise these. They appear directly — often without manipulation — in both CBSE and JEE papers.

The first two are the workhorses. In JEE Main, roughly 60–70% of limit problems reduce to one of these after the right substitution.

Methods for Evaluating Limits

Method 1: Direct Substitution

If $f$ is a polynomial or rational function and the denominator is non-zero at $x = a$, simply substitute.

Example: $\lim_{x \to 2} (3x^2 - 4x + 1) = 3(4) - 4(2) + 1 = 5$

Method 2: Factorisation (for $\frac{0}{0}$ forms)

When direct substitution gives $\frac{0}{0}$, factor the numerator and denominator and cancel the offending term.

Example: $\lim_{x \to 3} \frac{x^2 - 9}{x - 3}$

Direct substitution gives $\frac{0}{0}$. Factor: $\frac{(x-3)(x+3)}{x-3} = x+3$, so the limit is $3 + 3 = 6$.

Method 3: Rationalisation

Used when the expression involves square roots and gives $\frac{0}{0}$.

Multiply numerator and denominator by the conjugate to clear the radical.

Method 4: Standard Limit Substitution

When you see $\sin(\text{something})/\text{something}$, force it into the standard form $\frac{\sin\theta}{\theta} \to 1$ by identifying $\theta$.

Method 5: L’Hôpital’s Rule

For Class 12 and JEE: if $\lim_{x \to a} \frac{f(x)}{g(x)}$ gives $\frac{0}{0}$ or $\frac{\infty}{\infty}$, then:

provided the right-hand limit exists.

Differentiation Rules

Solved Examples (Easy to Hard)

Example 1 — Easy (CBSE Class 11)

Find $\lim_{x \to 0} \frac{\sin 3x}{x}$

We need to match the standard form $\frac{\sin\theta}{\theta}$. Multiply and divide by 3:

Let $\theta = 3x$. As $x \to 0$, $\theta \to 0$, so $\frac{\sin\theta}{\theta} \to 1$.

Answer: 3

Example 2 — Easy (CBSE Class 11)

Find $\lim_{x \to 5} \frac{x^2 - 25}{x - 5}$

Factor: $\frac{(x-5)(x+5)}{x-5} = x + 5 \to 5 + 5 = 10$

Answer: 10

Example 3 — Medium (CBSE Class 12 / JEE Main)

Find $\lim_{x \to 0} \frac{1 - \cos 2x}{x^2}$

Use $1 - \cos 2x = 2\sin^2 x$:

Answer: 2

This exact result — $\lim_{x \to 0} \frac{1-\cos x}{x^2} = \frac{1}{2}$ — is a standard result. The factor of 2 here comes from the $2x$ inside cosine, which is why we used the double-angle identity first.

Example 4 — Medium (JEE Main level)

Differentiate $f(x) = x^2 \sin x$ using the first principles.

This is algebra-heavy. The product rule gives the same answer faster:

Example 5 — Hard (JEE Main)

Evaluate $\lim_{x \to 0} \frac{e^x - 1 - x}{x^2}$

Direct substitution gives $\frac{0}{0}$. Apply L’Hôpital’s twice:

First application: $\lim_{x \to 0} \frac{e^x - 1}{2x}$ — still $\frac{0}{0}$

Second application: $\lim_{x \to 0} \frac{e^x}{2} = \frac{1}{2}$

Answer: $\frac{1}{2}$

This can also be seen from the Taylor expansion $e^x = 1 + x + \frac{x^2}{2} + \ldots$, so $\frac{e^x - 1 - x}{x^2} \to \frac{1}{2}$.

Example 6 — Hard (JEE Main 2023 pattern)

Find all values of $k$ for which $f(x)$ is continuous at $x = 0$:

For continuity: $\lim_{x \to 0} f(x) = f(0)$

Set $k = 4$.

Answer: $k = 4$

This type — finding parameter values for continuity — has appeared in JEE Main multiple times, including January 2023 Shift 2.

Exam-Specific Tips

CBSE Class 11 and 12

The marking scheme rewards method, not just the answer. Always write:

• What form you get on direct substitution ($\frac{0}{0}$, defined, etc.)
• The algebraic step that resolves the indeterminate form
• The final evaluation with the limit value written clearly

In CBSE Class 12, Chapter 5 (Continuity and Differentiability) has 10–12 marks weightage in the board exam. The common question types are: proving continuity at a point, finding $k$ for continuity, and differentiating composite functions using the chain rule.

JEE Main

For JEE Main, develop a decision tree:

1. Try direct substitution first
2. If $\frac{0}{0}$ or $\frac{\infty}{\infty}$: identify whether it’s algebraic (factorise/rationalise) or transcendental (standard limits / L’Hôpital)
3. For trig limits: force into $\frac{\sin\theta}{\theta}$ form by multiplying and dividing

Each MCQ here is worth 4 marks. A 90-second investment in getting these right is excellent time management.

Common Mistakes to Avoid

Practice Questions

Q1

Find $\lim_{x \to 0} \frac{\sin 5x}{\sin 3x}$

Q2

Find $\lim_{x \to 1} \frac{x^3 - 1}{x^2 - 1}$

Q3

Differentiate $f(x) = \frac{x^2 + 1}{\sin x}$

Q4

For what value of $a$ is $f(x) = \begin{cases} ax + 3 & x \leq 2 \\ x^2 - 1 & x > 2 \end{cases}$ continuous at $x = 2$?

Q5

Find $\lim_{x \to 0} \frac{e^{3x} - 1}{x}$

Q6

Find $\lim_{x \to \infty} \frac{3x^2 + 2x - 1}{5x^2 - 4}$

Q7

Differentiate $f(x) = e^{\sin x}$

Q8

Find the derivative of $f(x) = \ln(\cos x)$

Q9 (JEE level)

Show that $f(x) = |x - 2|$ is not differentiable at $x = 2$.

FAQs

What is the difference between a limit and the value of a function?

The limit $\lim_{x \to a} f(x)$ describes what $f(x)$ approaches as $x$ gets near $a$ — regardless of what $f(a)$ actually is. The function might not be defined at $a$, or it might have a different value there. The limit only looks at the neighbourhood of $a$, never at $a$ itself.

When does a limit not exist?

A limit fails to exist in three situations: (1) the left-hand and right-hand limits are different, (2) the function oscillates without settling (like $\sin(1/x)$ near 0), or (3) the function goes to $\pm\infty$.

Can a function be continuous but not differentiable?

Yes. The absolute value function $|x|$ is continuous at $x = 0$ but has a sharp corner there, so the derivative does not exist. Differentiability is a stricter condition than continuity.

Why is $\lim_{x \to 0} \frac{\sin x}{x} = 1$ and not 0?

Because both $\sin x$ and $x$ go to 0 at the same rate. The ratio stabilises at 1. A geometric proof uses the squeeze theorem — the area of a triangle, a circular sector, and another triangle sandwich $\frac{\sin x}{x}$ between values that both approach 1. CBSE occasionally asks for this proof.

What is the $\frac{0}{0}$ form and why can’t we just say the answer is 1?

$\frac{0}{0}$ is an indeterminate form — it gives no information about the actual limit. Different functions can give $\frac{0}{0}$ at a point but have completely different limits there. We must do further work (factorisation, L’Hôpital, standard limits) to resolve it.

How many questions from limits and derivatives appear in JEE Main?

Limits typically contribute 2–3 questions. Derivatives appear directly in 1–2 questions and indirectly through application chapters (maxima-minima, tangents and normals, rate of change) for 4–6 more. The entire calculus sequence (Limits → Derivatives → Applications → Integrals) carries about 35–40% weightage in JEE Main Mathematics.

Is L’Hôpital’s Rule allowed in CBSE?

It is not in the CBSE Class 11 syllabus but appears in Class 12 (Chapter 5, under indeterminate forms). Using it in Class 11 won’t cause mark deduction, but the expected method is algebraic. For board exams, show your steps — the examiner must see the working to award step marks.

What is the first principles method for finding derivatives?

First principles means using the definition directly: $f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$. CBSE Class 11 asks this explicitly. The process: substitute $x + h$ into $f$, subtract $f(x)$, simplify the numerator (most terms cancel), then take the limit as $h \to 0$. Practice this for $x^2$, $\sin x$, and $\frac{1}{x}$ — these three cover 90% of what CBSE asks.