Limits Solved Examples Made Easy

The topic "Limits Solved Examples" comprises fully worked solutions for evaluating limits of functions in calculus, illustrating standard techniques such as direct substitution, algebraic manipulation, rationalization, and application of limit theorems.

Evaluation of Limits Using Direct Substitution, Factoring, and Rationalization

If a function is continuous at a point $x=a$, then the limit $\displaystyle\lim_{x \to a} f(x)$ can be found by direct substitution; that is, by evaluating $f(a)$.

Example: Find $\displaystyle\lim_{x \to 2} (x^2 + 3x)$.

Solution: Substitute $x=2$ into the function:

$= (2)^2 + 3 \times 2$

$= 4 + 6$

$= 10$

If substitution results in an indeterminate form such as $\dfrac{0}{0}$, algebraic methods like factoring are employed.

Example: Evaluate $\displaystyle\lim_{x \to 1} \dfrac{x^2 - 1}{x - 1}$.

Solution: Substitute $x=1$:

Numerator: $(1)^2 - 1 = 0$

Denominator: $1 - 1 = 0$

This is an indeterminate form. Factor the numerator:

$x^2 - 1 = (x - 1)(x + 1)$

The expression becomes $\dfrac{(x-1)(x+1)}{x-1}$ where $x \neq 1$.

Cancel $(x - 1)$ in numerator and denominator:

$= x + 1$

Now take the limit as $x \to 1$:

$= 1 + 1 = 2$

When the limit involves radicals and substitution produces indeterminate forms, rationalization is an effective technique. For more on indeterminate forms, refer to L'Hôpital's Rule for Indeterminate Limits.

Example: Evaluate $\displaystyle\lim_{x \to 9} \dfrac{\sqrt{x} - 3}{x - 9}$.

Solution: Substitution gives numerator $= 0$, denominator $= 0$.

Multiply numerator and denominator by the conjugate of the numerator, $\sqrt{x} + 3$:

$\dfrac{\sqrt{x} - 3}{x - 9} \cdot \dfrac{\sqrt{x} + 3}{\sqrt{x} + 3} = \dfrac{(\sqrt{x} - 3)(\sqrt{x} + 3)}{(x - 9)(\sqrt{x} + 3)}$

Expand numerator using $(a-b)(a+b) = a^2 - b^2$:

$\sqrt{x}^2 - 3^2 = x - 9$

So the expression simplifies to:

$\dfrac{x - 9}{(x - 9)(\sqrt{x} + 3)} = \dfrac{1}{\sqrt{x} + 3}$, $x \neq 9$.

Now, take $x \to 9$:

$\sqrt{9} + 3 = 3 + 3 = 6$

Hence, the limit is $\dfrac{1}{6}$.

Solved Examples Involving Special Limits and Trigonometric Limits

Many standard limits, especially involving trigonometric functions, have established results which are frequently used in JEE-level problems.

Example: Find $\displaystyle\lim_{x \to 0} \dfrac{\sin x}{x}$.

Solution: The standard result $\displaystyle\lim_{x \to 0} \dfrac{\sin x}{x} = 1$ applies. No further simplification is required.

For composite arguments, manipulation using substitution is necessary.

Example: Evaluate $\displaystyle\lim_{x \to 0} \dfrac{\sin 3x}{x}$.

Solution: Rewrite as $\dfrac{\sin 3x}{3x} \times 3$ by multiplying and dividing by $3$:

$= 3 \cdot \dfrac{\sin 3x}{3x}$

As $x \to 0$, $3x \to 0$, thus $\dfrac{\sin 3x}{3x} \to 1$.

So the limit is $3 \times 1 = 3$.

For further concepts on limits and their continuity, visit Limit, Continuity, and Differentiability.

Worked Limits with Factorization and Algebraic Manipulation

When both numerator and denominator of a rational expression become zero at the limit point, factoring is crucial to simplify and resolve the limit.

Example: Evaluate $\displaystyle\lim_{x \to 5} \dfrac{x^2 - 25}{x - 5}$.

Solution: Substitute $x=5$ yields $\dfrac{25 - 25}{0}$, an indeterminate form.

Factor numerator: $x^2 - 25 = (x-5)(x+5)$.

The expression is $\dfrac{(x - 5)(x + 5)}{x - 5}$, $x \neq 5$.

After cancellation, it simplifies to $x + 5$.

Taking the limit as $x \to 5$ gives $5 + 5 = 10$.

For a comprehensive collection of limit problems, including exam-style questions, refer to Important Questions on Limits and Continuity.

Limits Involving Rationalization in Square Root Expressions

Rationalization allows simplification of differences involving square roots that yield indeterminate forms.

Example: Evaluate $\displaystyle\lim_{x \to 4} \dfrac{\sqrt{x} - 2}{x - 4}$.

Solution: Substitute $x=4$ gives numerator $= 0$, denominator $= 0$.

Multiply numerator and denominator by $\sqrt{x} + 2$:

$\dfrac{\sqrt{x} - 2}{x - 4} \cdot \dfrac{\sqrt{x} + 2}{\sqrt{x} + 2} = \dfrac{x - 4}{(x - 4)(\sqrt{x} + 2)}$

After cancellation, $\dfrac{1}{\sqrt{x} + 2}$.

Take the limit as $x \to 4$:

$\sqrt{4} + 2 = 2 + 2 = 4$

The limit is $\dfrac{1}{4}$. For additional solved examples in PDF format, see Limits Solved Examples.

Composite Functions and Piecewise Definitions in Limit Problems

For piecewise functions, left-hand and right-hand limits must be evaluated separately to determine the existence and value of the limit at points of definition change.

Example: Let $f(x) = \begin{cases} x^2, & x < 1 \\ 2x, & x \geq 1 \end{cases}$. Find $\displaystyle\lim_{x \to 1} f(x)$ if it exists.

Solution: Left-hand limit: $\displaystyle\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} x^2 = 1^2 = 1$.

Right-hand limit: $\displaystyle\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} 2x = 2 \times 1 = 2$.

Since the left-hand limit ($1$) and right-hand limit ($2$) are not equal, the limit does not exist at $x=1$.

For definitions and problem types on function limits, refer to Limit of a Function.