Limits of Trigonometric Functions

Trigonometric Limits

Trigonometric limits are fundamental in calculus for evaluating the behavior of trigonometric functions like $\sin x$, $\cos x$, and $\tan x$ as $x$ approaches a specific value. These limits are crucial for solving derivatives, integrals, and oscillating function problems, and they form a key part of NCERT, CBSE, JEE, and CUET exam preparations.

Trigonometric equations involve trigonometric functions and are satisfied only for specific angle values, either finite or infinite. Calculating trigonometric limits often requires standard formulas and theorems from calculus.

Theorem 1: Monotonic Functions Limit Property

Let $f$ and $g$ be two real-valued functions with the same domain such that:

$f(x) \le g(x) \quad \text{for all } x \text{ in the domain}$.

If for some $a$, both $\lim_{x \to a} f(x)$ and $\lim_{x \to a} g(x)$ exist, then:

$\lim_{x \to a} f(x) \le \lim_{x \to a} g(x)$

This property is particularly useful when bounding oscillating trigonometric functions while solving limits.

Theorem 2: Standard Trigonometric Limits

Apart from direct substitution, factorization, and rationalization, the following standard formulas are essential for calculating trigonometric limits:

1. $\lim_{x \to 0} \frac{\sin x}{x} = 1$

2. $\lim_{x \to 0} \frac{\tan x}{x} = 1$

Since:

$\lim_{x \to 0} \frac{\tan x}{x} = \lim_{x \to 0} \frac{\sin x}{x} \times \frac{1}{\cos x} = 1 \times 1 = 1$

3. $\lim_{x \to a} \frac{\sin(x-a)}{x-a} = 1$

Proof using substitution $h = x - a$:

$\lim_{x \to a} \frac{\sin(x-a)}{x-a} = \lim_{h \to 0} \frac{\sin h}{h} = 1$

4. $\lim_{x \to a} \frac{\tan(x-a)}{x-a} = 1$

5. $\lim_{x \to a} \frac{\sin(f(x))}{f(x)} = 1$, if $\lim_{x \to a} f(x) = 0$

Similarly:

$\lim_{x \to a} \frac{\tan(f(x))}{f(x)} = 1$, if $\lim_{x \to a} f(x) = 0$

6. $\lim_{x \to 0} \cos x = 1$

7. $\lim_{x \to 0} \frac{\sin^{-1} x}{x} = 1$

Since $\sin^{-1} x = y \implies x = \sin y$, we have:

$\lim_{x \to 0} \frac{\sin^{-1} x}{x} = \lim_{y \to 0} \frac{y}{\sin y} = 1$

8. $\lim_{x \to 0} \frac{\tan^{-1} x}{x} = 1$

Important Standard Trigonometric Limits

Learn the key standard trigonometric limits like $\lim_{x \to 0} \frac{\sin x}{x}$ and $\lim_{x \to 0} \frac{\tan x}{x}$, which are essential for solving derivatives, integrals, and oscillating function problems in calculus.

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

The limit $\lim_{x \to 0} \frac{\sin x}{x} = 1$ is one of the most fundamental trigonometric limits in calculus. It is widely used to evaluate derivatives of sine functions and to solve oscillating function problems. Understanding this limit is crucial for NCERT, CBSE, JEE, and CUET exams, as many competitive problems are based on it.

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

The limit $\lim_{x \to 0} \frac{1 - \cos x}{x} = 0$, while $\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}$. These formulas help in simplifying trigonometric expressions and are essential in derivative and integral calculations in calculus.

$\lim_{x \to 0} \frac{\tan x}{x}$

For small values of $x$, $\lim_{x \to 0} \frac{\tan x}{x} = 1$, which can be derived using:

$\lim_{x \to 0} \frac{\tan x}{x} = \lim_{x \to 0} \frac{\sin x}{x} \cdot \frac{1}{\cos x} = 1 \cdot 1 = 1$

This limit is extensively applied in solving derivatives of tangent functions and evaluating indeterminate forms in calculus problems.

Other Key Limits for $\cot x$, $\sec x$, and $\csc x$

Other essential trigonometric limits include:

• $\lim_{x \to 0} \frac{\cot x}{x} = \infty$ (or undefined directly, often handled via $\frac{1}{\tan x}$)

• $\lim_{x \to 0} (\sec x - 1)/x = 0$

• $\lim_{x \to 0} (\csc x - 1/x) = 0$

These limits are important for solving advanced oscillating function problems in NCERT, JEE, and CBSE calculus exercises.

Techniques to Evaluate Limits of Trigonometric Functions

Explore effective techniques including direct substitution, algebraic identities, the Squeeze Theorem, and L’Hospital’s Rule to evaluate trigonometric limits.

Direct Substitution Method

If the function is continuous at the point of interest, the direct substitution method is the fastest way to evaluate the limit. For example, $\lim_{x \to \pi/4} \sin x = \sin(\pi/4) = \frac{\sqrt{2}}{2}$. This method is essential for quick problem-solving in exams.

Using Algebraic Identities

Algebraic and trigonometric identities such as $\sin^2 x + \cos^2 x = 1$, $1 - \cos^2 x = \sin^2 x$, and double-angle formulas can simplify limits like $\lim_{x \to 0} \frac{1 - \cos 2x}{x^2} = 2$. This technique is critical for competitive exam-level trigonometric limits.

Applying the Sandwich (Squeeze) Theorem

For oscillating functions like $\lim_{x \to 0} x \sin(1/x)$, the Sandwich Theorem provides a reliable method. If a function is bounded between two functions with the same limit, the Squeeze Theorem guarantees the limit of the target function.

Using L’Hospital’s Rule for Indeterminate Forms

When a trigonometric limit results in $0/0$ or $\infty/\infty$, L’Hospital’s Rule can be applied:

$\lim_{x \to 0} \frac{\sin x}{x} = \lim_{x \to 0} \frac{\cos x}{1} = 1$

This is highly useful in JEE, CUET, and advanced CBSE problems involving indeterminate forms.

Step-by-Step Method for Solving Trigonometric Limits

Follow a structured step-by-step approach to identify, simplify, and calculate trigonometric limits accurately, ensuring mastery of NCERT and competitive exam problems.

Identifying the Form of the Limit

Determine if the limit is directly solvable, indeterminate, or oscillating. Recognizing the form is essential for choosing the correct solution technique.

Simplifying the Function Using Trigonometric Identities

Use trigonometric simplifications, such as $\sin^2 x = 1 - \cos^2 x$ or double-angle formulas, to rewrite the function in a solvable form.

Applying Standard Limit Formulas

Apply fundamental trigonometric limits like $\lim_{x \to 0} \frac{\sin x}{x} = 1$ and $\lim_{x \to 0} \frac{\tan x}{x} = 1$ to evaluate the simplified function.

Concluding the Final Limit Value

Combine the results from simplification and formula application to conclude the limit accurately, ensuring correctness for NCERT exercises and competitive exams.

Example 1: $\lim_{x \to 0} \frac{\sin 3x}{x}$

Rewrite using standard formula:

$\lim_{x \to 0} \frac{\sin 3x}{x} = \lim_{x \to 0} \frac{\sin 3x}{3x} \cdot 3 = 1 \cdot 3 = 3$

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

Using identity $1 - \cos 2x = 2 \sin^2 x$:

$\lim_{x \to 0} \frac{2 \sin^2 x}{x^2} = 2 \cdot \lim_{x \to 0} \left(\frac{\sin x}{x}\right)^2 = 2 \cdot 1^2 = 2$

Solved Examples Based On Trigonometric Limits

Example 1: $\lim\limits _{x \rightarrow 0} \frac{(1-\cos 2 x)^2}{2 x \tan x-x \tan 2 x}$ is [JEE Main 2016]
1) $-2$
2) $-\frac{1}{2}$
3) $\frac{1}{2}$
4) $2$

Solution:

$\begin{aligned}
& \lim\limits _{x \rightarrow 0} \frac{(1-\cos 2 x)^2}{2 x \tan x-x \tan 2 x} \\
& \lim\limits _{x \rightarrow 0} \frac{4 \sin ^4 x}{x(2 \tan x-\tan 2 x)} \\
& \because 1-\cos 2 x=2 \sin ^2 x \\
& \lim\limits _{x \rightarrow 0} 4\left(\frac{\sin x}{x}\right)^4 \frac{x^3}{(2 \tan x-\tan 2 x)}=\lim\limits _{x \rightarrow 0} \frac{4 x^3}{(2 \tan x-\tan 2 x)} \\
& \because \lim\limits _{x \rightarrow 0} \frac{\sin x}{x}=1
\end{aligned}$

Now use series expansion

$\begin{equation}
\begin{aligned}
&\begin{aligned}
& \lim\limits _{x \rightarrow 0} \frac{4 x^3}{(2 \tan x-\tan 2 x)} \\
& \lim\limits _{x \rightarrow 0} \frac{4 x^2}{2\left(x+\frac{x^3}{3}+\frac{2 x^5}{15}+\ldots\right)-\left(2 x+\frac{(2 x)^3}{3}+2 \cdot \frac{(2 x)^5}{15}+\ldots\right)} \\
& \lim\limits _{x \rightarrow 0} \frac{4 x^3}{2 x+\frac{2 x^3}{3}+\frac{4 x^5}{15}+\ldots-2 x-\frac{8 x^3}{3}-\frac{64 x^5}{15}-\ldots} \\
& \lim\limits _{x \rightarrow 0} \frac{4}{\frac{2}{3}+\frac{4 x^2}{15}+\ldots-\frac{8}{3}-\frac{64 x^2}{15}-\ldots} \\
& \lim\limits _{x \rightarrow 0} \frac{4}{\frac{2}{3}-\frac{8}{3}}=-2
\end{aligned}\\
&\text { Hence, the answer is the option } 1.
\end{aligned}
\end{equation}$

Example 2:

$\lim\limits _{x \rightarrow 0} \frac{x \cot (4 x)}{\sin ^2 x \cot ^2(2 x)}$ is equal to :
[JEE Main 2019]
1) $1$
2) $2$
3) $4$
4) $0$

Solution:

Evalution of Trigonometric limit -

$
\begin{aligned}
& \lim\limits _{x \rightarrow a} \frac{\sin (x-a)}{x-a}=1 \\
& \lim\limits _{x \rightarrow a} \frac{\tan (x-a)}{x-a}=1
\end{aligned}
$

put $x=a+h$ where $h \rightarrow 0$
Then it comes

$
\begin{aligned}
& \lim\limits _{h \rightarrow 0} \frac{\sin h}{h}=\lim\limits _{h \rightarrow 0} \frac{\tan h}{h}=1 \\
& \therefore \lim\limits _{x \rightarrow 0} \frac{\sin x}{x}=1 \text { and } \\
& \therefore \lim\limits _{x \rightarrow 0} \frac{\tan x}{x}=1 \\
& \lim\limits _{x \rightarrow 0} \frac{x \cot 4 x}{\sin ^2 x \cot ^2 2 x}=\lim\limits _{x \rightarrow 0} \frac{x \tan ^2 2 x}{\sin ^2 x \tan 4 x} \\
& =\lim\limits _{x \rightarrow 0} \frac{x\left(\frac{\tan ^2 2 x}{4 x^2}\right) \times 4 x^2}{\frac{\sin ^2 x}{x^2} \times x^2\left(\frac{\tan 4 x}{4 x}\right) \times 4 x} \\
& =1
\end{aligned}
$
$\begin{aligned}
&\left(\lim\limits _{z \rightarrow 0}\right) \frac{\tan z}{z}=1 \quad\left(\lim\limits _{z \rightarrow 0}\right) \frac{\sin z}{z}=1\\
&\text { Hence, the answer is the option } 1.
\end{aligned}$

Example 3: $\lim\limits _{x \rightarrow \frac{\pi}{2}} \frac{\cot x-\cos x}{(\pi-2 x)^3}$ equals:
[JEE Main 2017]
1) $\frac{1}{16}$
2) $\frac{1}{8}$
3) $\frac{1}{4}$
4) $\frac{1}{24}$

Solution:

$\begin{equation}
\begin{aligned}
&\begin{aligned}
& \lim\limits _{x \rightarrow \frac{\pi}{2}} \frac{\cot x-\cos x}{(\pi-2 x)^3} \\
& \text { Put } x=\frac{\pi}{2}-h \\
& \Rightarrow \lim\limits _{h \rightarrow 0} \frac{\cot \left(\frac{\pi}{2}-h\right)-\cos \left(\frac{\pi}{2}-h\right)}{\left[\pi-2\left(\frac{\pi}{2}-h\right)\right]^3} \\
& \Rightarrow \lim\limits _{h \rightarrow 0} \frac{\tan h-\sin h}{(\pi-\pi+2 h)^3} \\
& \Rightarrow \lim\limits _{h \rightarrow 0} \frac{\tanh (1-\cos h)}{8 h^3} \\
& \Rightarrow \frac{1}{8} \lim\limits _{h \rightarrow 0} \frac{\tan h}{h} \cdot \frac{2 \sin ^2 \frac{h}{2}}{h^2} \\
& \Rightarrow \frac{1}{8} \lim\limits _{h \rightarrow 0} \frac{\tan h}{h} \cdot \frac{2 \sin ^2 \frac{h}{2}}{4 \cdot\left(\frac{h}{2}\right)^2} \\
& \Rightarrow \frac{1}{8} \times 1 \times \frac{1}{2}=\frac{1}{16}
\end{aligned}\\
&\text { Hence, the answer is the option } 1.
\end{aligned}
\end{equation}$

Example 4: $\lim\limits _{x \rightarrow 0} \frac{(1-\cos 2 x)(3+\cos x)}{x \tan 4 x}$ is equal to :

[JEE Main 2013]
1) $2$
2) $-0.25$
3) $0.5$
4) $1$

Solution:

$
\begin{aligned}
& \lim\limits _{x \rightarrow 0} \frac{(1-\cos 2 x)(3+\cos x)}{x \tan 4 x} \\
& \lim\limits _{x \rightarrow 0} \frac{2 \sin ^2 x(3+\cos x)}{4 x^2 \frac{\tan 4 x}{4 x}} \\
& =\frac{1}{2} \times 4=2
\end{aligned}
$

Hence, the answer is the option (1).

Example 5 : If $\alpha$ is the positive root of the equation, $p(x)=x^2-x-2=0$, then $\lim\limits _{x \rightarrow \alpha^{+}} \frac{\sqrt{1-\cos (p(x))}}{x+\alpha-4}$ is equal to: [JEE Main 2020]
$\frac{3}{2}$
2) $\frac{3}{\sqrt{2}}$
3) $\frac{1}{\sqrt{2}}$
4) $1$

Solution:

$\begin{equation}
\begin{aligned}
&\begin{aligned}
& x^2-x-2=0 \\
& \text { roots are } 2 \&-1 \\
& \Rightarrow \lim\limits _{x \rightarrow 2^{+}} \frac{\sqrt{1-\cos \left(x^2-x-2\right)}}{(x-2)} \\
& =\lim\limits _{x \rightarrow 2^{+}} \frac{\sqrt{2 \sin ^2 \frac{\left(x^2-x-2\right)}{2}}}{(x-2)} \\
& =\lim\limits _{x \rightarrow 2^{+}} \frac{\sqrt{2} \sin \left(\frac{(x-2)(x+1)}{2}\right)}{(x-2)} \\
& =\frac{3}{\sqrt{2}}
\end{aligned}\\
&\text { Hence, the answer is an option (2). }
\end{aligned}
\end{equation}$

List of topics related to Limits

Check out a comprehensive list of limit-related topics to strengthen your concepts for limits, which are also crucial for CBSE, JEE, and CUET preparation.

Indeterminate forms of limits

Algebra of Limits

Limits using expansion

Left and Right Hand Limits

Higher Order Derivatives

NCERT Resources

Access NCERT-aligned notes, exercises, and exemplar solutions for Limits and Derivatives to strengthen your understanding of trigonometric limits and build a strong foundation for calculus and competitive exams.

NCERT Notes for Class 11 Maths Chapter 13 - Limits and Derivatives

NCERT Solutions for Class 11 Maths Chapter 13 - Limits and Derivatives

NCERT Exempar Solutions for Class 11 Maths Chapter 13 - Limits and Derivatives

Practice Questions based on Limits of Trigonometric Functions

Practice well-structured questions and solved examples to master trigonometric limits, improve speed and accuracy, and prepare effectively for NCERT exercises, CBSE exams, and competitive tests like JEE and CUET.

Limits Of Trigonometric Functions - Practice Question MCQ

We have shared the links below to practice questions on the related topics of limits: