Limits of Trigonometric Functions Formulas | Solved Examples, Practice Questions, FAQs

Limits of Trigonometric Functions – Class 11 Mathematics
Written by Neeraj Anand | Published by ANAND TECHNICAL PUBLISHERS

📘 Introduction

In calculus, understanding the limits of trigonometric functions is crucial for solving complex mathematical problems. A limit defines the value a function approaches as the input gets closer to a particular point. In Class 11 Mathematics, this concept is essential for mastering differentiation and integration, especially when dealing with trigonometric functions like sine, cosine, and tangent.

🔍 Key Concepts of Trigonometric Limits

Trigonometric limits are often evaluated using standard results and specific algebraic techniques. The most common trigonometric functions involved are:

• Sine (sinx)
• Cosine (cosx)
• Tangent (tanx)

Limits of Trigonometric Functions Formulas

The finite limit of trigonometric functions, i.e. the limit of the trigonometric function for the finite value of the x = a is discussed in the table below:

Function	Limit of the function
sin x	limx⇢asin x = sin a
cos x	limx⇢acos x = cos a
tan x	limx⇢atan x = tan a
cosec x	limx⇢acosec x = cosec a
sec x	limx⇢asec x = sec a
cot x	limx⇢acot x = cot a

As seen from the table above it is shows that the finite limit of the trigonometric function results in a finite value except for some exceptions. We know that for sin x, and cos x their range is between -1 and 1 and so for the infinite limit their value oscillates between -1 and 1 finding their exact value is not possible and thus their lint at infinity is undefined.

For x reaching positive or negative infinity, the limit of the trigonometric functions is discussed below.

Function Limit of the function for ±∞

The infinite limit of the trigonometric function is discussed in the table below,

Function	Limit of the function
sin x	limx⇢±∞sin x = not defined
cos x	limx⇢±∞cos x = not defined
tan x	limx⇢±∞tan x = not defined
cosec x	limx⇢±∞cosec x = not defined
sec x	limx⇢±∞sec x = not defined
cot x	limx⇢±∞cot x = not defined

Apart from the above formulas, we can define the following theorems that used to calculate limits of some trigonometric functions.

Theorem 1:

Let f and g be two real valued functions with the same domain such that f(x) ≤ g(x) for all x in the domain of definition, For some a, if both \(\begin{array}{l}\lim_{x \rightarrow a} f(x)\end{array} \) and \(\begin{array}{l}\lim_{x \rightarrow a} g(x)\end{array} \) exist, then

\(\begin{array}{l}\lim_{x \rightarrow a} f(x)\leq \lim_{x \rightarrow a} g(x)\end{array} \).

Theorem 2: Sandwich theorem or Squeeze theorem

Let f, g and h be real functions such that f(x) ≤ g( x) ≤ h(x) for all x in the common domain of definition. For some real number a, if

\(\begin{array}{l}\lim_{x \rightarrow a} f(x)=l=\lim_{x \rightarrow a} h(x)\end{array} \), then

\(\begin{array}{l}\lim_{x \rightarrow a} g(x)=l\end{array} \).

This is also known as Sandwich theorem or Squeeze theorem.

Based on this, we can write the following two important limits.

(i) \(\begin{array}{l}\lim_{x \rightarrow 0} \frac{sin\ x}{x}=1\end{array} \)

(ii) \(\begin{array}{l}\lim_{x \rightarrow 0} \frac{1-cos\ x}{x}=0\end{array} \)

These results are frequently used while solving more complicated problems involving limits of trigonometric functions.

Example: Given: g(x) = x²sin(1/x), Find: lim_x→0 g(x) using Squeeze theorem.

Solution: 

We know,

-1≤ sin(1/x) ≤ 1 Under its domain       

Multiplying with x²

-x² ≤ x2sin(1/x) ≤ x²

Then let f(x) = -x² and h(x) = x²

f(x) ≤ g(x) ≤ h(x)  

Using sandwich theorem,

As lim_x→a h(x) = lim_x→a f(x) = L

Therefore,

lim_x→a g(x) = L

⇒ lim_x→0 f(x) = lim_x→0 -x² = 0 and  

⇒ lim_x→0 h(x) = lim_x→0 x² = 0

Thus, lim_x→0 g(x) = 0

Limits of Various Trigonometric Functions

The limit of each trigonometric function is discussed in detail as below.

Limit of Sine Function

The function f(x) = sin(x) is a continuous function over its entire domain, with its domain consisting of all the real numbers. The range of this function is [-1, 1] as can be seen in the graph added below

So, if the limit of the sine function is calculated at any given real number it’s always defined and lies between [-1, 1].

Let, f(x) = sin(x)

lim_x→a f(x) = lim_x→a sin(x)

⇒ lim_x→a f(x) = sin(a),                    (where a is a real number)

Example: Find lim_x→π/2 sin(x)

Solution: 

As we know sin(x) is defined at π/2

Therefore,

lim_x→π/2 sin(x) = sin(π/2) = 1

Limit of Cosine Function

The function f(x) = cos(x) is a continuous function over its entire domain, with its domain consisting of all the real numbers.

The range of this function is [-1, 1] as can be seen in the graph added below.

So, if the limit of the cosine function is calculated at any given real number it’s always defined and lies between [-1, 1].

Let g(x) = cos(x)

lim_x→a g(x) = lim_x→a cos(x)

⇒ lim_x→a g(x) = cos(a), where a is a real number

Example: Find lim_x→π/2 cos(x)

Solution:   

As we know cos(x) is continuous and defined at π/2

Therefore,

lim_x→π/2 cos(x) = cos(π/2) = 0

Limit of Tangent Function

The function f(x) = tan(x) is defined at all real numbers except the values where cos(x) is equal to 0, that is, the values π/2 + πn for all integers n. Thus, its domain is all real numbers except π/2 + πn, n € Z.

The range of this function is (-∞, +c) as can be seen in the graph added below.

So, if the limit of the tangent function is calculated in its domain it is always defined and lies between(-∞, +∞).

f(x) = tan(x)

lim_x→a f(x) = lim_x→a  tan(x)

⇒ lim_x→a f(x) = tan(a)                 (where a belongs to real no. except π/2 + πn, n € Z)

Example: Find lim_x→π/2tan(x)

Solution:   

As we can see in graph tan(x) is tending to +∞ as x→-π/2 and to -∞ as x→+π/2

So, the right-hand and left-hand limits of this function at π/2 are not same

Therefore,

lim_x→π/2 tan(x) = Not defined

Limit of Cosec Function

The function f(x) = cosec(x) is defined at all real numbers except the values where sin(x) is equal to 0, that is, the values πn for all integers n. Thus, its domain is all real numbers except πn, n € Z.

The range of this function is (-∞,-1] U [1,+∞) as can be seen in the graph added below.

So, if the limit of the cosine function is calculated in its domain it is always defined and lies between its range.

Let f(x) = cosec(x)

lim_x→a f(x) = lim_x→a cosec(x)

⇒ lim_x→a f(x) = cosec(a)        (where a∈ R – nπ: n € Z)

Example: Find lim_x→π cosec(x)

Solution:  

As we know sin(x) is 0 as x→π           

So, 1/sin(x) i.e. cosec(x) at π cannot be defined

Therefore,

lim_x→π cosec(x) = Not defined

Limit of Secant Function

The function f(x) = sec(x) is defined at all real numbers except the values where cos(x) is equal to 0, that is, the values π/2 + πn for all integers n. Thus, its domain is all real numbers except π/2 + πn, n € Z.

The range of this function is (-∞, -1] U [1, +∞) as can be seen in the graph added below.

So, if the limit of the sec function is calculated in its domain it is always defined and lies between its range.

Let f(x) = sec(x)

lim_x→a f(x) = lim_x→a sec(x)

⇒ lim_x→a f(x) = sec(a)          (where a∈ R – {nπ + π/2}: n € Z)

Example: Find lim_x→π/2 sec(x)

Solution: 

As we know cos(x) is 0 as x→π/2

So, 1/cos(x) i.e. sec(x) at π/2 cannot be defined

Therefore,

lim_x→π/2 sec(x) = Not defined

Limit of Cot Function

The function f(x) = cot(x) is defined at all real numbers except the values where tan(x) is equal to 0, that is, the values πn for all integers n. Thus, its domain is all real numbers except πn, n € Z.

The range of this function is (-∞, +∞) as can be seen in the graph added below.

So, if the limit of the cot function is calculated in its domain it is always defined and lies between its range.

Let f(x) = cot(x)

lim_x→a f(x) = lim_x→a cot(x)

⇒ lim_x→a f(x) = cot(a)               (where a∈ R – nπ: n € Z)

Example: Find lim_x→π cot(x)

Solution: 

As we know tan(x) is 0   as x→π

So, 1/tan(x) i.e. cot(x) at π cannot be defined

Therefore,

lim_x→π cot(x) = Not defined

🧮 How to Solve Trigonometric Limits

To solve problems involving trigonometric limits:

1. Substitute the value directly and check if the limit exists.
2. If the substitution gives an indeterminate form like 0/0, apply:
   • Standard limits
   • L’Hôpital’s Rule (in advanced problems)
   • Trigonometric identities

🔑 Tips for Solving Trigonometric Limits

• Always try to convert the function into standard limit forms.
• Factor and simplify wherever possible to eliminate indeterminate forms.
• Use small-angle approximations:
  • sin⁡x ≈ x
  • tan⁡x ≈ x
  • cos⁡x ≈ 1−x²/2​

Solved Examples on Limits of Trigonometric Functions

Example 1:

Evaluate:

\(\begin{array}{l}\lim_{x \rightarrow 0} \frac{sin\ ax}{bx}\end{array} \)

Solution:

\(\begin{array}{l}\lim_{x \rightarrow 0} \frac{sin\ ax}{bx}\end{array} \)

Multiplying and dividing the function by “ax”,

\(\begin{array}{l}=\lim_{x \rightarrow 0} \frac{sin\ ax}{ax}\times \frac{ax}{bx}\\=\lim_{x \rightarrow 0} \frac{sin\ ax}{ax}\times \frac{a}{b}\end{array} \)

Now, take the constant term out of the limit.

\(\begin{array}{l}=\frac{a}{b}\times \lim_{x \rightarrow 0} \frac{sin\ ax}{ax}\end{array} \)

As we know,

\(\begin{array}{l}\lim_{x \rightarrow 0} \frac{sin\ x}{x}=1\end{array} \)

So, replacing “x” with “ax”,

\(\begin{array}{l}\frac{a}{b}\times \lim_{ax \rightarrow 0} \frac{sin\ ax}{ax} = \frac{a}{b}\times 1=\frac{a}{b}\end{array} \)

Therefore,

\(\begin{array}{l}\lim_{x \rightarrow 0} \frac{sin\ ax}{bx}=\frac{a}{b}\end{array} \)

Example 2:

Find the value of

\(\begin{array}{l}\lim_{x \rightarrow \frac{\pi}{2}} \frac{cos\ 2x}{sin\ x}\end{array} \).

Solution:

We know that,

\(\begin{array}{l}\lim_{x\rightarrow a}sin\ x = sin\ a\end{array} \)

and

\(\begin{array}{l}\lim_{x\rightarrow a}cos\ x = cos\ a\end{array} \)

Now,

\(\begin{array}{l}\lim_{x \rightarrow \frac{\pi}{2}} \frac{cos\ 2x}{sin\ x} = \frac{\lim_{x \rightarrow \frac{\pi}{2}} cos\ 2x}{\lim_{x \rightarrow \frac{\pi}{2}} sin x}=\frac{cos\ 2(\frac{\pi}{2})}{sin\ \frac{\pi}{2}}\end{array} \)

= cos π/ (sin π/2)

= -1/1

= -1

Therefore,

\(\begin{array}{l}\lim_{x \rightarrow \frac{\pi}{2}} \frac{cos\ 2x}{sin\ x}=-1\end{array} \)

Example 3: Given: g(x) = cos(x)/x, Find: lim_x→∞ g(x)

Solution: 

We know,

-1≤ cos(x)≤ 1 under its domain    

Dividing by x,

-1/x ≤ cos(x)/x ≤ 1/x

Then let f(x) = -1/x and h(x) = 1/x          

f(x) ≤ g(x) ≤ h(x)              

Using sandwich theorem,

As lim_x→a h(x) = lim_x→a f(x) = L

Therefore, 

lim_x→a g(x) = L

⇒ lim_x→∞ f(x) = lim_x→∞ -1/x = 0 and  

⇒ lim_x→∞ h(x) = lim_x→∞ 1/x = 0

Therefore, 

lim_x→∞ g(x) = 0 

Example 4: Find lim_x→π/2 cot(x)

Solution:

As we know cot(x) is continuous and defined at π/2

Therefore,   

lim_x→π/2 cot(x) = cot(π/2) = 0  

Example 5: Find lim_x→π sec(x)

Solution:   

As we know tan(x) is continuous and defined at π

Therefore,

lim_x→π cosec(x) = sec(π) = -1  

Example 6: Find lim_x→π/2 cosec(x)

Solution:   

As we know tan(x) is continuous and defined at π/2

Therefore,   

lim_x→π/2 cosec(x) = cosec(π/2) = 1  

Example 7: Find lim_x→πtan(x)

Solution:   

As we know tan(x) is continuous and defined at π

Therefore,  

lim_x→π tan(x) = tan(π) = 0   

Example 8: Find lim_x→π cos(x)

Solution:   

As we know cos(x) is continuous and defined at π

Therefore,  

lim_x→π cos(x) = cos(π) = -1 

Example 9: Find lim_x→πsin(x)

Solution:

As we know sin(x) is defined at π

Therefore,   

lim_x→π sin(x) = sin(π) = 0 

Limits of Trigonometric Functions – FAQs

What are trigonometric functions?

Trigonometric functions are the function that represents the relationship between the angles of the triangle with the sides of the triangle. They are also called circular functions as they repeat their value after some time in a circular manner.

What are six basic fundamental trigonometric functions?

The six basic trigonometric functions are,

• Sine Function
• Cosine Function
• Tangent Function
• Secant Function
• Cosecant Function
• Cotangent Function

Is the limit of the trigonometric function exist at infinity?

As we know that trigonometric functions are cyclic functions and generally their values oscillate between two values. So their values at infinity do not exist.

What is the limit of sin(x) as x approaches infinity?

The limit of sin(x) as x approaches infinity does not exist as the value of sin(x) oscillates between -1 and 1 without approaching a specific value.

What is the limit of cos(x) as x approaches infinity?

Similar to sin(x), the limit of cos(x) as x approaches infinity also does not exist as the value of sin(x) oscillates between -1 and 1 without approaching a specific value.

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🔔 Conclusion

Mastering the limits of trigonometric functions is essential for higher-level mathematics, especially in calculus and engineering exams. Practice regularly and focus on understanding the standard limits and their applications for better results in exams.