CBSE Class 11 - Mathematics - Limits and Derivatives Part-4- Limits of Trigonometric Functions #class11Maths #eduvictors #limits

CBSE Class 11 - Mathematics - Limits and Derivatives Part-4- Limits of Trigonometric Functions

#class11Maths #Limits

In the previous blog post Limits and derivatives Part-3, we learn about the algebra of limits. Now, let us study few theorems before we discuss about trignometric limits.

Theorem 1: Let f(x) and g(x) be two real valued functions with same domain such that f(x) ≤ g(x) for all x.

For some a, if $\lim_{x\rightarrow a} f(x)$ and $\lim_{x\rightarrow a} g(x)$ exist then $\lim_{x\rightarrow a} f(x) \leq  \lim_{x\rightarrow a} g(x)$

Theorem 2: Sandwich Theorem or Squeeze Theorem

Let f(x) and g(x) be two real valued functions with same domain such that f(x) ≤ g(x) ≤ h(x) for all x in common domain.

For some a, if $\lim_{x\rightarrow a} f(x) = l = \lim_{x\rightarrow a} h(x)$, then $\lim_{x\rightarrow a} g(x) = l$

Theorem 3:$\lim_{x\rightarrow 0} sin x = 0$ and $\lim_{x\rightarrow 0} cos x = 1$ where x is measured in radians.

Theorem 4:  $\lim_{x\rightarrow 0} \frac{sin x}{x} = 1  $

and 

$\lim_{x\rightarrow 0} \frac{tan x}{x} = 1  $

Theorem 5: $\lim_{x\rightarrow 0} \frac{1 - cos x}{x} = 0  $

Let us solve some problems.

Q1: Evaluate $\lim_{x\rightarrow 0} \frac{sin 3x}{x}$

Answer: 

=$\lim_{x\rightarrow 0} \left ( 3 \times \frac{sin 3x}{3x}  \right )$

=$ 3\lim_{x\rightarrow 0} \left (  \frac{sin 3x}{3x}  \right )$

= 3(1)

= 3

Q2: Evaluate $\lim_{x\rightarrow 0} \left ( \frac{\sin ax}{ \sin bx}  \right )$

Answer:

=$\lim_{x\rightarrow 0} \left ( \frac{\sin ax}{ \sin bx}  \right )$

= $\lim_{x\rightarrow 0} \left ( \frac{(\frac{\sin ax}{ax})ax}{ (\frac{\sin bx}{bx})bx}  \right )$

= $\frac{\lim_{x\rightarrow 0} (\frac{\sin ax}{ax})ax}{ \lim_{x\rightarrow 0} (\frac{\sin bx}{bx})bx}$

= $\frac{a(1)}{b(1)}$

= $\frac{a}{b}$

Q3: Evaluate $\lim_{x\rightarrow 0} \frac{\sin 5x}{\tan 3x}$

Answer: 

=$\lim_{x\rightarrow 0} \frac{\sin 5x}{\tan 3x}$

= $\lim_{x\rightarrow 0} \frac{\frac{\sin 5x}{x}}{\frac{\tan 3x}{3x}}$

= $\frac{\lim_{x\rightarrow 0} \frac{\sin 5x}{x}}{\lim_{x\rightarrow 0}\frac{\tan 3x}{3x}}$

=$\frac{5}{3}.\frac{1}{1}$

=$\frac{5}{3}$

Q4: Evaluate $\lim_{x\rightarrow \frac{\pi}{2}} \frac{\cos x}{\frac{\pi}{2} - x}$

Answer: Let $(x - \frac{\pi}{2}) = y$ so that when $x \rightarrow \frac{\pi}{2}$ then $y \rightarrow 0$

$ \therefore \lim_{x\rightarrow 0} \frac{\cos (\frac{\pi}{2} + y)}{- y}$

= $\lim_{x\rightarrow 0} (\frac{-\sin y}{-y})$

= $\lim_{x\rightarrow 0} (\frac{\sin y}{y})$

= 1

Hope it helps you get a fair idea about limits of Trigonometric Functions. In the next post, we shall discuss about other problems

related to limits.

👉See Also:

Ch13 Limits and Derivatives - Q & A (Part-1)

Ch 13 Limits and Derivatives - Q & A (Part - 2)

Ch 13 Limits and Derivatives - Q & A (Part-3)

SETS (NCERT Ex 1.1)
SETS (NCERT Ex 1.2)
Different Types of Sets
SETS (NCERT Ex 1.3)

SETS (Unit Test Paper)
SETS (VENN DIAGRAMS)
SETS (Operations of Sets)
SETS (NCERT Ex 1.4 Q1-Q5)
SETS (NCERT Ex 1.4 Q6 - Q8)
SETS (NCERT Ex 1.4 Q9 - Q12)
SETS (NCERT Ex 1.5)
SETS (NCERT Ex 1.6) 
Laws of Set Operations
Ch2: Relations and Functions (1 Mark Q & A) Part-1
Ch2: Cartesian Product of Two Sets (Important Points)
Ch2: Relations - Domain, Range and Co-Domain (Solved Problems)

Ch5: Complex Numbers (Part 1) - Solved Problems

Maths Annual Test Paper (2018-19)
Maths Annual Test Paper (2020-21) 
Maths Term1 MCQs (2021-22)