6.3 Series Expansions:

Maths > Limits > 6.0 Some Standard Methods of Evaluation of Limits:

  Limits
    1.0 Introduction
    2.0 Definition of Limit - In a different form:
    3.0 Conditions for existence of Limit
    4.0 Some Standard Limits
    5.0 Algebra of limits
    6.0 Some Standard Methods of Evaluation of Limits:
    7.0 Indeterminate Forms:
    8.0 Sandwich Theorem / Squeeze Play Theorem:
    9.0 L'Hospital's Rule for evaluation of limits:
6.3 Series Expansions:
  • $^{{a^x} = 1 + (\ln a)x + {{(\ln a)}^2}\tfrac{{{x^2}}}{{2!}} + \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot }$ where, $a \in {R^ + }$
  • ${e^x} = 1 + x + \frac{{{x^2}}}{{2!}} + \frac{{{x^3}}}{{3!}} +..........$
  • $\sin x = x - \tfrac{{{x^3}}}{{3!}} + \tfrac{{{x^5}}}{{5!}} - \tfrac{{{x^7}}}{{7!}} + \cdot \cdot \cdot \cdot \cdot \cdot $
  • $\cos x = 1 - \tfrac{{{x^2}}}{{2!}} + \tfrac{{{x^4}}}{{4!}} - \tfrac{{{x^6}}}{{6!}} + \cdot \cdot \cdot \cdot \cdot \cdot $
  • $\tan x = x + \tfrac{{{x^3}}}{3} + \tfrac{{2{x^5}}}{{15}} + \tfrac{{17{x^7}}}{{315}} + \cdot \cdot \cdot \cdot \cdot \cdot $
  • $ln(1 + x) = x - \tfrac{{{x^2}}}{2} + \tfrac{{{x^3}}}{3} - \tfrac{{{x^4}}}{4} + \cdot \cdot \cdot \cdot \cdot \cdot $,where $ - 1 \leqslant x \leqslant 1$
  • ${(1 + x)^n} = 1 + nx + \tfrac{{n(n - 1)}}{{2!}}{x^2} + \tfrac{{n(n - 1)(n - 2)}}{{3!}}{x^3} + \cdot \cdot \cdot \cdot \cdot \cdot $ n $ \in R$ and |$x$|<1

Question 17.


Prove that $\mathop {\lim }\limits_{x \to 0} \tfrac{{\tan x}}{x} = 1$, Where $x$ in radians.

Solution:

$\tan x = x + \tfrac{{{x^3}}}{3} + \tfrac{{2{x^5}}}{{15}} + \tfrac{{17{x^7}}}{{315}} + \cdot \cdot \cdot \cdot \cdot \cdot $

$ \Rightarrow \frac{{\tan x}}{x} = 1 + \tfrac{{{x^2}}}{3} + \tfrac{{2{x^4}}}{{15}} + \tfrac{{17{x^6}}}{{315}} + \cdot\cdot\cdot\cdot\cdot\cdot$

$\therefore \mathop {\lim }\limits_{x \to 0} \frac{{\tan x}}{x} = \mathop {\lim }\limits_{x \to 0} 1 + \tfrac{{{x^2}}}{3} + \tfrac{{2{x^4}}}{{15}} + \tfrac{{17{x^6}}}{{315}} + \cdot\cdot\cdot\cdot\cdot\cdot = 1 + 0 + 0 + 0 + \cdot\cdot\cdot\cdot\cdot\cdot = 1$

Hence Proved $\mathop {\lim }\limits_{x \to 0} \frac{{\tan x}}{x} = 1$

similarly we can prove $\mathop {\lim }\limits_{x \to 0} \tfrac{{\sin x}}{x} = 1$, Where $x$ in radians

$\mathop {\lim }\limits_{x \to 0} \tfrac{{{a^x} - 1}}{x} = \ln (a),a \in {R^ + }$

$\mathop {\lim }\limits_{x \to 0} \tfrac{{{e^x} - 1}}{x} = 1$

$\mathop {\lim }\limits_{x \to 0} \tfrac{{{{(1 + x)}^n} - 1}}{x} = n$

$\mathop {\lim }\limits_{x \to 0} \tfrac{{\ln (1 + x)}}{x} = 1$

Improve your JEE MAINS score
10 Mock Test
Increase JEE score
by 20 marks
Detailed Explanation results in better understanding
Exclusively for
JEE MAINS and ADVANCED
9 out of 10 got
selected in JEE MAINS
Lets start preparing
Read Complete Chapter
DIFFICULTY IN UNDERSTANDING CONCEPTS?
TAKE HELP FROM THINKMERIT DETAILED EXPLANATION..!!!
9 OUT OF 10 STUDENTS UNDERSTOOD
START LEARNING