Evaluate lim x → α x sin α - α sin x x - α .

Given, lim x → α x sin α - α sin x x - α Let h = x - α ⇒ x = h + α x - α ⇒ h - 0 ⇒ lim x → α x sin α - α sin x x - α = lim h → 0 ( h + α ) sin α - α sin ( h + α ) h = lim h → 0 h sin α + α sin α - α ( sin h cos α + cos h sin α ) h = lim h → 0 h sin α h + α sin α ( 1 - cos h ) h - α cos α sin h h = lim h → 0 sin α + lim h - ∞ α sin α 2 sin 2 h 2 h - lim h → 0 α cos α sin h h = sin α + 2 α sin α lim h → 0 sin 2 h 2 h 2 2 · h 2 2 h + α cos α ∵ lim x → 0 sin a x x = a = sin α + α cos α + 2 α sin α lim h → 0 h 4 = sin α + α cos α + 0 ∴ lim x → α x sin α - α sin x x - α = sin α + α cos α

E M B I B E

Bureau's Higher Secondary Elements of Mathematics Vol.1>Chapter 14 - Limit and Differentiation>EXERCISE 14(c)

Odisha Board Solutions for Chapter: Limit and Differentiation, Exercise 3: EXERCISE 14(c)

Author:Odisha Board

Odisha Board Mathematics Solutions for Exercise - Odisha Board Solutions for Chapter: Limit and Differentiation, Exercise 3: EXERCISE 14(c)

Attempt the free practice questions on Chapter 14: Limit and Differentiation, Exercise 3: EXERCISE 14(c) with hints and solutions to strengthen your understanding. Bureau's Higher Secondary Elements of Mathematics Vol.1 solutions are prepared by Experienced Embibe Experts.