Find the limits: lim x → π 4 tan x tan 2 x

Given lim x → π 4 tan x tan 2 x Let L = lim x → π 4 tan x tan 2 x Taking logarithm on both sides log L = lim x → π 4 tan 2 x log tan x log L = lim x → π 4 log tan x cot 2 x We put the limit value its form 0 0 then L Hospital rule to be used log L = lim x → π 4 sec 2 x tan x - 2 cosec 2 2 x log L = lim x → π 4 sec 2 x - 2 tan x cosec 2 2 x log L = - 2 2 2 × 1 × 1 2 log L = - 1 L = e - 1 L = 1 e

Find the limits: lim x → π 2 sin x tan x

Given lim x → π 2 sin x tan x Let K = lim x → π 2 sin x tan x Taking logarithm on both sides log K = lim x → π 2 tan x log sin x log K = lim x → π 2 log sin x c o t x We put limit value its form 0 0 Then we apply ' L ' Hospital rule log K = lim x → π 2 1 sin x cos x - cos e s 2 x log K = lim x → π 2 - cos x sin 2 x sin x Then we put limit value log K = 0 K = e 0 K = 1 lim x → π 2 sin x tan x = 1

Find the limits: lim x → ∞ 3 x 2 + 2 x + 1 x 2 + x + 2 6 x + 1 3 x + 2

Given lim x → ∞ 3 x 2 + 2 x + 1 x 2 + x + 2 6 x + 1 3 x + 2 lim x → ∞ x 2 3 + 2 x + 1 x 2 x 2 1 + 1 x + 2 x 2 x 6 + 1 x x 3 + 2 x lim x → ∞ 3 + 2 x + 1 x 2 1 + 1 x + 2 x 2 6 + 1 x 3 + 2 x x 2 and x Cancel out We put the limit value = 3 + 2 ∞ + 1 ∞ 2 1 + 1 ∞ + 2 ∞ 2 6 + 1 ∞ 3 + 2 ∞ = 3 + 0 + 0 1 + 0 + 0 6 + 0 3 + 0 = 3 2 = 9

Find the limits: lim x → 0 arccos 1 - x x

Given, lim x → 0 cos - 1 1 - x x 0 0 form By applying L'Hospital rule, we have lim x → 0 - 1 1 - 1 - x 2 · d d x 1 - x 1 2 x lim x → 0 - 1 1 - 1 - x 2 + 2 x · - 1 1 2 x lim x → 0 2 x 2 x - x 2 lim x → 0 2 x x 2 - x lim x → 0 2 x x 2 - x lim x → 0 2 2 - x Put x = 0 = 2 2 = 2 .

Find the limits: lim x → π 4 lntan x 1 - cot x

Given, f x = lim x → π 4 lntan x 1 - cot x 0 0 form Here, by applying L'Hospital rule lim x → π 4 1 tan x · d d x tan x 0 - - csc 2 x = lim x → π 4 1 tan x · sec 2 x csc 2 x = lim x → π 4 cos x sin x · sin 2 x cos 2 x = lim x → π 4 sin x cos x = lim x → π 4 tan x By putting the value of limit, we get = tan π 4 = 1 .

Find the limits: lim x → 0 1 sin x ln 1 + a sin x

Let L = lim x → 0 1 sin x ln 1 + a sin x = lim x → 0 a ln 1 + a sin x a sin x = a × 1 ∵ lim x → 0 1 x ln 1 + x = 1 = a Alternate solution: Let L = lim x → 0 1 sin x ln 1 + a sin x 0 0 form Applying L-Hospital rule = lim x → 0 1 cos x × 1 1 + a sin x × a cos x = 1 1 + a sin 0 × a = a

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PROBLEMS IN CALCULUS OF ONE VARIABLE>Chapter 1 - Introduction to Mathematical Analysis>Calculation of Limits of Functions

I A Maron Solutions for Chapter: Introduction to Mathematical Analysis, Exercise 10: Calculation of Limits of Functions

Author:I A Maron

I A Maron Mathematics Solutions for Exercise - I A Maron Solutions for Chapter: Introduction to Mathematical Analysis, Exercise 10: Calculation of Limits of Functions

Attempt the practice questions on Chapter 1: Introduction to Mathematical Analysis, Exercise 10: Calculation of Limits of Functions with hints and solutions to strengthen your understanding. PROBLEMS IN CALCULUS OF ONE VARIABLE solutions are prepared by Experienced Embibe Experts.