Limit

Answer :

lim_x→(π/2) [{[2x sin(4k + 1)(π/2) + x] + π cosec[(4k – 1)(π/2) + x]

                                   ∙ sin[(4k – 1)(π/2) – x]} / {sec(2kπ – x) ∙ cos[(4k – 1)(π/2) + x]}]

                             = lim_x→(π/2) [{2x cos x + π(– sec x) ∙ (– sin x)} / (sec x ∙ sin x)]

                             = lim_x→(π/2) [(2x cos² x) / (sin x)] + [(π sec x sin x) / (sec x sin x)]

                             = lim_x→(π/2) 2x[(cos² x) / (sin x)] + π

                             = 2 × (π/2) × 0 + π

                             = π.

Answer :

given: lim_x→1 {10(1 – x¹⁰)^–1 – 9(1 – x⁹)^–1}

                                      = lim_x→1 {[10 / (1 – x¹⁰)] – [9 / (1 – x⁹)]}                        (1)

                   As lim_x→1 [m / (1 – x^m)] – [n / (1 – xⁿ)] = [(m – n) / 2] .... (m, n ∈ N)

                   hence from (1), expression is

                                      = [(10 – 9) / 2]

                                      = 0.5.

Answer :

lim_x→α [(m sin x – n cos x) / (x – α)]

                             = m ∙ lim_x→α [{sin x – (n/m)cos x} / (x – α)]

                             = m ∙ lim_x→α [(sin x – tan α ∙ cos x) / (x – α)] ---- m sin α = n cos α

                             = m ∙ lim_x→α [(cos x + tan α ∙ sin x) / 1]

                             = m ∙ lim_x→α cos x + [(sin α) / (cos α)] ∙ sin x

                             = m ∙ [cos x + {(sin² α) / (cos α)}]

                             = m ∙ [1 / (cos α)] = [n / (sin α)]                                               (1)

                   Consider     m² + n² = m² + m² [(sin² α) / (cos² α)]

                                                  = m² [(cos² α + sin² α) / (cos² α)]

                                                  = [m² / (cos² α)]

                   ∴       √(m² + n²) = [m / (cos α)]

                   ∴ from (1) value of given limit is √(m² + n²)

Answer :

lim_x→1 [(x³⁶⁵ – 365x + 364) / (x – 1)²]

                             = lim_x→1 [(x³⁶⁵ – 1 – 365x + 365) / (x – 1)²]

                             = lim_x→1 [{(x – 1)(x³⁶⁴ + x³⁶³ + .... 1) – 365(x – 1)} / (x – 1)²]

                             = lim_x→1 [{(x³⁶⁴ + x³⁶³ + .... 1) – (1 + 1 + 1+ .... 365 times)} / (x – 1)]

                             = lim_x→1 [{(364x³⁶³ + 363x³⁶² + 362x³⁶¹ + ..... 0) – 0} / (1 – 0)]

                                                (∵ lim_x→a [{f(x)} / {g(x)}] = lim_x→a [{f'(x)} / {g'(x)}])

                             = [(364 + 363 + 362 + ..... 0) / 1]

                             = sum of A.P.

                             = (365 / 2) (364 + 0)

                             = 66,430.   

Answer :

lim_x→0 [{(1 + 99x)¹⁰⁰ – (1 + 100x)⁹⁹} / x²]

                             = lim_x→0 [{100(1 + 99x)0⁹⁹(99) – (1 + 100x)⁹⁸(99)(100)} / 2x]

                             = (9900 / 2) [lim_x→0 {[(1 + 99x)⁹⁹ – (1 + 100x)⁹⁸] / x}]

                             = (9900 / 2) [lim_x→0 {[99(1 + 99x)⁹⁸(99) – (98)(1 + 100x)⁹⁷(100)] / 1}]

                             = (9900 / 2) [99 × 99 – 98 × 100]

                             = (9900 / 2)

                             = 4950.