Limit%20

Answer :

in case of (0/0), lim_x→a [{f(x)} / {g(x)}] = lim_x→a [{f'(x)} / {g'(x)}]

                   ∴ given is lim_x→2 [(0 – 8 + 10x – 3x²) / (6x² – 18x + 12)]

                                                = lim_x→2 [(10 – 6x) / (12x – 18)]

                                                = [{10 – (6 × 2)} / {(12 × 2) – 18}]

                                                = [(– 2) / 6]

                                                = – (1/3).

Answer :

lim_x→3 [{√(x² + x + 3) – √(4x + 3)} / (x⁴ – 81)]

           = [{(lim_x→3[{(x² + x + 3)^(1/2) – (15)^(1/2)} / {(x² + x + 3) – 15}]) × lim_x→3(x + 4)

                                      – (lim_x→3[{(4x + 3)^(1/2) – (15)^(1/2)} / {(4x + 3) – 15}]) × 4}

                                                                                      / {lim_x→3[(x⁴ – 3⁴) / (x – 3)]}]

                       = [{(1/2)(15)^–(1/2)(3 + 4) – (4)(1/2)(15)^(1/2)} / (4 ∙ 3³)]

                       = [{(7/2) ∙ (1 / √15) – (2 / √15)} / (4 × 27)]

                       = [3 / (2√15)] × [1 / (4 × 27)]

                       = [1 / {72√(15)}].

Answer :

lim_x→(π/3) [{sin[(π/3) – x]} / (2 cos x – 1)]    [(0/0) form]

                             = lim_x→(π/3) [{– cos[(π/3) – x]} / (– 2 sin x)]

                             = [(cos 0) / {2 sin(π/3)}]

                             = [1 / {2 ∙ (√3 / 2)}]

                             = (1 / √3).   

Answer :

given: lim_x→0 [(sin 2x – tan 2x) / x³]

                                      = lim_x→0 [{tan 2x(cos 2x – 1)} / x³]

                                      = lim_x→0 ∙ [(tan 2x) / x] ∙ [(– 2 sin² x) / x²]

                                      = lim_x→0 ∙ [(tan 2x) / 2x] ∙ (– 4) ∙ lim_x→0 [(sin x) / x]²

                                      = (– 4)(1)(1)

                                      = – 4.

Answer :

lim_{x→–(π/4)} [{sin x ∙ cos(5π / 4) – cos(7π / 4) ∙ cos x} / (π + 4x)]

                             = lim_{x→–(π/4)} [{sin x ∙ [– (1 / √2)] – (1 / √2)cos x} / (π + 4x)]

                             = lim_{x→–(π/4)} [{– sin x ∙ (1 / √2) – (1 / √2)cos x} / {4[(π/4) + x]}]

                             = (– 1) lim_{x→–(π/4)} [{sin[(π/4) + x]} / {4[(π/4) + x]}]

                             = [– (1/4)] lim_{x→–(π/4)} [{sin[(π/4) + x]} / {(π/4) + x}]

                             = – (1/4).