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lim┬(x→0)⁡ (x cot⁡ (4x))/sin^2⁡ x cot^2⁡ (2x) is equal to | lim┬(x→π/4)⁡ cot^3⁡ x-tan⁡x /cos⁡(x+π/4)

#jee_maths #jee_pyq #jeepyq #jeemains #jeepapersolution #jeepaperanalysis ,
lim┬(x→0)⁡〖(x cot⁡〖(4x)〗)/sin^2⁡〖x cot^2⁡〖(2x)〗〗〗 is equal to
(a) 0 (b) 2 (c) 4 (d) 1
Ans: d
Sol.
Here,
lim┬(x→0)⁡〖(x cot⁡〖(4x)〗)/sin^2⁡〖x cot^2⁡〖(2x)〗〗〗=lim┬(x→0)⁡〖(x tan^2⁡2x)/tan⁡〖4x sin^2⁡x 〗〗
=lim┬(x→0)⁡〖(x(tan^2⁡2x/(4x^2 ))4x^2)/((tan⁡4x/4x)4x(sin^2⁡x/x^2 ) x^2 )〗=1

lim┬(x→π/4)⁡〖cot^3⁡〖x-tan⁡x 〗/cos⁡(x+π/4) 〗 is
(a) 4√2 (b) 8√2
(c) 4 (d) 8
Ans: d
Sol.
lim┬(x→π/4)⁡〖cot^3⁡〖x-tan⁡x 〗/cos⁡(x+π/4) 〗
=lim┬(x→π/4)⁡〖(1-tan^4⁡x)/cos⁡(x+π/4) 〗=2 lim┬(x→π/4)⁡〖(1-tan^2⁡x)/cos⁡(x+π/4) 〗
=2 lim┬(x→π/4)⁡〖cos^2⁡〖x-sin^2⁡x 〗/(1/√2 (cos⁡〖x-sin⁡x 〗 ) )〗∙1/cos^2⁡x
=4√2 lim┬(x→π/4)⁡〖(cos⁡〖x+sin⁡x 〗〗)=8
---------------------------
lim┬(x→π/4)⁡〖cot^3⁡〖x-tan⁡x 〗/cos⁡(x+π/4) 〗 is
(a) 4√2 (b) 8√2
(c) 4 (d) 8
Ans: d
Sol.
lim┬(x→π/4)⁡〖cot^3⁡〖x-tan⁡x 〗/cos⁡(x+π/4) 〗
=lim┬(x→π/4)⁡〖(1-tan^4⁡x)/cos⁡(x+π/4) 〗=2 lim┬(x→π/4)⁡〖(1-tan^2⁡x)/cos⁡(x+π/4) 〗
=2 lim┬(x→π/4)⁡〖cos^2⁡〖x-sin^2⁡x 〗/(1/√2 (cos⁡〖x-sin⁡x 〗 ) )〗∙1/cos^2⁡x
=4√2 lim┬(x→π/4)⁡〖(cos⁡〖x+sin⁡x 〗〗)=8
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Видео lim┬(x→0)⁡ (x cot⁡ (4x))/sin^2⁡ x cot^2⁡ (2x) is equal to | lim┬(x→π/4)⁡ cot^3⁡ x-tan⁡x /cos⁡(x+π/4) канала Math Master - Impetus Gurukul

iit jee previous year maths questions solutions iit maths paper solution nimcet maths paper solution nda maths paper solutions iit topic wise maths paper solutions impetus gurukul sharad gupta ncert class 11 maths solutions ncert class 12 maths paper solutions #jee_maths #jee_pyq #jeepyq #jeemains #jeepapersolution #jeepaperanalysis lim┬(x→0)⁡〖(x cot⁡〖(4x)〗)/sin^2⁡〖x cot^2⁡〖(2x)〗〗〗 is equal to lim┬(x→π/4)⁡〖cot^3⁡〖x-tan⁡x 〗/cos⁡(x+π/4) 〗 is

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