Please solve question no 17 and 18 Please solve question no 17 and 18 Ill IV. lim 1 —cosx In(3x) II. V. 3 tan x lim x2—3x lim In 1 lim x.0 sin(ßx) Q14. Evaluate lim[x — 31, where l.] is the greatest integer function Q15. Evaluate lim x4—5X3+27x-27 tan(x—5) sin(3x—15) Q16. lim x2-12X+35 LEVEL 3 -1 I 18. lim Evaluate lim 311 1—2 cos2 x B) Share with your friends Share 2 Zuhaib Ul Zamann answered this Dear student, Q no 17 Divide Numerator and Denominator by sin2x we getl=limx→0cos(sinx)-1x2×sin2xsin2x=limx→0cos(sinx)-1sin2x×limx→0sin2xx2=limx→0cos(sinx)-1sin2x×1Put sinx =t so that x→0⇒t→0=limt→0cos(t)-1t2=limt→0-2sin2(t2)t2=limt→0-2sin2(t2)(t2)2×(t2)2t2=-2×14=-12Option CQ no 18put tanx =t as x→3π4 t→-11-2cos2x=-cos2x=-1-tan2x1+tan2x=-1-t21+t2So we havel=-limt→-11+t131-t2×(1+t2)Applying LH rule once we get=-limt→-113t-23-2t(2t)=13Hope that helpsRegards 0 View Full Answer
