In this solution isnt (-h)^(1/2) an imaginary number??? Also how is sin(1/h) as h->0 = h as h->0??? Lim h ->0 (sinh/h) = 1 is applicable on if h tends to 0 But if h->0 (1/h) tends to infinity so this rule isnt applicable Also in this question 0Then (-h)^(1/3) is a real number Isnt this question a liitle incomplete??? In this solution isnt (-h)^(1/2) an imaginary number??? Also how is sin(1/h) as h->0 = h as h->0??? Lim h ->0 (sinh/h) = 1 is applicable on if h tends to 0 But if h->0 (1/h) tends to infinity so this rule isnt applicable Also in this question 0Then (-h)^(1/3) is a real number Isnt this question a liitle incomplete??? iPad lim 2:01 AM Page 10.10 sin lim sin Exit 96% lim Share with your friends Share 0 Aarushi Mishra answered this Dear studentYes -h12 is imaginary when h apporaches zero from right and hand side as h>0 and -h<0 and square root of which shall be imaginaryHowever when h approaches 0 from left hand side, h<0 and -h>0 and square root of -h will be realStill the limit will not exist asLeft hand limit≠ Right hand limitThere is some mistake in the solution you sent. It should be limh→0-h12sin10-h=limh→0-h12sin-1h=limh→0-h12-sin1h=limh→0h12sin1hlimh→0sin1h does not exist as h tends to 0 ,1h tends to ∞, and sine of inifnity cannot be determined. It will keep oscillating as you mentionedBut note -1≤limh→0sin1h≤1 as range of sin x is -1,1limh→0h12sin1hNow h12 tends to 0 as h tends to 0, and sin1h always lies between -1 and 1. Therefore sin1h is finite though can't be determine0 multiplied by some finite quantity shall result in zerolimh→0h12sin1h=0Rest the solution is wrongh12sin1h≠h32No matter how small is the value of h 0 View Full Answer
