1Yes, Sky has posted my method (ie khan's clue)....
@Eurekha , i think i hav seen similar type of prob in TMH....
lim_{x\to0}\frac{(1+tan^{-1}3x)^{-3}-(1+sin^{-1}3x)^{-3}}{(1+sin^{-1}2x)^{-2}-(1+tan^{-1}2x)^{-2}}
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20 Answers
11nahin sir mere saath bhi aisa hi hua tha but
bahut maatha maarne ke baad mujhe yeh mila
i solved it because it was a CHALLANGE
i was asking as if there could be a shorter method to this 1
1yuppy !!!...
thank god ye strike hua !!...varna pure teen ghante ye sum hi karne mein chala jataa,,,
1
1lim x->0 [....... ]
= lim x->0 [(1-3tan-13x) - (1-2sin-13x)]/[-(1-2tan-12x) - (1-2sin-12x)]
= lim x->0 3/2 [sin-13x - tan-13x]/[tan-12x-sin-12x]
11dude answer to abhi mystery hai
but sri post ur method !!!!!!!!!
1sum1 plzz chk naa...c in all cases as it is, second terms tend to zero..so mujhe kuchh galat nahi lagta use karne mein...
1hey guys, can v use binomial expansion ?? thn it becomes gr8 easy...
62bhai mani ye dekh kar hi mujhe dar lag raha hai..
tu bhahut din se pooch raha hai.. lekin yaar isko dekh kar hi haath utha dete hain... :)
11THIS QUESTION ANSWER : -81/16
THIS QUESTION BECOMES DIRTY LATER ON
IF U HAVE A BETTER AND SHORTER METHOD PLEASE POST IT !!!!!!!!!!!!!
1@manipal: kindly see that the powers are negative. even if we try to differentiate, we get negative powers going on increasing, but not able to get a constant. that is from -3, u get -4, -5 .......... even if u try to bring positive powers by taking reciprocals, then there is a big mess out there.
this is not simple as it might look initially, but this can be made simple with a very nice method.
11IT IS 0/0 FORM
APPLY L HOSPITAL
IF U DONT LIKE THAT THEN APPLY THE EXPANSIONS OF tan-1x AND sin-1x TO GET UR ANSWER
I HOPE MYSTERY IS SOLVED
IF U INSIST I WILL GIVE U THE ANSWER
AND KEEP UR TOPIC NAME SIMPLE
NO CHALLANGES!!!!!!!!!
1@MANIPAL,
Instead of posting meaningless messages,try to solve it!!!