IIT JEE , AIEEE Forum - Mathematics , Differential Calculus - help me out, man!!! i m near but can't get it.

pratikanand

Dudes,

what i wan't is solution of this q.

_{[x ]}

_{[ 0]} [ (sinx)^1/x+ (1/x)^{sinx .}

i think it is from iit-jee 2006. check it out plz.................

karthik2007

Let the limit be equal to some L

1. L = (sinx)^1/x + (1/x)^sinx

2. Let the two limits be equal to L1 and L2.

3. L1 = (sinx)^1/x => logL1 = log(sinx) / x. as x->0, log(sinx) will tend to

-infinity, and -infinity divided by a small number tending to zero will remain -infinity. So, L1 = e^-infinity = 0.

4. L2 = (1/x)^sinx=> logL2 = (sinx) x log(1/x) which is equal to 0 x a finite number = 0 (After applying the limit x->0). logL2 = 0 => L2 = e⁰= 1.

5. Thus, the required limit is equal to L1+L2 = 0+1 = 1.

Answer is 1.

karthik2007

Is my answer right?

pratikanand

YUP, man u r right although i got the answer on that very day. thanks for ur effort. have got u rated..........