IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

karthik2007

I think I have answered this question already in a different post. Anyway, let me go again.

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

let L1 = (sinx)^1/x.
LogL1 = log(sinx)/x. Log(0) = -infinity. Now, -infinity divided by a number tending to 0 will remain -infinity.

Thus, LogL1 = -infinity, thus, L1 = e^-infinity = 0.

now, L2 = (1/x)^sinx, thus LogL2 = log(1/x)/1/sinx.
This is infinity/infinity form.

Differentiating and applying the Limit, we get LogL2 = 0 => L2 = 1.

This, the required limit is L1+L2 = 0+1 = 1.

spideyunlimited

@karthik2007
let L1 = (sinx)1/x.
LogL1 = log(sinx)/x. Log(0) = -infinity. --> i didnt get this step ! where did log 0 come from?

spideyunlimited

u cant take log of terms separetely mate

spideyunlimited

u cant take log of terms separetely mate

aankurverma

well he is solving 2 terms taking one at a time coz d bracket is simple not fractional

karthik2007

@spider unlimited: It was Log(0)/x. I forgot the x. So, Log0 = -infinity, and when its divided by a small number tending to 0, it remains -infinity.

ambareesh13

hmm we are not having a full proof method but ok still the answer comes out to be one

waterdemon

See this one is simple and full proof

We will use L ' Hospitals rule

Differentiating we get :

The Term (Sinx)^1/x = 0

Now we have :

0 + e^limx0^{Sinx ln(1/x)}

Now this is nothing but equal to 0 +1

= 1

Hope you all find it useful.

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Cheers !!!!!!!!!!!

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