IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

prateek.agarwal

for x>0,

lmt {(sinx)^1/x + (1/x)^sinx} is

x

0

Plz give me a detailed solution of this...

spideyunlimited

are those curly braces for fractional part function or have u just used them for clarity?

bumba

if the curvy brackets r for clarity then is the answer 2.

spideyunlimited

and if they are fractional part function then is the answer 0 ?

kishan12

The answer is 1
i dont know the official method
But this will be helpful to you.
See any where you get 0 raise to inf , it will always be zero and also anything raise to zero is 1.
The different books do not except this but belive me i have solved TMH, arihant, Narayana's module and till now never faced a problem except one or two sum
See use this only when you have no other altenative.it works almost always.
If you get any sum which doesnot follow this plzz post it we can create our own shortcuts right naa....

spideyunlimited

yes i got 1 that way kishan12.. but after that see that it is contained in the function of fractional part which is { }
so we get 0

prateek.agarwal

naah thats just 4 clarity...itsnot 4 fractional part...but plz i want da detailed method...

spideyunlimited

sin x for x as a very small fraction will also be a fraction and in the power there will be infinity so fraction to power infinity is 0.
also 1/x is a fraction and it is raised to power 0 so it will be 1. (1/x is not taken as infinity here as x>0 is mentioned so jus greater than 0)

so, limit is 1

prateek.agarwal

naah...u can't do it like that...these are indeterminant forms and u can't directly find their limits...

jatinroxx

The answer is e+1...

For the solution, plz confirm the answer first...

ambareesh13

i am not able to understand exactly but we can apply the expansion formula for sinx ....still we get the answer as e/ indeterminate form having no formula (1^0) and the other part as 0^infi so i think only method left is the assumption method and the answer would come as e+1... tell me i f i am correct

jatinroxx

@ambareesh

Thats exactly wat i've done too and its correct...

As for 1^inf is concerned...

U can take log.

U get a 0 * inf form which can easily be converted into 0/0 form and then L-Hospitalz rule can b used to solve the limit....

vishal_tg

try to use L hospital rule

swati.07

Let _{[ x]}

_{[ 0]} {1/x}^sinx = y

Taking log on both the sides,we get

log y = _{[x ]}

_{[ 0]} sinxlog(1/x)

= _{[ x]}

_{[ 0]} -logx/cosecx ( infinity/infinity form )

= _{[x ]}

_{[ 0]} -1/x / -cosecxcotx = 0

Therefore y = 1

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