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llimit                  1  /  ( 1 +  (sin x / x )     )                // can we directly put sin x / x  =1  is this valid     because its not in product form  ?

x ->0

2 Answers

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YES , we can put directly ..there is no need to be in product form etc...BUT if you have doubt then you can use an alternative approach :

simply take LCM first in denominator then you will get  

lim     x / ( x + sinx ) 

x->0

 now since its 0/0 form so we can apply hopital's method hence just differenciate in numerator and denominator independently with respect to x , so we will get ,

lim    1 / (1 + cosx )

x->0

now put the limit i.e x=0 in the above expression :

ans = 1/2  ( since cos0 =1 )

1 votes
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Whenever there is something like (Sin x) / x ; where x is tending to 0 then you can directly place it with 1.

You can get the reason , after expanding the Sin series.

Examiner can play the game for those who just mug the things like

Sin 2x / 2x ; where x is tending to infinity , here you can't just place it with 1 because 2x will chase infinity when x tends to infinity. SO the value for this one is 0 because sin value will always between range [-1,1] and denominator is infinity so finite/infinity ~= 0

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