IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

bumba

lim sin x

---------- =????

1/x->0 x

i have solved by sandwich theorem,hoping to get any other process.

arnold

it is not as difficult as u taking it

replace 1/x by t

_{[t ]}

_{[ 0]} t sin(1/t)

as t tends to 0, 1/t tends to infinty but sin(1/t) is any real no. lying b/w -1 &+1

so the answer is simply 0

iitkgp_bipin

Let 1/x = z

Then limit becomes

_[z]

_[0] z.sin(1/z)

Now sin(1/z) is an oscillating function whose values lies between -1 and 1.

Hence as z tends to 0 the limit also becomes zero.

Hence the answer is 0.

Best Wishes