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![[Post New]](/templates/default/images/icon_minipost_new.gif) 24 Apr 2008 14:51:24 IST
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hey guys can anybody tell me how to find the limit of f(x)=x sin(1/x) when x approaches to zero?
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24 Apr 2008 14:52:41 IST
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it will be zero itself
i think
am i right
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24 Apr 2008 15:07:48 IST
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yes the answer is zero since sin(1/0)is some finite value
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they guys you canfind it by using the sandwich law here x is approching to zero and here f(x)=xsin1/x so domain of sin is [-1,1] so -1<= six 1/x <=1 -x<= xsin 1/x <=x so lim (-x) = 0 & lim x = 0 x-->0 x-->0 so according to sandwich law lim xsin1/x = 0 x-->0 so i think so your methord is might be wrong
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24 Apr 2008 19:25:56 IST
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yes ..... rathin is right ........
the above limit has to b found out by the sandwich rule .....
the same holds for (x^2)sin(1/x) ..
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24 Apr 2008 21:46:23 IST
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yes guys,the answer is zero.thnxs for replying
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24 Apr 2008 21:53:45 IST
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but thr is no need to apply sandwhich theorem...
we know that for any value of x even if it is infinite sin x
lies b/w [-1,1]
and so 0xsome value of sin x = 0
hence the answer will be 0
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25 Apr 2008 00:15:56 IST
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yes aditi is right
0 multiply by a number which oscilates b\w -1 to +1
will always yeild 0
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25 Apr 2008 10:59:45 IST
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x.sin(1/x)
here max value of sin(1/x) is 1 ...so 1 X 0 =0....hence lt is 0....
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