This is a confusion...
\lim_{x\rightarrow 0}\left[\frac{5sinx}{x} \right] = ? \left[. \right] \; is \; gif
my vote is 5. opponent is 4. what's your vote ? (also explain)
62My vote 4
Reason....
sin x / x <1 if x ≠0
so 5 sin x / x < 5 if x ≠0
near 1, it will tend to 5
Hence the limit is 4
21nishant sir, x>sin x when x≠0, this is not true, what about negative x? further
we say limx→o sin xx = 1 not limx→0sin xx →1
So [ limx→o sin xx ] will be equal to one only.. No?
1i go with nishant sir's explanation
infact i think
limx→0sin xx→1
is right
See the epsilon-delta definition of limits
1from wiki pedia
"
The function has a limit L at an input p if f(x) is "close" to L whenever x is "close" to p. In other words, f(x) becomes closer and closer to L as x moves closer and closer to p. More specifically, when f is applied to each input sufficiently close to p, the result is an output value that is arbitrarily close to L. If the inputs "close" to p are taken to values that are very different, the limit is said to not exist.
"
1it approaches to 5,and it will be almost equal to 5.(slightly less than
5).
so i vote 5.
66@Praveen, probably you would like to look at the definition of box function. Its true that 5 sin x/ x will be very close to 5 but it is also true that it will be less than 5 (but very close). So the [] applied to it will return the integer just less which, in this case, would be 4.
21I perfectly agree with you nishant sir, and kaymant sir. but please answer
[limx→0 5sin x/x] = ?
and explain..
btw ,
please see this http://www.targetiit.com/iit-jee-forum/posts/di-electric-19520.html
21kreyszig,
you siad infact i think
limx→0 sin x/x →1
is right
I can bet my life that THIS IS WRONG.
341i am not sure whether you intended to discuss
\lim_{x \rightarrow 0} \left[\frac{5\sin x}{x} \right]
(which is 4)
or
\left[\lim_{x \rightarrow 0} \frac{5\sin x}{x} \right]
(which is 5)
21Sorry sorry sorry sorry.......
It is not a doubt anymore, but i would like to hear something on why the results are different.
11@ nasiko, think on this :
lim_{x\rightarrow a} f(g(x))
&
f(lim_{x\rightarrow a} g(x)) .
Are these two equal for any functions f and g. What's the exact requirement for equality?
1sin(x)=x ONLY for x=0 (strictly). for every other x in the neighborhood of 0,
/x/>/sin(x)/
your sin(x)/x is not defined at x=0., as we are taking limit x→0, we consider a point close to 0 but not zero.
so, LH limit= RH limit =[5*k] k here is definitely<1 (very close to 1 as x approaches 0)
so the answer 4.
cheers!