Question

prove, f(x)=xcos1/x (when x not = 2)
=0 (when x=0)
f(x) is continuous at the point x=0

Answer 1

is your function
f(x)=xcos(1/x) (when x not 0)
=0 when x=0 ?

Answer 2

yes yes. i am wrong xnot 0

Answer 3

\[\lim_{x \rightarrow 0+}xcos(1/x)=\lim_{x \rightarrow 0-}xcos(1/x)=f(0)=0\]
hence continuous at x=0.

Answer 4

if you are wondering why
\[\lim_{x \rightarrow 0+}xcos(1/x)=0\]
then note that \[-1\leq \cos(1/x)\leq 1\ for\ all\ x.\]
So \[\lim_{x \rightarrow 0}xcos(1/x)=\lim_{x \rightarrow 0}x\cdot a=a \lim_{x \rightarrow 0}x=0\]
here a is any real value between -1 and 1.

Answer 5

thanks

Answer 6

in which class u are?

Answer 7

class 11.u?

Answer 8

me in college 1st year:) which city u?

Answer 9

kolkata

Answer 10

whic school..i am in kolkata

Answer 11

khidirpur st.thomas

Answer 12

cbse? if u need help regarding math problems u can ask me! i can suggest a great math book for classes 11 and 12!

Answer 13

oh , say the name.can u suggest some books for mathematical olympiad?

Answer 14

your college?

Answer 15

yeah! I am in Jadavpur University (Mech Engg)
for school to IIT level problems: Problems Plus In IIT Mathematics by Asit Dasgupta

Answer 16

the book by asit dasgupta is great for the basics but for RMO,INMO u need to try some other books...

Answer 17

did u appear for RMO/INMO any time.

Answer 18

yes one time in class 11:was not well prepared:( didnt get through...u need tough preparations for RMO...however if u r a genius u dont need much preparataions!

Answer 19

are u good in probability?it is tough to me.

Answer 20

u can try me. :)

Answer 21

i have no hard problem at present.when i found a very hard problem,i would ask u .

Answer 22

u can call me (if u want my cell no. u can ask me now) or u can mail me to saubhik.mukherjee@gmail.com...i dont come to openstudy often..so u can take my phone no. or gmail

Answer 23

ok.i will mail u.however do u have interest in astrophysics.it is my most fav.

Answer 24

about me: i got 100 in maths in class 12 and 100 in class 10 CBSE. I love olympiad mathematics and want to help anyone interested in olympiad maths.
about astrophysics:there is an olympiad for that and I dont know anything about it :)

Answer 25

wow.u r so brilliant!

Answer 26

thanks.hope to help u soon :)

Answer 27

in triangle ABC,AB=AC,A is right angle.M & N are 2points on BC satisfying BM^2+CN^2=MN^2.
prove that angle MAN=45.

Answer 28

I proved it using coordinate geometry. Did not get a good pure geometry solution though. I can show you the proof. You can post this as a separate question.

Answer 29

show

Answer 30

wait...i will attach a picture wait 2 min.

Answer 31

see

Answer 32

did u see the picture? The side AB is on y-axis and AC is on x-axis...

Answer 33

yes i did and understand.thanks a lot !!!but whenever u get pure geometry solution please post,30 MEDALS for u.

Answer 34

i did not explain anything..u got it? u have to use the distance formula to obtain a relation and then use this relation in another equation which u will get by considering the angle between the two lines AM and AN, which u can get by considering the slopes of the two lines AM and AN. did u get this angle as 45 degrees?

Answer 35

what's your e-mail id?

Answer 36

yes my elder sister helped me

Answer 37

it is chowdhurysaugata6@gmail.com
ok

Answer 38

ok...any more problems u can always ask me:)

Answer 39

i will.buy