## anonymous 5 years ago prove, f(x)=xcos1/x (when x not = 2) =0 (when x=0) f(x) is continuous at the point x=0

1. anonymous

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

2. anonymous

yes yes. i am wrong xnot 0

3. anonymous

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

4. anonymous

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.

5. anonymous

thanks

6. anonymous

in which class u are?

7. anonymous

class 11.u?

8. anonymous

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

9. anonymous

kolkata

10. anonymous

whic school..i am in kolkata

11. anonymous

khidirpur st.thomas

12. anonymous

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

13. anonymous

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

14. anonymous

15. anonymous

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

16. anonymous

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

17. anonymous

did u appear for RMO/INMO any time.

18. anonymous

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!

19. anonymous

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

20. anonymous

u can try me. :)

21. anonymous

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

22. anonymous

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

23. anonymous

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

24. anonymous

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 :)

25. anonymous

wow.u r so brilliant!

26. anonymous

thanks.hope to help u soon :)

27. anonymous

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.

28. anonymous

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.

29. anonymous

show

30. anonymous

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

31. anonymous

see

32. anonymous

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

33. anonymous

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

34. anonymous

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?

35. anonymous

36. anonymous

yes my elder sister helped me

37. anonymous

it is chowdhurysaugata6@gmail.com ok

38. anonymous

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

39. anonymous