Ques.   {(x/x+1)^a + sin(1/x)}^x

Solution given: e^1-a

My solution : e

Ques . {x} denotes the fractional part of x. then   ({x}/tan{x})

solution given : not in the option (none of these )

my solution :1

Ques. if f(x)=1/3{f(x+1)+[5/f(x+2)]} and f(x)>0 for all  x belongs to R ,  then f(x)

Solution given : (5/2)^1/2

^{Specially do the third Question ,}

For the 1st question :

put h = 1/x

  {(1+h)^-a + sinh}^1/h 

=    (1-ah+h)^1/h   (after expanding both of the terms and neglecting higher powers)

=    {1 + h(1-a)}^1/h

it is in the form of 1^infinity which is evaluated as f(x)^g(x) = e^{{f(x) - 1}.g(x)}

so above limit becomes e^h(1-a)/h = e^1-a

1)It is of the form 1^infinity.So,the limit becomes

e^{lim(x)[(x/x+1)^a+sin(1/x)-1]} .But xsin(1/x) tends to 1 as x tends to infinity and hence it becomes e^1+lim[-x/].

Now,In the numerator x^a+1 cancels and coeff. of x^a is -a.Hence the limit of this becomes -a and hence the ans is e^1-a

ANSWER TO QS. 2      

                  AS    SO SMETIMES X IS AN INTEGER AND SOMETIME A FRACTIONAL NUMBER SO {X}

           IS AN OSCILLATING FUNCTION BETWEEN       [0,1)  HENCE WE CAN NEVER PREDICT WHAT IT IS

       NOW YOUR QUESTION    {X }   /   TAN {X}          FOR SOMEVALUE  {X} IS 0 THEN ANSWER IS O

         WHILE FOR SOME VALUE IT IS SAY 0.3 THEN IT IS O.3 /  TAN 0.3          HENCE THE FUNCTION IF PLOTTED

     IS PERIODIC  WITH PERIOD 1 AND WOULD BE MOVING UP AND DOWN WITH UNDEFINED AND INTEGERS

      SO LIMIT DOESN'T EXIST ............WE CAN'T PREDICT WHAT THE LIMIT IS

In accordance to the second Ques.

Friend , I am not satisfied with your answer . What you are saying that the given function is an oscillating one but i reall disagree that when {x}=0 then the given function would be zero . Actually it is assuming 0/0 indeterminate form.

For 3rd question,as x tends to infinity,x+1,x+2 also tend to infinity.Let limf(x) as xinfinity be k,then limf(x+1) and limf(x+2) as xinfinity also are k and hence,

3k=k+5/k2k=5/k and hence,k=