Conference rusure::math

>                                   ___
>    The answer is given as:  1/2  V 6   

        					 ---
    				1		V 6
    Presumably you mean    ---------    not     ----
    				 ---              2
			    2 * V 6

    Remember that the question isn't "What is the value of the function
    at the limit point?", which is what you'd get if you just plugged
    in the value (0 in this case).  The question is "What one value
    does the function approach, as x --> 0?".  Or "What is the one value
    that you can get arbitrarily close to simply by taking x close enough
    to 0?"  Even though the function has no value at 0, it gets real
    close to the limit value in the neighborhood of 0.
    
    Does the name L'H�pital sound familiar?  That's an easy way of solving
    these kinds of problems.
    
      John

    I'm not a math guru, but I think the answer is along these lines:
    
    When you let x=0 you would get 0/0.  In that case to find the limit
    use L'Hopital's rule which says take the derivative of the numerator
    and the denominator and then let x->0.  This gives 1/(2*SQRT(6)).
                                                                     
    

    
    
    This is from a calc 1 class; L'Hopital's Rule isn't touched on until
    calc 3.  I looked at it though, and while it did make sense, it
    didn't help me correctly solve the problem.  I understand what you
    said about the limit not actually being 0, but that still doesn't
    change my answer.  Any other suggestions?
    
    Cathy

     If you expand sqrt(x + 6) using the binomial theorem
     you get [that is, one way of doing it gets]
     
                 1/2    1/2         -1/2                -3/2 2
          (6 + x)    = 6    + (1/2)6    x + (1/2)(-1/2)6    x  + ...
                                  
                                     m
     where the terms in ... involve x  for m >= 3.  This
     infinite sum converges for small enough values of x [and
     converges to the expected value].
     
                          1/2
     If you now subtract 6    and divide by x you get
     
                     -1/2         -3/2
        f(x) = (1/2)6     - (1/4)6    x + ...
                                          m
     where here the terms in ... involve x  for m >= 2.
     
     Now as x -> 0, this last expression shows that the function
     approaches the value given by others,
     
                     -1/2
               (1/2)6
     
     Dan           

    
    
    Now I understand.  Guess it was easy after all!
    
    Cathy
    

    Another way to do it is to return to the definition of limit.  Suppose
    that [sqrt(x+6)-sqrt(6)]/x = L + e for x's near zero and small e's
    (positive or negative).  Simply solve for x to get:
    
    	[1 - 2 sqrt(6) (L+e)] / (L+e)^2 = x.
    
    (To get there from the previous equation, multiply both sides by x,
    square, cancel the 6's on each side, divide by x, and you now have a
    linear equation in x.) 
    
    Since (L+e)^2 is positive, to get the above expression near zero, you
    have to make 1 - 2 sqrt(6) (L+e) close to zero, and it is pretty
    obvious what value of L does that. 
    
    This can also be converted directly into a proof, since any desired
    maximum error, e, gives you a range of x's near zero that produce that
    error or less. 
    
    
    				-- edp

    
    
    Another way to obtain the solution is to rationalize the numerator.
     In this case that requires multiplying the numerator and denominator
    by sqrt(x + 6) + sqrt(x).  The result of this would be:
    
        Limit            x
        x->0  ---------------------
              x(sqrt(x+6) + sqrt(x))
    
    which simplifies to  Limit         1
                         x->0 ------------------
                              sqrt(x+6) + sqrt(x)
    
    This limit is 1/(2*sqrt(x)).
    

    Geometrically, you have two curves which happen to intersect at the
    origin... so the limiting behaviour can be visualized in that neighborhood
    as a pair of intersecting tangent lines with (possibly) different slopes.

    I find this easier to remember than a specific formula.

    - Jim

>    
>                                ______      ___
>    Here's the problem:  Lim   V x + 6  -  V 6
>    			 x->0  ----------------
>                                       x
>                                        
>                                   ___
>    The answer is given as:  1/2  V 6   
>                                      
>    

Multiply top and bottom by
	________     ____
       V x + 6   +  V 6

and divide numerator and denominator by x.

The resulting expression can be evaluated at x= 0.