Correct solutions came from Will Cragoe, Lena Folwaczny, Trevor Butenhoffi.  Other correct solutions were submitted by Juan Carlos Marivela, Elie Ghosn, Al Zimmermann, Bill Webb, Nancy Scharzkopf, Lou Cairoli, Siupor Lam.

The limit is l =1/2. 

I'll work out the general case, which, as these things often go, is easier than the special case.  Start with the Mean Value Theorem, 

f (a + h) = f (a) + h f '(a + l h), for some 0 < l  < 1.

Apply this to the function f ' to get  f '(a + l h) = f '(a) + l h f ''(a + d l h), for some 0 < d < 1, so  f (a + h) = f (a) + h f '(a) + l h² f ''(a + d l h).  On the other hand, the Taylor remainder formula states that f (a + h) = f (a) + h f '(a) + h²/2 f ''(a + m h), for some 0 < m  < 1.  Equate the two and cancel and get l f ''(a + d l h) = 1/2 f ''(a + m h).  Assume that f '' is continuous and let h goes to 0, then both  f ''(a + d l h) and f ''(a + m h) approach the (non-zero) value of f ''(a).  So l = 1/2.   

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