Introduction to Mathematical Analysis I - Second Edition

161 Exercise 4.4.6 . We first consider the case where n = 1 to get ideas for solving this problem in the general case. From the standard derivative theorems we get that the function is differentiable at any x 6 = 0 with f 0 ( x ) = 2 x − 3 e − 1 x 2 = 2 x 3 e − 1 x 2 . Consider the limit lim x → 0 f ( x ) − f ( 0 ) x − 0 = lim x → 0 e − 1 x 2 x . Letting t = 1 / x and applying L’Hospital rule yields lim x → 0 + e − 1 x 2 x = lim t → ∞ t e t 2 = lim t → ∞ 1 2 te t 2 = 0 . Similarly, lim x → 0 − e − 1 x 2 x = 0 . It follows that f is differentiable on R with f 0 ( x ) =   2 x 3 e − 1 x 2 , if x 6 = 0; 0 , if x = 0 . In a similar way, we can show that f is twice differentiable on R with f 00 ( x ) =   6 x 4 + 2 x 6 e − 1 x 2 , if x 6 = 0; 0 , if x = 0 . Based on these calculations, we predict that f is n times differentiable for every n ∈ N with f ( n ) ( x ) =   P 1 x e − 1 x 2 , if x 6 = 0; 0 , if x = 0 , where P is a polynomial. Now we proceed to prove this conclusion by induction. The conclusion is true for n = 1 as shown above. Given that the conclusion is true for some n ∈ N , for x 6 = 0 we have f ( n + 1 ) ( x ) = − x − 2 P 0 1 x + 2 x 3 P 1 x e − 1 x 2 = Q 1 x e − 1 x 2 , where Q is also a polynomial. It is an easy exercise to write the explicit formula of Q based on P . Moreover, successive applications of l’Hôpital’s rule give lim x → 0 + f ( n ) ( x ) − f ( n ) ( 0 ) x − 0 = lim x → 0 + 1 x P 1 x e − 1 x 2 = lim t → ∞ tP ( t ) e t 2 = 0 . In a similar way, we can show that lim x → 0 − f ( n ) ( x ) − f ( n ) ( 0 ) x − 0 = 0 .