x → 0 lim f ′ ′ ′ ( x ) + 3 x sin x + e x f ′ ( x )
Let f be a differentiable function on R with continuous derivatives til order 4. Suppose f ( − x ) = f ( x ) ∀ x ∈ R . Compute the limit above.
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Good approach.
After showing that the numerator is zero, we still have to show that the denominator is non-zero, in order to conclude that the limit is zero.
Did the same
@Kishore S Shenoy The question said f is an even function : it needn't be a polynomial at all.
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Ya... That's true... Thanks. I need to think...
Hey, but all even functions can be represented like this using McLaren Series... So it should be right... Point me out if I'm wrong...
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http://math.stackexchange.com/questions/553080/can-any-continuous-function-be-represented-as-an-infinite-polynomial
Polynomials of finite degree are different from formal power series. Furthermore, MacLaren is far from representing all possible even functions.
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Given f ( x ) = f ( − x ) That means, f ( x ) is an even function. f ( x ) = a 2 n x 2 n + a 2 n − 2 x 2 n − 2 + ⋯ + a 4 x 4 + a 2 x 2 f ′ ( x ) = ⋯ + 2 a 2 x f ′ ′ ′ ( x ) = ⋯ + 2 4 a 4 x ⇒ f ′ ( 0 ) = 0 f ′ ′ ′ ( 0 ) = 0
Also, x → 0 lim 3 x sin x = x → 0 lim x 3 x 2 sin x = 0 × 1 = 0 x → 0 lim e x = 1
Therefore, x → 0 lim f ′ ′ ′ ( x ) + 3 x sin x + e x f ′ ( x ) = 0 + 0 + 1 0 = 0