Engineering Calculus Notes 244

Engineering Calculus Notes 244 - 232 CHAPTER 3 REAL-VALUED...

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232 CHAPTER 3. REAL-VALUED FUNCTIONS: DIFFERENTIATION Furthermore, since ( −→ x ) is a polynomial, it is continuous, so that lim −→ x −→ 0 ( −→ x ) = 0 and lim −→ x −→ x 0 p f ( −→ x ) T −→ x 0 f ( −→ x ) P = lim −→ x −→ x 0 [ f ( −→ x ) f ( −→ x 0 ) ( −→ x )] = lim −→ x −→ x 0 [ f ( −→ x ) f ( −→ x 0 )] lim −→ x −→ 0 ( −→ x ) = lim −→ x −→ x 0 [ f ( −→ x ) f ( −→ x 0 )] . But since the denominator in Equation ( 3.4 ) goes to zero, so must the numerator, which says that the limit above is zero. This shows Remark 3.3.2. If f ( −→ x ) is diFerentiable at −→ x = −→ x 0 then it is continuous there. To calculate the “linear part” ( −→ x ) of T −→ x 0 f ( −→ x ) (if it exists), we consider the action of f ( −→ x ) along the line through −→ x 0 with a given direction vector −→ v : this is parametrized by
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