Engineering Calculus Notes 235

# Engineering Calculus Notes 235 - denominator can be much...

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3.1. CONTINUITY AND LIMITS 223 This carries over verbatim to functions of several variables: a function f : R 3 R is continuous at a point −→ x 0 in its domain if lim −→ x −→ x 0 f ( −→ x ) = f ( −→ x 0 ) . If a function has a limit at −→ x 0 but fails to be continuous at −→ x 0 either because the limit as −→ x −→ x 0 diFers from the value at −→ x = −→ x 0 , or because f ( −→ x 0 ) is unde±ned, then we can restore continuity at −→ x = −→ x 0 simply by rede±ning the function at −→ x = −→ x 0 to equal its limit there; we call this a removable discontinuity . If on the other hand the limit as −→ x −→ x 0 fails to exist, there is no way (short of major revisionism) of getting the function to be continuous at −→ x = −→ x 0 , and we have an essential discontinuity . Our divergent examples above show that the behavior of a rational function (a ratio of polynomials) in several variables near a zero of its
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Unformatted text preview: denominator can be much more complicated than for one variable, if the discontinuity is essential. Exercises for Â§ 3.1 Practice problems: 1. Â²or each function below, Â±nd its limit as ( x,y ) â†’ (0 , 0): (a) sin( x 2 + y 2 ) x 2 + y 2 (b) x 2 r x 2 + y 2 (c) x 2 x 2 + y 2 (d) 2 x 2 y x 2 + y 2 (e) e x y (f) ( x + y ) 2 âˆ’ ( x âˆ’ y ) 2 xy (g) x 3 âˆ’ y 3 x 2 + y 2 (h) sin( xy ) y (i) e xy âˆ’ 1 y (j) cos( xy ) âˆ’ 1 x 2 y 2 (k) xy x 2 + y 2 + 2 (l) ( x âˆ’ y ) 2 x 2 + y 2 2. Â²ind the limit of each function as ( x,y,z ) â†’ (0 , , 0): (a) 2 x 2 y cos z x 2 + y 2 (b) xyz x 2 + y 2 + z 2 Theory problems: 3. Prove Remark 3.1.2 . 4. Prove Proposition 3.1.4 ....
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