Page 136 - Schaum's Outline of Theory and Problems of Advanced Calculus
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CHAP. 6]                       PARTIAL DERIVATIVES                              127

                                                             x 2  z 2
                              Trace on xz plane ðy ¼ 0Þ is the hyperbola     ¼ 1, y ¼ 0.
                                                             a 2  c 2
                              Trace on any plane z ¼ p parallel to the xy plane is the ellipse
                                            x 2         y 2
                                         2    2  2  þ  2  2  2  ¼ 1
                                        a ð1 þ p =c Þ  b ð1 þ p =c Þ
                           As j pj increases from zero, the elliptic cross section increases in size.
                              The surface is a hyperboloid of one sheet (see Fig. 6-8).

                     LIMITS AND CONTINUITY
                                        2
                      6.4. Prove that lim ðx þ 2yÞ¼ 5.
                                    x!1
                                    y!2                                                  Fig. 6-8
                           Method 1,using definition of limit.
                                                                             2
                              We must show that given any  > 0, we can find  > 0suchthat jx þ 2y   5j <  when 0 < jx   1j < ,
                           0 < j y   2j < .
                              If 0 < jx   1j <  and 0 < j y   2j < ,then1    < x < 1 þ   and 2    < y < 2 þ  ,excluding
                           x ¼ 1; y ¼ 2.
                                                    2
                                          2
                                      2
                           Thus, 1   2  þ   < x < 1 þ 2  þ   and 4   2 < 2y < 4 þ 2 .  Adding,
                                           2
                                               2
                                                                            2
                                                                                2
                                    5   4  þ   < x þ 2y < 5 þ 4  þ   2  or    4  þ   < x þ 2y   5 < 4  þ   2
                                                                   2
                                                                                    2
                              Now if   @ 1, it certainly follows that  5 < x þ 2y   5 < 5 , i.e., jx þ 2y   5j < 5  whenever
                           0 < jx   1j < ,0 < j y   2j < .  Then choosing 5  ¼  , i.e.,   ¼  =5(or   ¼ 1, whichever is smaller), it
                                                                                   2
                                     2
                           follows that jx þ 2y   5j <  when 0 < jx   1j < ,0 < j y   2j < , i.e., lim ðx þ 2yÞ¼ 5.
                                                                                x!1
                                                                                y!2
                           Method 2,using theorems on limits.
                                                      2
                                                                2
                                                  lim ðx þ 2yÞ¼ lim x þ lim 2y ¼ 1 þ 4 ¼ 5
                                                  x!1        x!1   x!1
                                                  y!2        y!2   y!2
                                            2
                      6.5. Prove that f ðx; yÞ¼ x þ 2y is continuous at ð1; 2Þ.
                                                                   2
                              By Problem 6.4, lim f ðx; yÞ¼ 5.  Also, f ð1; 2Þ¼ 1 þ 2ð2Þ¼ 5.
                                          x!1
                                          y!2
                              Then lim f ðx; yÞ¼ f ð1; 2Þ and the function is continuous at ð1; 2Þ.
                                  x!1
                                  y!2
                              Alternatively, we can show, in much the same manner as in the first method of Problem 6.4, that given
                           any  > 0wecan find  > 0suchthat j f ðx; yÞ  f ð1; 2Þj <  when jx   1j < ; j y   2j < .
                                                     2
                                                     x þ 2y;
                                                           ðx; yÞ 6¼ð1; 2Þ .
                                                   0;      ðx; yÞ¼ ð1; 2Þ
                      6.6. Determine whether f ðx; yÞ¼

                           (a) has a limit as x ! 1and y ! 2,  (b)is continuous at ð1; 2Þ.
                           (a)By Problem 6.4, it follows that lim f ðx; yÞ¼ 5, since the limit has nothing to do with the value at ð1; 2Þ.
                                                     x!1
                                                     y!2
                           (b)Since lim f ðx; yÞ¼ 5 and f ð1; 2Þ¼ 0, it follows that lim f ðx; yÞ 6¼ f ð1; 2Þ.  Hence, the function is
                                   x!1                                x!1
                                   y!2                                y!2
                              discontinuous at ð1; 2Þ:
                                                         8  2   2
                                                           x   y
                                                         <
                                                            2
                      6.7. Investigate the continuity of f ðx; yÞ¼  x þ y 2  ðx; yÞ 6¼ð0; 0Þ  at ð0; 0Þ.
                                                         :
                                                           0       ðx; yÞ¼ ð0; 0Þ
                              Let x ! 0 and y ! 0insucha way that y ¼ mx (a line in the xy plane). Then along this line,
                                                           2
                                                               2 2
                                                 2
                                                x   y 2   x   m x       2    2  1   m 2
                                             lim     ¼ lim        ¼ lim  x ð1   m Þ
                                                           2
                                                               2 2
                                                 2
                                             x!0 x þ y 2  x!0 x þ m x   2    2  ¼  1 þ m 2
                                             y!0                    x!0 x ð1 þ m Þ
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