Page 62 - Schaum's Outline of Theory and Problems of Advanced Calculus
P. 62

CHAP. 3]                FUNCTIONS, LIMITS, AND CONTINUITY                        53


                                  Then the graph, shown in the adjoining Fig. 3-12,  f (x)
                              consists of the lines y ¼ 1, x > 3; y ¼ 1, x < 3 and
                              the point ð3; 0Þ.
                           (b)As x ! 3 from the right, f ðxÞ! 1, i.e., lim f ðxÞ¼ 1,
                                                                                                 1
                                                           x!3þ
                              as seems clear from the graph. To prove this we must
                                                                                                      x
                              show that given any  > 0, we can find  > 0suchthat       (3, 0)
                              j f ðxÞ  1j <  whenever 0 < x   1 < .        1
                                  Now since x > 1, f ðxÞ¼ 1 and so the proof con-
                              sists in the triviality that j1   1j <  whenever
                              0 < x   1 < .
                                                                                    Fig. 3-12
                           (c)  As  x ! 3  from  the  left,  f ðxÞ!  1,  i.e.,
                           lim f ðxÞ¼  1.  A proof can be formulated as in (b).
                           x!3
                           (d)Since lim f ðxÞ 6¼ lim f ðxÞ, lim f ðxÞ does not exist.
                                                   x!3
                                   x!3þ     x!3
                     3.13. Prove that lim x sin 1=x ¼ 0.
                                    x!0
                              We must show that given any  > 0, we can find  > 0suchthat jx sin 1=x   0j <  when
                           0 < jx   0j < .
                              If 0 < jxj < ,then jx sin 1=xj¼jxjj sin 1=xj @ jxj <  since j sin 1=xj @ 1for all x 6¼ 0.
                              Making the choice   ¼  ,we see that jx sin 1=xj <  when 0 < jxj < ,completing the proof.
                                         2
                     3.14. Evaluate lim    1=x .
                                  x!0þ 1 þ e
                              As x ! 0þ we suspect that 1=x increases indefinitely, e 1=x  increases indefinitely, e  1=x  approaches 0,
                           1 þ e  1=x  approaches 1; thus the required limit is 2.
                              To prove this conjecture we must show that, given  > 0, we can find  > 0suchthat

                                                       2


                                                             2 <  when  0 < x <
                                                     1 þ e  1=x


                                                     2          2   2   2e  1=x   2
                           Now                             2 ¼               ¼

                                                   1 þ e  1=x        1 þ e  1=x     e 1=x  þ 1
                              Since the function on the right is smaller than 1 for all x > 0, any  > 0will work when e   1.  If
                                        2          e 1=x  þ 1  1  2    1     2                1
                           0 < < 1, then    <  when      > , e 1=x  >    1,  > ln    1 ;or0 < x <  ¼  .
                                      e 1=x  þ 1      2               x                   lnð2=    1Þ
                                                                       1
                     3.15. Explain exactly what is meant by the statement lim  ¼1 and prove the validity of this
                                                                          4
                           statement.                             x!1 ðx   1Þ
                              The statement means that for each positive number M,wecan find a positive number   (depending on
                           M in general) such that
                                                     1
                                                       4  > 4  when  0 < jx   1j <
                                                  ðx   1Þ
                                                  1                      1              1
                                                                      4
                              To prove this note that  > M when 0 < ðx   1Þ <           ffiffiffiffiffi.
                                                    4                    M  or 0 < jx   1j < p 4
                                               ðx   1Þ                                  M
                                          p ffiffiffiffiffi
                                          4
                              Choosing   ¼ 1= M,the required results follows.
                                                       sin
                     3.16. Present a geometric proof that lim  ¼ 1.
                                                     !0
                              Construct a circle with center at O and radius OA ¼ OD ¼ 1, as in Fig. 3-13 below. Choose point B on
                           OA extended and point C on OD so that lines BD and AC are perpendicular to OD.
                              It is geometrically evident that
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