Page 289 - Calculus Demystified
P. 289

276

















                                          Fig. S1.19(c)                  Solutions to Exercises
                                                                               Fig. S1.19(e)

                                        (g)  Not invertible.
                                                               √
                                 20.    (a)  Invertible, f  −1 (t) =  t.
                                                                t
                                        (b)  Invertible, g −1 (t) = e .
                                        (c)  Invertible, h −1 (t) = Sin −1  t.
                                        (d)  Invertible, f  −1 (t) = Cos −1  t.
                                        (e)  Invertible, g −1 (t) = Tan −1  t.
                                        (f)  Not invertible because h(−1) = h(1) = 1.
                                                                    √
                                        (g)  Invertible, f  −1 (t) =[3 +  9 + 4t]/2.


                               Chapter 2
                                                    x
                                                                                 x
                                  1.    (a)  lim x · e = 0 because x tends to 0 and e tends to 1.
                                             x→0
                                                  2
                                                 x − 1
                                        (b)  lim       = lim x + 1 = 2.
                                             x→1 x − 1    x→1
                                        (c)  lim (x − 2) · cot(x − 2) = lim [(x − 2)/ sin(x − 2)]· cos(x − 2) =
                                             x→2                    x→2
                                             1 · 1 = 1. [Here we use the non-trivial fact, explored in Chapter 5,
                                             that lim (sin h/h) = 1.]
                                                h→0
                                                                 x
                                        (d)  lim x · ln x = lim ln x = ln 1 = 0. [Here we use the non-trivial
                                             x→0          x→0
                                                                              x
                                             fact, explored in Chapter 5, that lim x = 1.]
                                                                         x→0
                                                 2
                                                t − 7t + 12
                                        (e)  lim            = lim(t − 4) =−1.
                                             t→3    t − 3      t→3
   284   285   286   287   288   289   290   291   292   293   294