Cracking The Ap Calculus ab Exam 2018

(Marvins-Underground-K-12) #1
because:     =  x^2 :    =   .   Next,   simplify    the     top     and     bottom:    

. Now, if we take the limit as x goes to infinity, we get = .


4. −1

Here    we  have    to  think   about   what    happens when    we  plug    in  a   value   that    is  very    close   to  0,  but

a    little  bit     less.   The     top     expression  will    be  negative,   and     the     bottom  expression  will    be

positive,   so  we  get     =   −1.

5. + ∞

In  order   to  evaluate    the limit   as  x   approaches  7,  we  find    the limit   as  it  approaches  7+  (from   the

right)  and the limit   as  it  approaches  7−  (from   the left).  If  the two limits  approach    the same

value,  or  both    approach    positive    infinity    or  both    approach    negative    infinity,   then    the limit   is  that

value,  or  the appropriately-signed    infinity.   If  the two limits  do  not agree,  the limit   “Does   Not

Exist.” Here,   we  see that    as  x   approaches  7+, the top expression  will    approach    7.  The bottom

expression  will    approach    0,  but will    be  a   little  bit positive.   Thus,   the limit   will    be   ,  which

is   +∞.     As x    approaches  7−,     the     top     expression  will    again   approach    7.  The     bottom  will

approach    0   but will    be  a   little  bit positive.   Thus,   the limit   will    be   ,  which   is  +∞. Because

the two limits  are the same,   the limit   is  +∞.

6.

Remember    Rule    No. 4,  which   says    that     .  Here     .  If  we  want    to

evaluate     the     limit   the     long    way,    first   we  divide  the     numerator   and     the     denominator     of  the

expression  by  x:   .  Next,   we  multiply    the numerator   and the denominator of  the top
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