Cracking The Ap Calculus ab Exam 2018

(Marvins-Underground-K-12) #1
expression  by  3   and the numerator   and the denominator of  the bottom  expression  by  8.  We  get 

. Now, we can evaluate the limit: = .


7. 0

Here,   use the trigonometric   identity    sin^2 x =   1   −   cos^2 x  to  rewrite     the     bottom  expression:    

. Next, we can break up the limit into . Remember that


=    1   and     that    =   1   as  well.   Now     we  can     evaluate    the     limit: 

= (1)(1)(0) = 0.


  1. cos x


Notice   that    if  we  plug    in  0   for    h,   we  get     ,   which   is  indeterminate.  Recall  that    the

trigonometric   formula sin(A   +   B)  =   sin A   cos B   +   cos A   sin B.  Here    we  can rewrite the top

expression  as:     =    .  We  can break   up

the limit   into        +    .  Next,   factor  sin x   out of  the top of  the left-

hand     expression:     +   .   Now,    we  can     break   this    into    separate

limits:  +   .   The     left-hand   limit   is  =  

    =   sin x   •   0   =   0.  The right-hand  limit   is      =   cos x   •   1   =   cos x.

Finally,     combine     the     left-hand   and     right-hand  limits:     sin     x +    

    =   0   +   cos x   =   cos x.


  1. No. It fails condition 1.


In  order   for a   function    f(x)    to  be  continuous  at  a   point   x   =   c,  it  must    fulfill all three   of  the
following conditions:

Condition   1:  f(c)    exists.
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