Math 1300: Calculus I

Partial Solutions to Review #1

Here are some partial solutions to the questions on Part II of the Review for Midterm #1.  For most of the problems, I’ve just provided the answer, so that you can check to make sure that you are doing the problem correctly.  In some cases, I’ve provided some of the intermediate steps to help guide you.


1.     See back of book.


2.     See back of book.


3.     The domains of  and  are .  For , the domain is .


4.     Use the limit laws to get 0.


5.     (a)  

(b)  Multiply by  and use the fact that  to get  

(c)   Get common denominators and simplify to get  

(d)  Does not exist since the limit from the right does not agree with the limit from the left.

(e)   Multiply by conjugate and then use the fact that  to get  

(f)   Multiply by conjugate and simplify to get  

(g)    (Look at graph)

(h)  Evaluate each term and look at graph for 3rd term:  

(i)    5

(j)    Limit does not exists because the limit from the left is 1 and the limit from the right is 1.


  1. See back of book.


  1. No,
     is not continuous at
     since the limit of
     does not exist, and so can’t be equal to


  1.  is continuous on
    , therefore, there must exist at  least one value c in the interval
     such that


  1. Factor and see what cancels.  Vertical asymptote at


  1.  and
    , which are not equal, and so
     does not exist.


  1. Just do it:


  1. Take the derivative and get
    , which is undefined at
    .  So,
     does not exist.


  1. (a) Use Power Rule:

(b) Use Quotient Rule: .

(c) Rewrite 3rd term and use Power Rule: .

(d) Rewrite and use Product Rule: .

(e) Use Chain Rule: .

(f) Rewrite and use Power Rule and Chain Rule: .

(g) Use Quotient Rule and Chain Rule: .

(h) Use Product Rule and Chain Rule:  


  1. First, note that
    .  Taking the derivative, we get
    .  Then the slope of the tangent line is
    .  So, the equation of the line is
    , or equivalently,