MATH 150
Midterm 1, October 6, 2005 Solutions
1) (4 marks) Evaluate the following limit if it exists. If it does not exist, explain why. You
must provide your reasoning which CANNOT be simply “this is what the calculator tells
me.”
lim
x→∞ xh1
xi
Here, [x] is the greatest integer less than or equal to x(slightly different notation than the
text). (So h1
xiis the greatest integer less than or equal to 1
x)
Solution:
If x > 1,1
x<1, so h1
xi= 0. Thus, if x > 1, xh1
xi=x·0 = 0, and so
lim
x→∞ xh1
xi= 0
2a) (2 marks) State the Intermediate Value Theorem.
Solution:
If f(x) is continuous on [a, b] with f(a)6=f(b) and if Nis any number between f(a) and
f(b), then there is a c∈(a, b) such that f(c) = N.
2b) (4 marks) Use the Intermediate Value Theorem to find an interval of length less than
or equal to 1
4that contains a root of the equation
f(x) = 2x3−3x+ 4
Explain your answer completely.
Solution:
First, f(x) is a continuous function (so it will be continuous on any interval [a, b]). Here,
N= 0. So to use the IVT, we need to find aand bsuch that f(a)·f(b)<0 (then Nis
between f(a) and f(b)).
f(0) >0 and f(−2) <0, so by IVT there is a root of f(x) in (−2,0). This interval is of
length 2, though. So we need to refine our estimate (find a smaller interval). The easiest
thing to do is to divide our initial interval (−2,0) in half. So we look at x=−1.
f(−1) >0, so by the IVT there is a root in (−2,−1). This interval is of length 1.
f(−3
2)>0, so by IVT there is a root in (−2,−3
2). This interval is of length 1
2.
f(−7
4)<0, so by IVT there is a root in (−7
4,−3
2). This interval is of length 1
4.
Our estimate for the root is then x=−13/8 (the midpoint of (−7
4,−3
2)) with an error of no
more than ±1
8.
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