Math 1500 Fall 2010
Exam 2 Solutions
October 19, 12:30 PM
1. Evaluate lim
x→1
Name:−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
√
x+
√1
x
6
and justify each step by indicating the appropriate Limit
Laws.
First use the law that the limit of a power of a function is the power of the limit of that function.
6
√
√
1 6
1
lim
x+ √
= lim ( x + √ )
x→1
x→1
x
x
Next the limit of a sum is the sum of a limit.
√
1 6
√
= lim x + lim
x→1
x→1
x
The limit can be brought under the square root and the limit of a quotient is the quotient of a
limit.

6
lim 1
q
=  lim x + qx→1 
x→1
lim x
x→1
And finally the limit of a constant is the constant and the limit as x approaches a of x is a.
√
1 6
1+ √
= 26 = 64.
=
1
2. Evaluate the limit, if it exists. If the limit does not exist, explain why.
1−x
2
x→1 x +x−2
(a) lim
1−x
x2 +x−2
=
−(x−1)
(x−1)(x+2)
=
−1
x+2
1−x
2
x→1 x +x−2
when x 6= 1. So lim
−1
x→1 x+2
= lim
=
−1
3 .
|x|
x→0 x
(b) lim
|x|
x→0− x
This limit is undefined since lim
−x
x→0− x
= lim
|x|
x→0+ x
= −1 and lim
x
x→0+ x
= lim
= 1.
x+2
.
2
x→1+ x −4x+3
3. Find the infinite limit lim
As x approaches 1 from the right, x + 2 is still positive so the numerator is positive. The
denominator factors as x2 − 4x + 3 = (x − 3)(x − 1) and (x − 3) is negative near 1 while (x − 1)
is positive as we approach 1 from the right. So the denominator is negative. Hence the limit
approaches −∞.
4. Find the derivative of the function f (x) =
f (x + h) − f (x)
f (x) = lim
= lim
h→0
h→0
h
0
√
√
x+2 using the definition of the derivative.
x+h+2−
h
1
√
x−2
√
= lim
h→0
x+h−
h
√
x
√
√
x+h+ x
·√
√ .
x+h+ x
= lim
h→0
x+h−x
1
1
√
√ = lim √
√ = √ .
h→0
2
x
h( x + h + x)
x+h+ x
5. Sketch a graph of a function f (x) with all of the following properties:
(a) The function is not continuous at x = 2.
(b) The lim f (x) is undefined.
x→−1
(c) f (−1) = 4.
There are a number of different answers. Here is an example of one:
6. Find the limits.
2x3 +3x−1
3
2
x→∞ 1−5x +4x
(a) lim
This is the quotient of two polynomials with the same degree and so the limit is the ratio of the
coefficients of the leading terms so y = − 52 .
1−x
2
x→∞ 4x +1
(b) lim
Since the denominator is a larger degree polynomial, the limit is y = 0.
7. Use the Intermediate Value Theorem to show that there is a root of the function
f (x) = −x5 − x4 + 6x2 + 2x − 3 on the interval (−1, 0).
f (−1) = −(−1)5 − (−1)4 + 6(−1)2 + 2(−1) − 3 = 1 − 1 + 6 − 2 − 3 = 1 and f (0) = −3. Since one
is positive and one is negative, there is some x = c between x = −1 and x = 0 so that f (c) = 0,
in other words x = c is a root of the function f .
8. (a) What is the formal definition of a function being differentiable at a?
A function is differentiable at a if lim
h→0
f (a+h)−f (a)
h
2
exists.
(b) Find the value for m which would make this function differentiable at x = 2.
(
x2 + 4, if x ≥ 2;
f (x) =
mx,
if x < 2.
For this function to be differentiable, the limit above must exist and the function must be continuous. In order for this to be continuous, the two one-sided limits must equal the function value
at x = 2 In particular, lim f (x) = f (2) and lim mx = 2m while f (2) = 8 so m = 4.
x→2−
x→2−
We need to check that this makes the function differentiable though. Since the function is
piecewise, we need to compare the two one sided limits from the definition of the derivative and
make sure they are the same value at a = 2.
f (2 + h) − f (2)
(2 + h)2 + 4 − 22 − 4
4 + 4h + h2 − 4
h(4 + h)
= lim
= lim
= lim
=4
h→0+
h→0+
h→0+
h→0+
h
h
h
h
lim
4(2 + h) − 8
4h
f (2 + h) − f (2)
= lim
= lim
=4
h→0−
h→0− h
h→0−
h
h
lim
So m = 4 makes the function differentiable.
9. Differentiate the following functions.
√
(a) f (x) = ex · x.
√
Product Rule: f 0 (x) = ex · 12 x−1/2 + x · ex
(b) f (x) =
x+2
x2 −1
Quotient Rule:
(x2 −1)−(x+2)(2x)
(x2 −1)2
10. Find an equation of the tangent line to the curve f (x) = −2x2 + 5x + 4 at the point
x = 1.
We find the slope of the tangent line by finding f 0 (1). f 0 (x) = −4x + 5 and so f 0 (1) = −4 + 5 = 1.
The point (1, f (1)) = (1, 7) is also on the tangent line so point-slope formula gives us y − 7 =
1(x − 1) or y = (x − 1) + 7 or y = x + 6.
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