Oct 4, 2011
MATH 140: Calculus I
Tutorial 3
Problem 1
Find the limit: limx→4+
Solution:
4−x
|4−x|
We can rewrite this function as a piecewise function in the following way:
4−x
=
|4 − x|
Now it’s clear that limx→4+
4−x
|4−x|
−1,
1,
x>4
x<4
= 1.
Problem 2
√
Find the limit: limx→3
Solution:
x+6−x
x3 −3x2
√
The square root is hard to work with, so let’s make a change of variables. y 2 = √ x + 6, or x = y 2 − 6. If we
√
find the limit as x → 3, then we are finding the limit as y approaches 3 + 6 = 9 = 3. Let’s write this out:
√
√
x+6−x
x+6−x
lim
= lim 2
x→3 x3 − 3x2
x→3 x (x − 3)
y − (y 2 − 6)
= lim 2
y→3 (y − 6)2 (y 2 − 6 − 3)
−y 2 + y + 6
= lim 2
y→3 (y − 6)2 (y 2 − 9)
−(y + 2)(y − 3)
= lim 2
y→3 (y − 6)2 (y − 3)(y + 3)
−(y + 2)
= lim 2
y→3 (y − 6)2 (y + 3)
−(3 + 2)
= 2
(3 − 6)2 (3 + 3)
−5
=
54
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Oct 4, 2011
MATH 140: Calculus I
Tutorial 3
Problem 3
Find the limit: limx→1
Solution:
x1/3 −1
.
x1/2 −1
Again, it’s hard to work with fractional powers (the cube root and square root, in this problem). So let’s
make a change of variable. The most convenient change of variable will leave us with powers instead of
roots. We can do this by finding the least common denominator of the fractional powers: the least common
denominator of 1/3 and 1/2 is 6. (Recall if we were to add 1/3 and 1/2 we would transform the fractions to
the same denominator: 1/3 + 1/2 = 2/6 + 3/6). Therefore our change of variables is
y = x1/6 .
Let’s simplify.
x1/3 − 1
y2 − 1
= lim 3
1/2
x→1 x
− 1 y→1 y − 1
(y − 1)(y + 1)
= lim
y→1 (y − 1)(y 2 + y + 1)
y+1
= lim 2
y→1 y + y + 1
2
=
3
lim
Problem 4
Let f (x) = 3x2 − 15x + 6. The point P = (2, −12) is a point along this curve. Define Q = (x, f (x)). Find
the slope of the line segment PQ where x = 2.01, x = 2.001, x = 1.99, and x = 1.999. Then guess the value
of the tangent to the curve at x = 2.
Solution:
The slope of a line is equal to the rise (in y) over the run (in x). At x = 2.01, the slope of PQ is:
f (2.01) − f (2)
= −2.97
2.01 − 2
At x = 2.001, the slope of PQ is:
f (2.001) − f (2)
= −2.997
2.001 − 2
At x = 1.99, the slope of PQ is:
f (2) − f (1.99)
= −3.03
2 − 1.99
And at x = 1.999, the slope of PQ is:
f (2) − f (1.999)
= −3.003
2 − 1.999
A reasonable guess for the tangent to f (x) at x = 2 is -3.
Page 2 of 3
Oct 4, 2011
MATH 140: Calculus I
Tutorial 3
Problem 5
Let f (x) be defined as follows:

3x,


 2
x ,
f (x) =

5,


3,
x ≤ −2
−2 < x < 2
x=2
x>2
Find the following limits: limx→2+ f (x), limx→2− f (x), limx→2 f (x), limx→−2+ f (x), limx→−2− f (x), and
limx→−2 f (x).
Solution:
limx→2+ f (x) = 3, since f (x) = 3 for x > 2.
limx→2− f (x) = limx→2− x2 = 4, since f (x) = x2 for |x| < 2.
But since the right and left-sided limits at x = 2 disagree, we know that limx→2 f (x) does not exist.
Similarly, limx→−2+ f (x) = 4, limx→−2− f (x) = −6, and limx→−2 f (x) does not exist. It’s helpful to plot
the function.
Problem 6
Find the vertical asymptotes of f (x) =
Solution:
x2 +3x+1
4x2 −9 .
The vertical asymptotes of this function will be where the denominator is equal to zero. That is, when
4x2 − 9 = 0. Hence the vertical asymptotes are at x = ±3/2.
Problem 7
Find the horizontal asymptotes of f (x) =
Solution:
x2 +3x+1
4x2 −9 .
The horizontal asymptotes of this function can be found by taking the limit as x → ∞ and as x → −∞:
lim f (x) = 1/4
x→∞
lim f (x) = 1/4
x→−∞
Therefore, the only horizontal asymptote of this function is y = 1/4.
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