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Test 1 Solutions Calculus I June 11, 2013
- The graph of a function y = f (x) is shown below. Determine the following limits or state that the limit does not exist:
a) limx→− 1 −^ f (x) = −∞
b) limx→− 1 + f (x) = +∞
c) limx→− 1 f (x) =DNE
d) limx→ 0 − f (x) = 2
e) limx→ 0 + f (x) = 2
f) limx→ 0 f (x) = 2
g) limx→ 2 −^ f (x) = 2
h) limx→ 2 +^ f (x) = 1
i) limx→ 2 f (x) =DNE
- Evaluate the following limit using the limit laws, along with the facts that limx→c k = k and limx→c x = c (show your work and/or explain):
lim x→ 1 cos(πx)(x^2 + 3x − 2) x^3 + 2
limx→ 1
cos(πx)(x^2 + 3x − 2)
limx→ 1
x^3 + 2
limx→ 1 cos(πx)
limx→ 1 (x^2 + 3x − 2)
limx→ 1
x^3 + 2
cos
π(limx→ 1 x)
limx→ 1 x
- 3(limx→ 1 x) − limx→ 1 2 (limx→ 1 x)^3 + limx→ 1 2
cos(π)(1^2 + 3 − 2) 13 + 2
- Calculate the following limit, given that limx→ 2 f (x) = 5 and limx→ 2 g(x) = 3 (show your work and/or explain):
lim x→ 2
x^2 f (x) − [g(x)]^2 cos(f (x) − 2)
cos(5 − 2)
cos(3)
What property of the cosine function are you assuming is true in your computations above?
We are using the fact that cosine is continuous everywhere, so that we can find limits by just plugging in.
- (a) For what value(s) of c is the following function continuous at x = 5 (show your work and/or explain):
f (x) =
x + 3 x > 5 c − x^2 x ≤ 5
Note that f (5) = c− 25 , using the second formula. But using the first, we have that lim x→ 5 +^
f (x) = lim x→ 5 +^
x + 3 = 8.
So to ensure that f (x) is continuous at x = 5 we need to make sure that c − 25 = 8; i.e., we need to make c = 33.
(b) What should we define f (3) to be in order to make the following function continuous at x = 3 (show your work and/or explain):
f (x) = x^2 − x − 6 x − 3
Note that f (x) =
(x − 3)(x − 2) x − 3
. Thus as long as x 6 = 3, we have that f (x) = x − 2. It follows that
x^ lim→ 3 f^ (x) = 3^ −^ 2 = 1, so we need to define f (3) = 1 to make everything continuous.
- (a) Complete the following definition: We say that the limit of a function f (x) as x approaches c equals L if...
for any > 0 we can find a δ > 0 so that |f (x) − L| < whenever |x − c| < δ.
= lim x→∞
3 x^2 − 2 x + 4 2 x^2 + x − 1
1 /x^2 1 /x^2
= lim x→∞
3 − (^2) x + (^) x^42 2 + (^1) x − (^) x^12
where this last equality holds because all the terms with xs in the denominator go to zero as x → ∞.
e) lim x→ 3 x + 1 x − 3
Trying to plug in directly gives 4 / 0 , which is undefined. Thus there is a vertical asymptote at x = 3. When x is slightly larger than (i.e., to the right of) three, the fraction (x + 1)/(x − 3) is approximately four over a tiny positive number. This equals a huge positive number. We deduce that lim x→ 3 +
x + 1 x − 3
Plugging in a number slightly less than (i.e., to the left of) three, we get a fraction which is approximately four over a tiny negative number. This equals a hge negative number. We deduce that
lim x→ 3 −
x + 1 x − 3
As these are different limits, we deduce that the limit as x → 3 does not exist.
- (a) Complete the following definition: A function f (x) is continuous at x = c if three things are true:...
(1) f (c) is defined; (2) limx→c f (x) exists; and (3) limx→c f (x) = f (c).
(b) Use the Intermediate Value Theorem to show that cos(x) = x has a solution.
Consider the function f (x) = cos(x)−x. Note that f (x) is continuous everywhere. We also have f (0) = 1 − 0 = 1 > 0 , while f (π/2) = 0 − π/ 2 < 0. It follows that f (x) has a zero between x = 0 and x = π/ 2. Thus cos(x) = x has a solution between x = 0 and x = π/ 2.
