7.1 Define: f(x) is continuous at x - a.
f(a) is defined, exists, and
7.2 Find the points of discontinuity (if any) of the function f(x) whose graph is shown in Fig. 7-1.
Fig. 7-1
x = 0 is a point of discontinuity because lim f(x) does not exist, x = 1 is a point of discontinuity
because lim f ( x ) * f ( l ) [since lim/(jt) = 0 and /(I) = 2].
7.3 Determine the points of discontinuity (if any) of the function f(x) such that f(x) = x2 if x =£ 0 and f(x) - x
if x>0.
f(x) is continuous everywhere. In particular, f(x) is continuous at x = 0 because /(O) = (O)2 = 0 and
lim f(x) = 0.
*-»0
7.4 Determine the points of discontinuity (if any) of the function/(*) such that f(x) = 1 if x^O and /(jt)=-l
if x<0. (See Fig. 7-2.)
Fig. 7-2
/(*) is not continuous at x = 0 because lim f(x) does not exist.
7.5 Determine the points of discontinuity (if any) of the function f(x) such that f(x) = if and
fix) = 0 if x=-2. (See Fig. 7-3.)
Since x2 -4 = (x -2)(x + 2), f(x) = x-2 if x *-2. So, /(*) is not continuous at x=-2 because
lim_^f(x)*f(-2) [since /(-2) = 0 but jmi2/(A:) =-4]. [However, j: =-2 is called a removable dis-
continuity, because, if we redefine f(x) at x= -2 by setting /(-2) = -4, then the new function is
continuous at x = -2. Compare Problem 7.2.]
43
, 44 CHAPTER 7
Fig. 7-3
7.6 Find the points of discontinuity of the function
Since x2 — 1 = (x — l)(x + 1), f(x) = x + l wherever it is defined. However,/(or) is not defined when
x = \, since (x2 - l ) / ( x - 1) does not make sense when x = l. Therefore, f(x) is not continuous at
x=l.
7.7 Find the points of discontinuity (if any) of the function f(x) such that FOR AND
for x = 3.
Since x2 -9 = (x -3)(* + 3), /(*) = *+ 3 for x ^3. However, f(x) = x + 3 also when x = 3, since
/(3) = 6 = 3 + 3. Thus, f(x) = x + 3 for all x, and, therefore, f(x) is continuous everywhere.
7.8 Find the points of discontinuity (if any) of the function /(*) such that
(See Fig. 7-4.)
Fig. 7-4
f(x) is discontinuous at x = 1 because lim f(x) does not exist. f(x) is continuous at x = 2 because
/(2) = 2 + 1 = 3 and lim /(*) = 3. Obviously f(x) is continuous for all other x.
7.9 Find the points of discontinuity (if any) of , and write an equation for each vertical and
horizontal asymptote of the graph of /.
See Fig. 7-5. f(x) is discontinuous at x = 4 and x = -1 because it is
\ S\ f
not defined at those points. [However, x = —1 is a removable discontinuity, If we let the
new function is continuous at *=—!.] The only vertical asymptote is x = 4. Since
the jc-axis, y = 0, is a horizontal asymptote to the right and to the left.
The benefits of buying summaries with Stuvia:
Guaranteed quality through customer reviews
Stuvia customers have reviewed more than 700,000 summaries. This how you know that you are buying the best documents.
Quick and easy check-out
You can quickly pay through credit card for the summaries. There is no membership needed.
Focus on what matters
Your fellow students write the study notes themselves, which is why the documents are always reliable and up-to-date. This ensures you quickly get to the core!
Frequently asked questions
What do I get when I buy this document?
You get a PDF, available immediately after your purchase. The purchased document is accessible anytime, anywhere and indefinitely through your profile.
Satisfaction guarantee: how does it work?
Our satisfaction guarantee ensures that you always find a study document that suits you well. You fill out a form, and our customer service team takes care of the rest.
Who am I buying these notes from?
Stuvia is a marketplace, so you are not buying this document from us, but from seller jureloqoo. Stuvia facilitates payment to the seller.
Will I be stuck with a subscription?
No, you only buy these notes for £6.47. You're not tied to anything after your purchase.