Page 54 - Calculus Workbook For Dummies

38 Part II: Limits and Continuity

_
^ h
^ h
h lim f x = 4 because f 5 = 4 and f is continuous from the left at 5 4i.
,
x " 5 -
i lim f x = 6. This question is just like problem 8 except that there’s a hollow dot — instead
^ h
x " 5 +
_
_
of a solid one — at ,5 6i. But the hollow dot at ,5 6i is irrelevant to the limit question — just
like with question 7 where the hole was irrelevant.
j List the x-coordinates of all points of discontinuity of f and state whether the points of disconti-
nuity are removable or non-removable, and state the type of discontinuity — removable, jump,
or infinite.
At x = –7, the vertical asymptote, there is a non-removable, infinite discontinuity.
At x = 5 there’s a non-removable, jump discontinuity.
At x = 13 and x = 18 there are holes which are removable discontinuities. Though infinitely
small, these are nevertheless discontinuities. They’re “removable” discontinuities because you
can “fix” the function by plugging the holes.

k lim sinx does not exist. There’s no limit as x approaches infinity because the curve oscillates —
x " 3
it never settles down to one exact y-value. (The three-part definition of a limit does not apply to
limits at infinity.)
1
l lim = 0. In contrast to sinx, this function does hone in on a single value; as you go out
x " 3 x
further and further to the right, the function gets closer and closer to zero, so that’s the limit.