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Math 19: Calculus Summer 2010
Practice Problems on Limits and Continuity
1 A tank contains 10 liters of pure water. Salt water containing 20 grams of salt per liter is pumped into the tank at 2 liters per minute.
- Express the salt concentration C(t) after t minutes (in g/L).
- What is the long-term concentration of salt, i.e., limtโโ C(t)?
Solution:
- The concentration is, in units of g/L,
C(t) = total salt total volume
20 ยท 2 ยท t 10 + 2 ยท t
20 t 5 + t
- The long-term concentration is, in units of g/L,
lim tโ+โ
20 t 5 + t = lim tโ+โ
20 t 5 + t
1/t 1/t = lim tโ+โ
5/t + 1
2 Find the values of a and b that make f (x) continuous for all real x.
f (x) =
bex^ + a + 1, x โค 0 ax^2 + b(x + 3 ), 0 < x โค 1 a cos( ฯ x) + 7 bx, x > 1
Solution: We note that the functions are continuous on their domains, so we check that the left- and right-hand limits agree at the boundary x-values. At x = 0,
lim xโ 0 โ^
f (x) = lim xโ 0 โ^
bex^ + a + 1 = b + a + 1,
lim xโ 0 +^
f (x) = lim xโ 0 +^
ax^2 + b(x + 3 ) = 3 b,
so b + a + 1 = 3 b, and a = 2 b โ 1. Next, at x = 1,
lim xโ 1 โ^ f (x) = lim xโ 1 โ^ ax^2 + b(x + 3 ) = a + 4 b, lim xโ 1 +^
f (x) = lim xโ 1 +^
a cos( ฯ x) + 7 bx = โa + 7 b,
so a + 4 b = โa + 7 b, and 2a = 3 b. Solving this linear system in a and b yields b = 2 and a = 3 as the only solution.
4 Show that the equation
x โ 5 =
x + 3 has at least one real solution.
Solution: Let f (x) =
x โ 5 โ
x + 3 , so that f (x) = 0 if and only if x is a solution to the
equation. Then f is defined and continuous for all x โฅ 5. Evaluating f at 5 and at 6, we see that
f ( 5 ) =
< 0 and f ( 6 ) =
By the Intermediate Value Theorem, there is some c in the interval (5, 6) so that f (c) = 0, so f has at least one root. (In fact, it is possible to reduce this equation to the cubic polynomial equation (x โ 5 )(x + 3 )^2 โ 1 = 0, and it is unpleasant but not impossible to find its roots exactly; the only valid root of the original equation is
c = โ
3
3
5 Consider the rational function
f (x) = x^5 โ x^4 โ 2 x^3 x^4 โ 3 x^3 โ x^2 + 3 x
- For what values of a does f have a removable discontinuity at a? What is limxโa f (x) at those a?
- For what values of a does f have an infinite discontinuity at a?
- What is limxโ+โ f (x)? (Hint: Factor the numerator and the denominator.)
Solution: We factor the numerator and denominator of f to obtain
f (x) = x^3 (x + 1 )(x โ 2 ) x(x + 1 )(x โ 1 )(x โ 3 )
Away from x = 0 and x = โ1, f (x) simplifies to
x^2 (x โ 2 ) (x โ 1 )(x โ 3 )
which is defined and continuous at these x-values. Thus, f has removable discontinuities at x = 0 and at x = โ1. From this form of f , we also compute that
lim xโ 0
f (x) = lim xโ 0
x^2 (x โ 2 ) (x โ 1 )(x โ 3 )
= 0 and
lim xโโ 1
f (x) = lim xโโ 1
x^2 (x โ 2 ) (x โ 1 )(x โ 3 )
(โ 1 )^2 (โ 3 )
At x = 1 and at x = 3, however, the numerator of the original function is not 0, so f has infinite singularities (and vertical asymptotes) at these x-values. Finally, we observe that
lim xโ+โ
x^5 โ x^4 โ 2 x^3 x^4 โ 3 x^3 โ x^2 + 3 x
= lim xโ+โ
x โ 1 โ (^2) x 1 โ (^3) x โ (^) x^12 + (^) x^33
