Limits and Continuity: practice solutions, College – download the PDF

1 Guess a limit from a table ★★★

\(f(2.9)=6.8\), \(f(2.99)=6.98\), \(f(2.999)=6.998\); \(f(3.1)=7.2\), \(f(3.01)=7.02\), \(f(3.001)=7.002\).

The values approach 7 from both sides, so we suggest \(\displaystyle\lim_{x\to3}(2x+1)=7\). (Substitution confirms it: \(2\cdot3+1=7\).)

2 Direct substitution ★★★

A polynomial is continuous, so we substitute: \((-2)^2+3(-2)-1=4-6-1=-3\).

\(\displaystyle\lim_{x\to-2}(x^2+3x-1)=-3\).

3 Applying the limit laws ★★★

1. \(3\cdot4+(-2)=10\).
2. \(4\cdot(-2)=-8\).
3. The denominator limit is \(-2\neq0\), so the limit is \(\dfrac{4}{-2}=-2\).
4. \(4^2-(-2)=16+2=18\).

4 Reading limits on a graph ★★★

1. On the left the curve approaches height 3; on the right it approaches height 2. So the left-hand limit is 3 and the right-hand limit is 2.
2. The one-sided limits differ (\(3\neq2\)), so the limit does not exist.
3. The filled point shows \(h(1)=5\).

5 Two sides of the absolute value ★★★

1. For \(x<0\), \(|x|=-x\), so \(\dfrac{|x|}{x}=-1\). For \(x>0\), \(|x|=x\), so the quotient is \(1\). The limits are \(-1\) (left) and \(1\) (right); the two-sided limit does not exist.
2. For \(x<2\), \(|x-2|=-(x-2)\), so the quotient equals \(-1\). The limit is \(-1\).

6 A simple limit at infinity ★★★

Divide numerator and denominator by \(x\): \(\dfrac{4+\frac7x}{2-\frac1x}\). As \(x\to+\infty\), \(\frac7x\to0\) and \(\frac1x\to0\), so the limit is \(\dfrac42=2\).

7 True or false? ★★★

False. The limit ignores the value at 2. Take \(f(x)=\dfrac{x^2-4}{x-2}\): it is undefined at 2, but \(f(x)=x+2\) for \(x\neq2\), so \(\lim_{x\to2}f(x)=4\).

8 Cancel a common factor ★★★

Substitution gives \(\frac00\). Factor: \(x^2+5x+6=(x+2)(x+3)\). For \(x\neq-3\) the quotient equals \(x+2\), so the limit is \(-3+2=-1\).

9 Rationalize the numerator ★★★

Substitution gives \(\frac00\). Multiply by the conjugate \(\sqrt{x+9}+3\): the numerator becomes \((x+9)-9=x\). So the quotient is \(\dfrac{x}{x(\sqrt{x+9}+3)}=\dfrac{1}{\sqrt{x+9}+3}\) for \(x\neq0\). The limit is \(\dfrac{1}{3+3}=\dfrac16\).

10 Find all asymptotes ★★★

Horizontal: divide by \(x^2\): \(\dfrac{6+\frac1{x^2}}{3-\frac1x}\to\dfrac63=2\) as \(x\to\pm\infty\). So \(y=2\) is a horizontal asymptote.

Vertical: \(3x^2-x=x(3x-1)=0\) at \(x=0\) and \(x=\frac13\). The numerator equals 1 at \(x=0\) and \(\frac23+1=\frac53\) at \(x=\frac13\); neither is 0. So \(x=0\) and \(x=\frac13\) are vertical asymptotes.

11 Square roots at infinity ★★★

We have \(\sqrt{4x^2+1}=|x|\sqrt{4+\frac1{x^2}}\).

For \(x\to+\infty\): \(|x|=x\), so the quotient is \(\dfrac{\sqrt{4+1/x^2}}{1+3/x}\to\dfrac{2}{1}=2\).

For \(x\to-\infty\): \(|x|=-x\), so the quotient is \(-\dfrac{\sqrt{4+1/x^2}}{1+3/x}\to-2\).

12 Is it continuous at 2? ★★★

(1) \(f(2)=3\cdot2-1=5\) is defined.

(2) Left limit: \(2^2+1=5\). Right limit: \(3\cdot2-1=5\). They agree, so the limit exists and equals 5.

(3) The limit equals \(f(2)=5\). All three conditions hold, so \(f\) is continuous at 2.

13 Repair a hole ★★★

For \(x\neq4\), \(f(x)=\dfrac{(x-4)(x+4)}{x-4}=x+4\), so \(\lim_{x\to4}f(x)=8\). Continuity requires \(k=f(4)=8\).

14 Parking garage fees ★★★

\(C(2)=4\cdot2=8\) dollars. For \(1

The one-sided limits differ, so \(C\) has a jump discontinuity at \(t=2\) (not continuous). The same jump of 4 dollars occurs at every integer \(t=1,2,3,4,5\) inside the interval.

15 Squeeze with a cosine ★★★

For all \(x\neq0\), \(-1\le\cos\frac4x\le1\). Multiplying by \(x^2\ge0\): \(-x^2\le x^2\cos\frac4x\le x^2\). Since \(\lim_{x\to0}(-x^2)=\lim_{x\to0}x^2=0\), the squeeze theorem gives \(\lim_{x\to0}x^2\cos\frac4x=0\).

16 Squeeze at infinity ★★★

1. For \(x>0\), \(-\frac1x\le\frac{\sin x}{x}\le\frac1x\). Both bounds tend to 0, so the limit is 0.
2. \(\dfrac{2x+\cos x}{x}=2+\dfrac{\cos x}{x}\). By the same squeeze, \(\frac{\cos x}{x}\to0\), so the limit is \(2\).

17 Locate three roots ★★★

\(f\) is a polynomial, hence continuous. \(f(-3)=-27+12+1=-14\) and \(f(-2)=-8+8+1=1\): opposite signs, so a root lies in \((-3,-2)\).

\(f(0)=1\) and \(f(1)=1-4+1=-2\): a root lies in \((0,1)\).

\(f(1)=-2\) and \(f(2)=8-8+1=1\): a root lies in \((1,2)\).

A cubic has at most three real roots, so these are all of them.

18 Classify the discontinuities ★★★

The denominator factors: \(x^2-3x+2=(x-1)(x-2)\), so \(f\) is undefined at \(x=1\) and \(x=2\). Also \(f(x)=\dfrac{(x-1)(x+1)}{(x-1)(x-2)}\).

At \(x=1\): for \(x\neq1\), \(f(x)=\dfrac{x+1}{x-2}\to\dfrac{2}{-1}=-2\). The limit exists, so it is a removable discontinuity.

At \(x=2\): the numerator tends to \(3\neq0\) while the denominator tends to 0, so \(|f|\to\infty\). It is an infinite discontinuity (vertical asymptote \(x=2\)).

19 Make a piecewise function continuous ★★★

At \(x=0\): the left limit is \(0+1=1\) and \(f(0)=b\), so \(b=1\).

At \(x=3\): the left value is \(3a+b=3a+1\) and the right limit is \(10-3=7\), so \(3a+1=7\) and \(a=2\).

With \(a=2\), \(b=1\), the three pieces connect, and each piece is continuous, so \(f\) is continuous everywhere.

20 Cooling coffee ★★★

(a) \(T(0)=72+98=170\) degrees F. \(T(10)=72+98e^{-1}\approx72+36.05=108.1\) degrees F.

(b) As \(t\to+\infty\), \(e^{-0.1t}\to0\), so \(T(t)\to72\). The line \(T=72\) is a horizontal asymptote: the coffee approaches room temperature. In Celsius, \((72-32)\cdot\frac59\approx22.2\) degrees C.

(c) Solve \(72+98e^{-0.1t}=73\): \(e^{-0.1t}=\frac1{98}\), so \(t=10\ln98\approx45.85\) minutes. The coffee never actually reaches 72.

21 When the IVT applies ★★★

(a) \(4\) lies between \(-3\) and \(5\), so yes, \(f\) must take the value 4. The value 6 is outside \([-3,5]\), and the IVT gives no information (it may or may not be reached).

(b) \(s(0)=-1<0\) and \(s(2)=1>0\), but \(s\) only takes the values \(-1\) and \(1\), so it never equals 0. The IVT requires continuity, and \(s\) has a jump at \(x=1\); the hypothesis fails.

(c) \(g\) is continuous. \(g(0)=1>0\) and \(g(1)=\cos1-1\approx0.540-1=-0.460<0\). By the IVT, there is a root in \((0,1)\).