1.1Evaluate a Polynomial Limit
Evaluate the limit:
$$
\lim_{x \to 4} (2x^2 - 3x + 1)
$$
Solution
Because polynomials are continuous, substitute $x=4$ directly:
$$
2(4^2) - 3(4) + 1 = 32 - 12 + 1 = 21
$$
So the limit is $21$.
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Difficulty
Direct
1.2Factor a Removable Discontinuity
Evaluate the limit:
$$
\lim_{x \to 1} \frac{x^2 - 1}{x - 1}
$$
Solution
Factor the numerator:
$$
x^2 - 1 = (x - 1)(x + 1)
$$
For $x \ne 1$,
$$
\frac{x^2 - 1}{x - 1} = x + 1
$$
Now substitute $x=1$:
$$
1 + 1 = 2
$$
So the limit is $2$.
1.3Read a Left-Hand Limit from a Piecewise Function
Let
$$
f(x)=
\begin{cases}
1, & x<2 \\
5, & x\ge 2
\end{cases}
$$
What is $\lim_{x \to 2^-} f(x)$?
Solution
Approaching $2$ from the left means $x<2$, so the rule $f(x)=1$ applies.
Therefore,
$$
\lim_{x \to 2^-} f(x)=1
$$
1.4Use the Sine Standard Limit
Evaluate the limit:
$$
\lim_{x \to 0} \frac{\sin x}{x}
$$
Solution
This is a standard limit:
$$
\lim_{x \to 0} \frac{\sin x}{x}=1
$$
1.5Identify an Oscillating Limit
Does the limit exist?
$$
\lim_{x \to 0} \sin\!\left(\frac{1}{x}\right)
$$
Solution
No. As $x$ gets close to $0$, the inside value $\frac{1}{x}$ grows without bound, so the sine function keeps oscillating between $-1$ and $1$.
Since it does not approach a single value, the limit does not exist.
1.6Evaluate a Rational Limit at Infinity
Evaluate the limit:
$$
\lim_{x \to \infty} \frac{5x^2 + 1}{2x^2 - 7}
$$
Solution
The numerator and denominator have the same degree, so the limit is the ratio of leading coefficients:
$$
\frac{5}{2}
$$
So the limit is $\frac{5}{2}$.
1.7Recognize an Infinite Limit
Evaluate the limit:
$$
\lim_{x \to 2} \frac{1}{(x-2)^2}
$$
Solution
As $x$ approaches $2$, the denominator $(x-2)^2$ approaches $0$ through positive values.
So the fraction grows without bound:
$$
\lim_{x \to 2} \frac{1}{(x-2)^2}=+\infty
$$
This means $x=2$ is a vertical asymptote.
1.8Use Continuity of a Polynomial
Evaluate the limit:
$$
\lim_{x \to 2} (x^3 - 4x + 1)
$$
Solution
Polynomials are continuous, so substitute $x=2$:
$$
2^3 - 4(2) + 1 = 8 - 8 + 1 = 1
$$
So the limit is $1$.
1.9Evaluate an Exponential Standard Limit
Evaluate the limit:
$$
\lim_{x \to 0} \frac{e^x - 1}{x}
$$
Solution
This is a standard limit:
$$
\lim_{x \to 0} \frac{e^x - 1}{x}=1
$$
1.10Rationalize a Root Limit
Evaluate the limit:
$$
\lim_{x \to 9} \frac{\sqrt{x} - 3}{x - 9}
$$
Solution
Multiply by the conjugate:
$$
\frac{\sqrt{x} - 3}{x - 9}\cdot \frac{\sqrt{x} + 3}{\sqrt{x} + 3}
= \frac{1}{\sqrt{x}+3}
$$
Now substitute $x=9$:
$$
\frac{1}{\sqrt{9}+3}=\frac{1}{6}
$$
So the limit is $\frac{1}{6}$.
Difficulty
Integrated
2.1Make a Removable Discontinuity Continuous
Let
$$
f(x)=
\begin{cases}
\frac{x^2-16}{x-4}, & x\ne 4 \\
k, & x=4
\end{cases}
$$
What value of $k$ makes $f$ continuous at $x=4$?
Solution
Factor the numerator:
$$
x^2-16=(x-4)(x+4)
$$
So for $x\ne 4$,
$$
\frac{x^2-16}{x-4}=x+4
$$
The limit as $x\to 4$ is
$$
4+4=8
$$
To make the function continuous, set
$$
k=8
$$
2.2Determine Whether a Piecewise Limit Exists
Let
$$
g(x)=
\begin{cases}
\frac{x^2-1}{x-1}, & x<1 \\
2x+1, & x\ge 1
\end{cases}
$$
Does $\lim_{x \to 1} g(x)$ exist?
Solution
Find the one-sided limits.
From the left:
$$
\frac{x^2-1}{x-1}=\frac{(x-1)(x+1)}{x-1}=x+1
$$
so
$$
\lim_{x \to 1^-} g(x)=2
$$
From the right:
$$
\lim_{x \to 1^+} (2x+1)=3
$$
Since the one-sided limits are different, the two-sided limit does not exist.
2.3Combine a Factored Limit with a Simple Substitution
Evaluate the limit:
$$
\lim_{x \to 2} \left(\frac{x^2-4}{x-2}+x\right)
$$
Solution
Factor the fraction first:
$$
\frac{x^2-4}{x-2}=\frac{(x-2)(x+2)}{x-2}=x+2
$$
So the expression becomes
$$
(x+2)+x=2x+2
$$
Now substitute $x=2$:
$$
2(2)+2=6
$$
So the limit is $6$.
2.4Use a Standard Trig Limit with a Constant
Evaluate the limit:
$$
\lim_{x \to 0} \frac{\sin(3x)}{x}
$$
Solution
Rewrite the expression so the standard limit appears:
$$
\frac{\sin(3x)}{x}=3\cdot \frac{\sin(3x)}{3x}
$$
Then apply the standard limit:
$$
\lim_{x \to 0} 3\cdot \frac{\sin(3x)}{3x}=3\cdot 1=3
$$
So the limit is $3$.
2.5Apply the Squeeze Theorem
Evaluate the limit:
$$
\lim_{x \to 0} x^2\sin\left(\frac{1}{x}\right)
$$
Solution
Since
$$
-1 \le \sin\left(\frac{1}{x}\right) \le 1,
$$
multiplying by $x^2 \ge 0$ gives
$$
-x^2 \le x^2\sin\left(\frac{1}{x}\right) \le x^2
$$
Both bounds go to $0$ as $x\to 0$, so by the squeeze theorem,
$$
\lim_{x \to 0} x^2\sin\left(\frac{1}{x}\right)=0
$$
2.6Find the Horizontal Asymptote
Find the horizontal asymptote of
$$
f(x)=\frac{4x^3-x}{2x^3+7}
$$
Solution
To find the horizontal asymptote, compute the limit as $x\to\infty$.
The numerator and denominator have the same degree, so the limit is the ratio of leading coefficients:
$$
\lim_{x \to \infty} \frac{4x^3-x}{2x^3+7}=\frac{4}{2}=2
$$
So the horizontal asymptote is
$$
y=2
$$
2.7Use the Logarithmic Standard Limit
Evaluate the limit:
$$
\lim_{x \to 0} \frac{\ln(1+2x)}{x}
$$
Solution
Rewrite the expression to match the standard limit:
$$
\frac{\ln(1+2x)}{x}
=2\cdot \frac{\ln(1+2x)}{2x}
$$
Since
$$
\lim_{u \to 0} \frac{\ln(1+u)}{u}=1,
$$
the limit is
$$
2\cdot 1=2
$$
2.8Choose the Value that Makes a Piecewise Function Continuous
Let
$$
h(x)=
\begin{cases}
x^2+1, & x<1 \\
k, & x=1 \\
2x^2-1, & x>1
\end{cases}
$$
What value of $k$ makes $h$ continuous at $x=1$?
Solution
Find the one-sided limits.
From the left:
$$
\lim_{x \to 1^-}(x^2+1)=1^2+1=2
$$
From the right:
$$
\lim_{x \to 1^+}(2x^2-1)=2(1^2)-1=1
$$
The one-sided limits are different, so no value of $k$ can make the function continuous at $x=1$.
Difficulty
Applied
3.1Model a Vertical Asymptote
A sensor reading is modeled by
$$
P(x)=\frac{1}{(x-5)^2}.
$$
What happens as $x \to 5$?
Solution
As $x$ approaches $5$, the denominator $(x-5)^2$ approaches $0$ through positive values.
So the reading grows without bound:
$$
\lim_{x \to 5} \frac{1}{(x-5)^2}=+\infty
$$
This means $x=5$ is a vertical asymptote.
3.2Interpret a Long-Run Ratio
For
$$
R(t)=\frac{7t^2-3t+1}{2t^2+5},
$$
find the value approached as $t \to \infty$.
Solution
The numerator and denominator have the same degree, so the limit is the ratio of leading coefficients:
$$
\lim_{t \to \infty} R(t)=\frac{7}{2}
$$
So the model approaches $\frac{7}{2}$ in the long run.
3.3Use L'Hopital's Rule Once
Evaluate the limit:
$$
\lim_{x \to 0} \frac{e^{2x}-1}{x}
$$
Solution
The expression has the indeterminate form $\frac{0}{0}$, so L'Hopital's rule applies.
Differentiate the numerator and denominator:
$$
\lim_{x \to 0} \frac{2e^{2x}}{1}
$$
Now substitute $x=0$:
$$
2e^0=2
$$
So the limit is $2$.
3.4Classify a Jump at a Pricing Threshold
A delivery fee is modeled by
$$
F(w)=
\begin{cases}
10+2w, & w<5 \\
w+9, & w\ge 5
\end{cases}
$$
Is $F$ continuous at $w=5$?
Solution
Compute the one-sided limits.
From the left:
$$
\lim_{w \to 5^-} F(w)=10+2(5)=20
$$
From the right:
$$
\lim_{w \to 5^+} F(w)=5+9=14
$$
The one-sided limits are different, so $F$ is not continuous at $w=5$. This is a jump discontinuity.
3.5Spot a Hole and Its Fill-In Value
The function
$$
g(x)=\frac{x^2-9}{x-3}
$$
is undefined at $x=3$.
What type of discontinuity is this, and what value would remove it?
Solution
Factor the numerator:
$$
x^2-9=(x-3)(x+3)
$$
So for $x\ne 3$,
$$
g(x)=x+3
$$
The limit as $x\to 3$ is
$$
3+3=6
$$
Since the limit exists but the original function is undefined at $x=3$, the discontinuity is removable.
To remove it, define $g(3)=6$.
Difficulty
Challenge
4.1Use the Epsilon-Delta Definition
Use the epsilon-delta definition to show that
$$
\lim_{x \to 2} (3x-1)=5.
$$
Give one valid choice of $\delta$ in terms of $\varepsilon$.
Solution
Start with the expression:
$$
|(3x-1)-5|=|3x-6|=3|x-2|
$$
To make this less than $\varepsilon$, it is enough to require
$$
3|x-2|<\varepsilon
$$
So choose
$$
\delta=\frac{\varepsilon}{3}
$$
Then whenever $0<|x-2|<\delta$,
$$
|(3x-1)-5|=3|x-2|<3\delta=\varepsilon
$$
which proves the limit.
4.2Solve for a Continuous Piecewise Rule
Let
$$
f(x)=
\begin{cases}
\frac{x^2-4}{x-2}, & x<2 \\
ax+b, & x\ge 2
\end{cases}
$$
If $f$ is continuous at $x=2$ and $f(3)=10$, find $a$ and $b$.
Solution
First find the limit from the left:
$$
\frac{x^2-4}{x-2}=\frac{(x-2)(x+2)}{x-2}=x+2
$$
So the value needed for continuity at $x=2$ is
$$
2+2=4
$$
That gives the equation
$$
2a+b=4
$$
The condition $f(3)=10$ gives
$$
3a+b=10
$$
Subtract the first equation from the second:
$$
a=6
$$
Then
$$
2(6)+b=4
$$
so
$$
b=-8
$$
4.3Combine Squeeze and a Standard Limit
Evaluate the limit:
$$
\lim_{x \to 0} \frac{x\sin(1/x)}{1+\cos x}
$$
Solution
Use the squeeze theorem first on $x\sin(1/x)$:
$$
-|x| \le x\sin(1/x) \le |x|
$$
So
$$
\lim_{x \to 0} x\sin(1/x)=0
$$
Also,
$$
\lim_{x \to 0} (1+\cos x)=2
$$
Since the denominator approaches a nonzero number, the quotient approaches
$$
\frac{0}{2}=0
$$
So the limit is $0$.
4.4Continuity and End Behavior Together
Let
$$
f(x)=
\begin{cases}
\frac{x^2-1}{x-1}, & x\ne 1 \\
m, & x=1
\end{cases}
$$
Find the value of $m$ that makes $f$ continuous at $x=1$, and decide whether $f$ has a horizontal asymptote.
Solution
Factor the numerator:
$$
x^2-1=(x-1)(x+1)
$$
So for $x\ne 1$,
$$
f(x)=x+1
$$
The limit as $x\to 1$ is
$$
1+1=2
$$
So continuity at $x=1$ requires
$$
m=2
$$
For end behavior, the function behaves like $x+1$ as $x\to\infty$, so it grows without bound.
Therefore, $f$ does not have a horizontal asymptote.