1.1 The Limit of a Function
Finding limits numerically with tables and visually with graphs — and learning when a limit fails to exist.
The idea of a limit is central to all of calculus. Early mathematicians used a limiting process to approximate the areas of circles more and more accurately, yet the formal definition of a limit did not appear until the late 19th century. We begin the way our mathematical ancestors did: with an intuitive approach.
Consider $f(x)=\dfrac{x^2-4}{x-2}$. This function is undefined at $x=2$, but saying only that gives a very incomplete picture. Look at the graph: as $x$ gets close to $2$ from either side, the $y$-values get close to $4$. We say that the limit of $f(x)$ as $x$ approaches $2$ is $4$ and write $\displaystyle \lim_{x\to2}f(x)=4$.
Figure 1.1 — $f(x)=\dfrac{x^2-4}{x-2}$ is the line $y=x+2$ with a hole at $(2,4)$.
📘 Definition — Limit (intuitive)
Let $f(x)$ be defined at all values in an open interval containing $a$, with the possible exception of $a$ itself, and let $L$ be a real number. If all values of $f(x)$ approach $L$ as the values of $x\;(\neq a)$ approach $a$, then we say the limit of $f(x)$ as $x$ approaches $a$ is $L$:
$$\lim_{x\to a} f(x)=L.$$
More succinctly: as $x$ gets closer to $a$, $f(x)$ gets closer and stays close to $L$.
🧭 Strategy — Evaluating a limit with a table of values
- Choose two sets of $x$-values: one approaching $a$ from below ($a-0.1,\ a-0.01,\ a-0.001,\dots$) and one from above ($a+0.1,\ a+0.01,\ a+0.001,\dots$).
- Compute $f(x)$ for each value and read down each column.
- If both columns approach the same number $L$, conclude $\displaystyle \lim_{x\to a}f(x)=L$. If they approach different numbers (or no number), the limit does not exist (DNE).
- Confirm with a graph when possible — zoom in near $x=a$.
Finding Limits with Tables
✏️ Example 1 — Evaluating a limit using a table
Evaluate $\displaystyle \lim_{x\to4}\frac{\sqrt{x}-2}{x-4}$ using a table of functional values.
💡 Solution ▶
Step 1Choose $x$-values approaching $4$ from both sides and compute $f(x)$.
| $x$ | $f(x)$ | $x$ | $f(x)$ |
|---|---|---|---|
| 3.9 | 0.251582342 | 4.1 | 0.248456731 |
| 3.99 | 0.250156446 | 4.01 | 0.249843945 |
| 3.999 | 0.250015627 | 4.001 | 0.249984377 |
| 3.9999 | 0.250001563 | 4.0001 | 0.249998438 |
| 3.99999 | 0.250000156 | 4.00001 | 0.249999844 |
Step 2Read down each column: the left values decrease toward $0.25$; the right values increase toward $0.25$.
Step 3Both sides agree, so $$\lim_{x\to4}\frac{\sqrt{x}-2}{x-4}=0.25=\tfrac14.$$
For a limit to exist, the functional values must approach a single real number. If they do not, the limit does not exist.
✏️ Example 2 — A limit that fails to exist
Evaluate $\displaystyle \lim_{x\to3}\frac{|x-3|}{x-3}$ using a table of functional values.
💡 Solution ▶
Step 1Build the table.
| $x$ | $f(x)$ | $x$ | $f(x)$ |
|---|---|---|---|
| 2.9 | −1 | 3.1 | 1 |
| 2.99 | −1 | 3.01 | 1 |
| 2.999 | −1 | 3.001 | 1 |
| 2.9999 | −1 | 3.0001 | 1 |
| 2.99999 | −1 | 3.00001 | 1 |
Step 2For $x<3$, $|x-3|=-(x-3)$, so $f(x)=-1$. For $x>3$, $|x-3|=x-3$, so $f(x)=1$.
Step 3The two columns approach different values ($-1$ and $1$), so $\displaystyle \lim_{x\to3}\frac{|x-3|}{x-3}$ does not exist.
Finding Limits with Graphs
✏️ Example 3 — Evaluating a limit using a graph
For $g(x)$ shown in Figure 1.2, evaluate $\displaystyle \lim_{x\to-1}g(x)$.
Figure 1.2 — The function $g(x)$: a hole at $(-1,3)$ and a filled point at $(-1,4)$.
💡 Solution ▶
Step 1Trace the curve toward $x=-1$ from the left and from the right. Both paths lead to the height $y=3$ (the hole).
Step 2The filled dot tells us $g(-1)=4$, but the limit ignores the value at $x=-1$.
Step 3Therefore $\displaystyle \lim_{x\to-1}g(x)=3$, even though $g(-1)=4$. We found this without knowing a formula for $g$.
⚠️ Watch out
A limit can exist at a point where the function is defined, yet the limit and the function value can be different. The limit describes where the function is heading, not where it is.
One-Sided Limits
Saying that a limit does not exist does not tell the whole story. For $g(x)=\dfrac{|x-2|}{x-2}$, every value to the left of $2$ equals $-1$ and every value to the right equals $1$. We describe this with one-sided limits: $$\lim_{x\to2^-}g(x)=-1\qquad\text{and}\qquad\lim_{x\to2^+}g(x)=1.$$
📘 Definition — Limit from the left
Let $f(x)$ be defined on an open interval $(c,a)$. If the values of $f(x)$ approach the real number $L$ as $x$ (with $x<a$) approaches $a$, then $L$ is the limit of $f(x)$ as $x$ approaches $a$ from the left: $\displaystyle \lim_{x\to a^-}f(x)=L$.
📘 Definition — Limit from the right
Let $f(x)$ be defined on an open interval $(a,c)$. If the values of $f(x)$ approach the real number $L$ as $x$ (with $x>a$) approaches $a$, then $L$ is the limit of $f(x)$ as $x$ approaches $a$ from the right: $\displaystyle \lim_{x\to a^+}f(x)=L$.
📜 Theorem 1.1 — Relating one-sided and two-sided limits
Let $f(x)$ be defined on an open interval containing $a$, except possibly at $a$, and let $L$ be a real number. Then
$$\lim_{x\to a}f(x)=L\quad\text{if and only if}\quad\lim_{x\to a^-}f(x)=L\ \text{ and }\ \lim_{x\to a^+}f(x)=L.$$
✏️ Example 4 — Evaluating one-sided limits
For $f(x)=\left\{\begin{array}{cl} x+1 & \quad x<2 \\[2pt] x^2-4 & \quad x\ge 2 \end{array}\right.$, evaluate (a) $\displaystyle \lim_{x\to2^-}f(x)$ and (b) $\displaystyle \lim_{x\to2^+}f(x)$.
💡 Solution ▶
Step 1For $x<2$ use $f(x)=x+1$; for $x>2$ use $f(x)=x^2-4$.
| $x$ | $x+1$ | $x$ | $x^2-4$ |
|---|---|---|---|
| 1.9 | 2.9 | 2.1 | 0.41 |
| 1.99 | 2.99 | 2.01 | 0.0401 |
| 1.999 | 2.999 | 2.001 | 0.004001 |
| 1.9999 | 2.9999 | 2.0001 | 0.00040001 |
| 1.99999 | 2.99999 | 2.00001 | 0.0000400001 |
Step 2(a) $\displaystyle \lim_{x\to2^-}f(x)=3$ (b) $\displaystyle \lim_{x\to2^+}f(x)=0$.
Step 3Since $3\neq0$, Theorem 1.1 says the two-sided limit $\displaystyle \lim_{x\to2}f(x)$ does not exist.
Figure 1.3 — The jump at $x=2$: open circle at $(2,3)$, filled point at $(2,0)$.
Exercises 1.1
Exercises 1–10 are below, followed by extra practice (11–14). Use the mini-lab in each card to explore before opening the step-by-step solution.
1
Estimate $\displaystyle \lim_{x\to1}\frac{\frac1x-1}{x-1}$ using a table of functional values.
💡 Step-by-step solution ▶
Step 1 · ChooseUse $x=0.9,\,0.99,\,0.999,\,0.9999,\,0.99999$ and $1.1,\,1.01,\,1.001,\,1.0001,\,1.00001$.
| $x$ | $f(x)$ | $x$ | $f(x)$ |
|---|---|---|---|
| 0.9 | −1.111111111 | 1.1 | −0.909090909 |
| 0.99 | −1.010101010 | 1.01 | −0.990099010 |
| 0.999 | −1.001001001 | 1.001 | −0.999000999 |
| 0.9999 | −1.000100010 | 1.0001 | −0.999900010 |
| 0.99999 | −1.000010000 | 1.00001 | −0.999990000 |
Step 2 · ReadLeft values increase toward $-1$; right values decrease toward $-1$.
Step 3 · Conclude$\displaystyle \lim_{x\to1}\frac{\frac1x-1}{x-1}=-1$. (Check: $\frac{1/x-1}{x-1}=\frac{(1-x)/x}{x-1}=-\frac1x\to-1$.)
2
Use a table of functional values to evaluate $\displaystyle \lim_{x\to2}\frac{|x^2-4|}{x-2}$, if possible.
💡 Step-by-step solution ▶
Step 1 · ChooseUse $x=1.9,\,1.99,\dots$ and $2.1,\,2.01,\dots$
| $x$ | $f(x)$ | $x$ | $f(x)$ |
|---|---|---|---|
| 1.9 | −3.9 | 2.1 | 4.1 |
| 1.99 | −3.99 | 2.01 | 4.01 |
| 1.999 | −3.999 | 2.001 | 4.001 |
| 1.9999 | −3.9999 | 2.0001 | 4.0001 |
| 1.99999 | −3.99999 | 2.00001 | 4.00001 |
Step 2 · ReadLeft column → $-4$; right column → $4$. (Near $2$ with $x<2$, $x^2-4<0$, so $f(x)=-(x+2)$; with $x>2$, $f(x)=x+2$.)
Step 3 · ConcludeThe columns disagree, so $\displaystyle \lim_{x\to2}\frac{|x^2-4|}{x-2}$ does not exist.
3
Use a table of functional values to evaluate $\displaystyle \lim_{x\to0}\frac{\sin x}{x}$, if possible. ($x$ in radians.)
💡 Step-by-step solution ▶
Step 1 · ChooseUse $x=\pm0.1,\ \pm0.01,\ \pm0.001,\ \pm0.0001$ (radians).
| $x$ | $\frac{\sin x}{x}$ | $x$ | $\frac{\sin x}{x}$ |
|---|---|---|---|
| −0.1 | 0.998334166 | 0.1 | 0.998334166 |
| −0.01 | 0.999983333 | 0.01 | 0.999983333 |
| −0.001 | 0.999999833 | 0.001 | 0.999999833 |
| −0.0001 | 0.999999998 | 0.0001 | 0.999999998 |
Step 2 · ReadBoth columns are identical (the function is even) and approach $1$.
Step 3 · Conclude$\displaystyle \lim_{x\to0}\frac{\sin x}{x}=1$.
4
Use the graph of $h(x)$ in Figure 1.1.1 to evaluate $\displaystyle \lim_{x\to2}h(x)$, if possible.
💡 Step-by-step solution ▶
Step 1 · LookThe graph of $h$ is a parabola with its vertex removed: an open circle at $(2,-1)$.
Step 2 · TraceMoving toward $x=2$ from either side, the $y$-values approach $-1$.
Step 3 · Conclude$\displaystyle \lim_{x\to2}h(x)=-1$ (even though $h(2)$ is undefined).
5
Use a table of functional values to estimate, if possible: (a) $\displaystyle \lim_{x\to1^-}\frac{|x^2-1|}{x-1}$ (b) $\displaystyle \lim_{x\to1^+}\frac{|x^2-1|}{x-1}$
💡 Step-by-step solution ▶
Step 1 · ChooseLeft: $x=0.9,\,0.99,\dots$ Right: $x=1.1,\,1.01,\dots$
| $x$ | $f(x)$ | $x$ | $f(x)$ |
|---|---|---|---|
| 0.9 | −1.9 | 1.1 | 2.1 |
| 0.99 | −1.99 | 1.01 | 2.01 |
| 0.999 | −1.999 | 1.001 | 2.001 |
| 0.9999 | −1.9999 | 1.0001 | 2.0001 |
| 0.99999 | −1.99999 | 1.00001 | 2.00001 |
Step 2 · WhyFor $x$ just below $1$, $x^2-1<0$, so $f(x)=-(x+1)$; just above $1$, $f(x)=x+1$.
Step 3 · Conclude(a) $\displaystyle \lim_{x\to1^-}f(x)=-2$ (b) $\displaystyle \lim_{x\to1^+}f(x)=2$. Hence the two-sided limit at $1$ does not exist.
6
Use the graph of $f(x)$ in Figure 1.1.2 to determine each value:
(a) $\displaystyle \lim_{x\to-4^-}f(x);\ \lim_{x\to-4^+}f(x);\ \lim_{x\to-4}f(x);\ f(-4)$
(b) $\displaystyle \lim_{x\to-2^-}f(x);\ \lim_{x\to-2^+}f(x);\ \lim_{x\to-2}f(x);\ f(-2)$
(c) $\displaystyle \lim_{x\to1^-}f(x);\ \lim_{x\to1^+}f(x);\ \lim_{x\to1}f(x);\ f(1)$
(d) $\displaystyle \lim_{x\to3^-}f(x);\ \lim_{x\to3^+}f(x);\ \lim_{x\to3}f(x);\ f(3)$
(a) $\displaystyle \lim_{x\to-4^-}f(x);\ \lim_{x\to-4^+}f(x);\ \lim_{x\to-4}f(x);\ f(-4)$
(b) $\displaystyle \lim_{x\to-2^-}f(x);\ \lim_{x\to-2^+}f(x);\ \lim_{x\to-2}f(x);\ f(-2)$
(c) $\displaystyle \lim_{x\to1^-}f(x);\ \lim_{x\to1^+}f(x);\ \lim_{x\to1}f(x);\ f(1)$
(d) $\displaystyle \lim_{x\to3^-}f(x);\ \lim_{x\to3^+}f(x);\ \lim_{x\to3}f(x);\ f(3)$
💡 Step-by-step solution ▶
(a) x = −4The graph passes smoothly through $(-4,0)$: left limit $=0$, right limit $=0$, $\displaystyle \lim_{x\to-4}f(x)=0$, $f(-4)=0$.
(b) x = −2Open circle at $(-2,3)$, and both sides head to it: left $=3$, right $=3$, $\displaystyle \lim_{x\to-2}f(x)=3$; $f(-2)$ is undefined.
(c) x = 1Filled dot at $(1,6)$ ends the left piece; the right piece starts at an open circle $(1,3)$: left $=6$, right $=3$, $\displaystyle \lim_{x\to1}f(x)$ DNE; $f(1)=6$.
(d) x = 3Both sides plunge down along the vertical asymptote $x=3$: left $=-\infty$, right $=-\infty$, so $\displaystyle \lim_{x\to3}f(x)=-\infty$; $f(3)$ is undefined.
7
Consider the graph of $y=f(x)$ in Figure 1.1.3. Which statements are true and which are false? Explain why a statement is false.
(a) $\displaystyle \lim_{x\to10}f(x)=0$ (b) $\displaystyle \lim_{x\to-2^+}f(x)=3$ (c) $\displaystyle \lim_{x\to-8}f(x)=f(-8)$ (d) $\displaystyle \lim_{x\to6}f(x)=5$
(a) $\displaystyle \lim_{x\to10}f(x)=0$ (b) $\displaystyle \lim_{x\to-2^+}f(x)=3$ (c) $\displaystyle \lim_{x\to-8}f(x)=f(-8)$ (d) $\displaystyle \lim_{x\to6}f(x)=5$
💡 Step-by-step solution ▶
TRUE(a) The last piece is continuous near $x=10$ and crosses the $x$-axis at $(10,0)$, so the limit is $0$.
FALSE(b) Just to the right of $-2$, the middle piece shoots upward along the asymptote $x=-2$: $\displaystyle \lim_{x\to-2^+}f(x)=+\infty$. (The value $3$ is $f(-2)$ and the left-hand limit.)
FALSE(c) The curve approaches the open circle $(-8,-6)$, so $\displaystyle \lim_{x\to-8}f(x)=-6$, but the filled dot gives $f(-8)=-3$. Since $-6\ne-3$, the statement is false.
FALSE(d) $\displaystyle \lim_{x\to6^-}f(x)=2$ (open circle) while $\displaystyle \lim_{x\to6^+}f(x)=5$, so $\displaystyle \lim_{x\to6}f(x)$ does not exist.
8
Use the graph of $y=f(x)$ in Figure 1.1.4 to find, if possible:
(a) $\displaystyle \lim_{x\to-2^-}f(x)$ (b) $\displaystyle \lim_{x\to-2^+}f(x)$ (c) $\displaystyle \lim_{x\to-2}f(x)$ (d) $\displaystyle \lim_{x\to2^-}f(x)$ (e) $\displaystyle \lim_{x\to2^+}f(x)$ (f) $\displaystyle \lim_{x\to2}f(x)$
💡 Step-by-step solution ▶
Step 1 · Read the piecesLeft piece (slope $1$) ends at the open circle $(-2,0)$. The middle piece (slope $-1$) runs from $(-2,2)$ to $(2,-2)$ with filled ends. The right piece (slope $1$) starts at the open circle $(2,2)$.
Step 2 · At x = −2(a) $0$ (b) $2$ (c) DNE, since $0\neq2$.
Step 3 · At x = 2(d) $-2$ (e) $2$ (f) DNE, since $-2\neq2$.
9
Use the graph of $y=f(x)$ in Figure 1.1.5 to find, if possible:
(a) $\displaystyle \lim_{x\to0^-}f(x)$ (b) $\displaystyle \lim_{x\to0^+}f(x)$ (c) $\displaystyle \lim_{x\to0}f(x)$ (d) $\displaystyle \lim_{x\to1}f(x)$ (e) $\displaystyle \lim_{x\to2}f(x)$
💡 Step-by-step solution ▶
Step 1 · Read the piecesFor $x<0$: a line of slope $-1$ ending at the open circle $(0,1)$. For $x\ge0$: the right half of an upward parabola with vertex (filled) $(0,-4)$ through $(2,0)$.
Step 2 · At x = 0(a) $1$ (b) $-4$ (c) DNE, since $1\neq-4$.
Step 3 · Inside the parabola pieceThe parabola is smooth for $x>0$: (d) $\displaystyle \lim_{x\to1}f(x)=-3$ (e) $\displaystyle \lim_{x\to2}f(x)=0$.
10
Using the graph in Figure 1.1.6, find each limit or explain why it does not exist:
(a) $\displaystyle \lim_{x\to-1}f(x)$ (b) $\displaystyle \lim_{x\to1}f(x)$ (c) $\displaystyle \lim_{x\to0^+}f(x)$ (d) $\displaystyle \lim_{x\to2}f(x)$
💡 Step-by-step solution ▶
(a) x = −1From the left the curve ends at the open circle $(-1,1)$ → $1$. From the right the line starts at $(-1,-2)$ → $-2$. Since $1\neq-2$, $\displaystyle \lim_{x\to-1}f(x)$ DNE.
(b) x = 1Left: the curve rises to the open circle $(1,1)$ → $1$. Right: the horizontal line $y=1$ starts at $(1,1)$ → $1$. So $\displaystyle \lim_{x\to1}f(x)=1$, even though $f(1)=-1$.
(c) x → 0⁺The curve to the right of $0$ starts at the open circle $(0,0)$, so $\displaystyle \lim_{x\to0^+}f(x)=0$.
(d) x = 2Near $x=2$ the graph is the flat line $y=1$, so $\displaystyle \lim_{x\to2}f(x)=1$.
11
Use a table of functional values to estimate each limit, then confirm your estimate algebraically.
(a) $\displaystyle \lim_{x\to0}\frac{\sqrt{x+4}-2}{x}$ (b) $\displaystyle \lim_{x\to0}\frac{1-\cos x}{x^2}$ ($x$ in radians) (c) $\displaystyle \lim_{x\to1}\frac{x^3-1}{x-1}$
(a) $\displaystyle \lim_{x\to0}\frac{\sqrt{x+4}-2}{x}$ (b) $\displaystyle \lim_{x\to0}\frac{1-\cos x}{x^2}$ ($x$ in radians) (c) $\displaystyle \lim_{x\to1}\frac{x^3-1}{x-1}$
💡 Step-by-step solution ▶
(a) TableWith $x=\pm0.1,\ \pm0.01,\ \pm0.001,\ \pm0.0001$:
| $x$ | $f(x)$ | $x$ | $f(x)$ |
|---|---|---|---|
| −0.1 | 0.2515823419 | 0.1 | 0.2484567313 |
| −0.01 | 0.2501564456 | 0.01 | 0.2498439450 |
| −0.001 | 0.2500156270 | 0.001 | 0.2499843770 |
| −0.0001 | 0.2500015625 | 0.0001 | 0.2499984375 |
(a) ConcludeBoth columns close in on $0.25$. Check: $\dfrac{\sqrt{x+4}-2}{x}\cdot\dfrac{\sqrt{x+4}+2}{\sqrt{x+4}+2}=\dfrac{x}{x(\sqrt{x+4}+2)}=\dfrac{1}{\sqrt{x+4}+2}\to\dfrac14$. So the limit is $\tfrac14$.
| $x$ | $\frac{1-\cos x}{x^2}$ | $x$ | $\frac{1-\cos x}{x^2}$ |
|---|---|---|---|
| −0.1 | 0.4995834722 | 0.1 | 0.4995834722 |
| −0.01 | 0.4999958333 | 0.01 | 0.4999958333 |
| −0.001 | 0.4999999583 | 0.001 | 0.4999999583 |
(b) ConcludeThe function is even, so the columns match, and they approach $0.5$. Hence $\displaystyle \lim_{x\to0}\frac{1-\cos x}{x^2}=\frac12$. (We prove this exactly in the limit-laws section.)
(c) Table and check$f(0.9)=2.71,\ f(0.99)=2.9701,\ f(0.999)=2.997001$; $f(1.1)=3.31,\ f(1.01)=3.0301,\ f(1.001)=3.003001$ → both sides approach $3$. Check: $x^3-1=(x-1)(x^2+x+1)$, so $f(x)=x^2+x+1\to1+1+1=3$ for $x\ne1$. The limit is $3$.
12
Let $f(x)=\left\{\begin{array}{cl} x+3 & \quad x<-1 \\[2pt] x^2 & \quad -1\le x<2 \\[2pt] 5 & \quad x=2 \\[2pt] 8-2x & \quad x>2 \end{array}\right.$ (Figure 1.1.7). Find each value, or explain why it does not exist.
(a) $\displaystyle \lim_{x\to-1^-}f(x),\ \lim_{x\to-1^+}f(x),\ \lim_{x\to-1}f(x),\ f(-1)$ (b) $\displaystyle \lim_{x\to2}f(x)$ and $f(2)$ (c) $\displaystyle \lim_{x\to0}f(x)$ (d) $\displaystyle \lim_{x\to3}f(x)$
(a) $\displaystyle \lim_{x\to-1^-}f(x),\ \lim_{x\to-1^+}f(x),\ \lim_{x\to-1}f(x),\ f(-1)$ (b) $\displaystyle \lim_{x\to2}f(x)$ and $f(2)$ (c) $\displaystyle \lim_{x\to0}f(x)$ (d) $\displaystyle \lim_{x\to3}f(x)$
💡 Step-by-step solution ▶
(a) x = −1For $x<-1$ use $x+3$: $\displaystyle \lim_{x\to-1^-}f(x)=-1+3=2$. For $-1\le x<2$ use $x^2$: $\displaystyle \lim_{x\to-1^+}f(x)=(-1)^2=1$. Since $2\ne1$, $\displaystyle \lim_{x\to-1}f(x)$ DNE. The value is $f(-1)=(-1)^2=1$.
(b) x = 2Left: $\displaystyle \lim_{x\to2^-}x^2=4$. Right: $\displaystyle \lim_{x\to2^+}(8-2x)=4$. Both sides agree, so $\displaystyle \lim_{x\to2}f(x)=4$, while $f(2)=5$. The limit ignores the single point $(2,5)$.
(c) x = 0Near $0$ (on both sides) $f(x)=x^2$, so $\displaystyle \lim_{x\to0}f(x)=0$.
(d) x = 3Near $3$ only the piece $8-2x$ is used, so $\displaystyle \lim_{x\to3}f(x)=8-6=2$.
13
Infinite limits and vertical asymptotes. Evaluate each limit (answer $+\infty$, $-\infty$, or DNE where appropriate).
(a) $\displaystyle \lim_{x\to3^-}\frac{1}{x-3}$ (b) $\displaystyle \lim_{x\to3^+}\frac{1}{x-3}$ (c) $\displaystyle \lim_{x\to2}\frac{x+1}{(x-2)^2}$ (d) $\displaystyle \lim_{x\to-1^-}\frac{x}{x+1}$ and $\displaystyle \lim_{x\to-1^+}\frac{x}{x+1}$
(e) Find all vertical asymptotes of $g(x)=\dfrac{x-1}{x^2-x-6}$ (Figure 1.1.8) and describe the behaviour of $g$ on each side of them.
(a) $\displaystyle \lim_{x\to3^-}\frac{1}{x-3}$ (b) $\displaystyle \lim_{x\to3^+}\frac{1}{x-3}$ (c) $\displaystyle \lim_{x\to2}\frac{x+1}{(x-2)^2}$ (d) $\displaystyle \lim_{x\to-1^-}\frac{x}{x+1}$ and $\displaystyle \lim_{x\to-1^+}\frac{x}{x+1}$
(e) Find all vertical asymptotes of $g(x)=\dfrac{x-1}{x^2-x-6}$ (Figure 1.1.8) and describe the behaviour of $g$ on each side of them.
💡 Step-by-step solution ▶
(a), (b)As $x\to3^-$, $x-3$ is a tiny negative number, so $\frac{1}{x-3}\to-\infty$. As $x\to3^+$, $x-3$ is tiny and positive, so $\frac{1}{x-3}\to+\infty$. (The two-sided limit does not exist; $x=3$ is a vertical asymptote.)
(c) Sign checkNumerator $\to3>0$; denominator $(x-2)^2\to0$ through positive values from both sides. Hence $\displaystyle \lim_{x\to2}\frac{x+1}{(x-2)^2}=+\infty$.
(d) Sign checkNumerator $\to-1$. For $x\to-1^-$, $x+1\to0^-$, so $\frac{-1}{0^-}\to+\infty$. For $x\to-1^+$, $x+1\to0^+$, so $\frac{-1}{0^+}\to-\infty$. The two-sided limit does not exist.
(e) Factor$x^2-x-6=(x-3)(x+2)$ and the numerator $x-1$ is nonzero at $x=3$ and $x=-2$, so both are vertical asymptotes. Signs of $\frac{x-1}{(x-3)(x+2)}$: as $x\to-2^-$: $\frac{(-)}{(-)(-)}\to-\infty$; as $x\to-2^+$: $\frac{(-)}{(-)(+)}\to+\infty$; as $x\to3^-$: $\frac{(+)}{(-)(+)}\to-\infty$; as $x\to3^+$: $\frac{(+)}{(+)(+)}\to+\infty$.
14
Applications and concepts.
(a) A car park charges 2 OMR for each hour or part of an hour, up to 4 hours, so $C(t)=2\lceil t\rceil$ for $0<t\le4$ (Figure 1.1.9). Find $\displaystyle \lim_{t\to2^-}C(t)$, $\displaystyle \lim_{t\to2^+}C(t)$, $\displaystyle \lim_{t\to2}C(t)$ and $\displaystyle \lim_{t\to1.5}C(t)$. What do the one-sided limits mean for a driver?
(b) Suppose $\displaystyle \lim_{x\to a^-}f(x)=4$, $\displaystyle \lim_{x\to a^+}f(x)=4$ and $f(a)=7$. What is $\displaystyle \lim_{x\to a}f(x)$? Can $f(a)$ be changed without changing the limit?
(c) Explain why $\displaystyle \lim_{x\to0}\sin\frac{\pi}{x}$ does not exist by evaluating the function at $x=\frac1n$ and at $x=\frac{2}{4n+1}$ for positive integers $n$.
(a) A car park charges 2 OMR for each hour or part of an hour, up to 4 hours, so $C(t)=2\lceil t\rceil$ for $0<t\le4$ (Figure 1.1.9). Find $\displaystyle \lim_{t\to2^-}C(t)$, $\displaystyle \lim_{t\to2^+}C(t)$, $\displaystyle \lim_{t\to2}C(t)$ and $\displaystyle \lim_{t\to1.5}C(t)$. What do the one-sided limits mean for a driver?
(b) Suppose $\displaystyle \lim_{x\to a^-}f(x)=4$, $\displaystyle \lim_{x\to a^+}f(x)=4$ and $f(a)=7$. What is $\displaystyle \lim_{x\to a}f(x)$? Can $f(a)$ be changed without changing the limit?
(c) Explain why $\displaystyle \lim_{x\to0}\sin\frac{\pi}{x}$ does not exist by evaluating the function at $x=\frac1n$ and at $x=\frac{2}{4n+1}$ for positive integers $n$.
💡 Step-by-step solution ▶
(a) Car parkFor $1<t\le2$, $C(t)=4$; for $2<t\le3$, $C(t)=6$. So $\displaystyle \lim_{t\to2^-}C(t)=4$, $\displaystyle \lim_{t\to2^+}C(t)=6$, and $\displaystyle \lim_{t\to2}C(t)$ DNE. Near $t=1.5$ the cost is constantly $4$, so $\displaystyle \lim_{t\to1.5}C(t)=4$. Meaning: leaving just before 2 hours costs 4 OMR, but staying even a moment past 2 hours costs 6 OMR.
(b) ConceptThe one-sided limits agree, so $\displaystyle \lim_{x\to a}f(x)=4$. The value $f(a)=7$ plays no role: changing (or deleting) $f(a)$ never changes the limit, because the limit only looks at $x$ near $a$, not at $a$.
(c) Two sequencesAt $x=\frac1n$: $\sin(n\pi)=0$. At $x=\frac{2}{4n+1}$: $\sin\!\left(\frac{(4n+1)\pi}{2}\right)=\sin\!\left(2n\pi+\frac{\pi}{2}\right)=1$.
(c) ConcludeBoth sequences of $x$-values approach $0$, yet the function values stay at $0$ along one and at $1$ along the other. The outputs do not settle on one number, so $\displaystyle \lim_{x\to0}\sin\frac{\pi}{x}$ does not exist.
🔬 Research–Teaching NexusResearch-oriented · optional, about 30 minutes
Research question. How many digits of a table do you need before a limit is clear, and when does the calculator give a wrong answer?
Why it matters. Numerical analysts study rounding error every day; engineers at Oman's refineries and power plants rely on software that must avoid exactly these mistakes.
What to do.
- Make a table of $\frac{\sin x}{x}$ for $x=0.1, 0.01, 0.001, 0.0001$ (radians). How many digits agree with $1$?
- Make a table of $(1+\frac1n)^n$ for $n=10, 10^2, \dots, 10^6$ and compare with $e\approx2.71828$.
- Now try $\frac{1-\cos x}{x^2}$ for $x=10^{-2},10^{-4},10^{-6},10^{-8}$. The true limit is $\frac12$. What does your calculator show?
- Try $(1+\frac1n)^n$ for $n=10^{15}$ and $10^{16}$ in a spreadsheet. Explain what went wrong.
Tools: Calculator, spreadsheet or Desmos
Hand in: A half-page note: your tables, the point where each table stops being reliable, and one reason why.
Going further. Anywhere: repeat the test on two different devices (phone and laptop) and compare where each one fails.
Instructor note: Students should see that very small $x$ causes cancellation ($1-\cos x$ rounds to $0$), so a table can suggest the wrong limit.
🧪 Self-check quiz
0/5 correct
Q1. If $f(2)=5$ and $\displaystyle \lim_{x\to2}f(x)=3$, which statement is correct?
Q2. If $\displaystyle \lim_{x\to a^-}f(x)=4$ and $\displaystyle \lim_{x\to a^+}f(x)=4$, then $\displaystyle \lim_{x\to a}f(x)$ is:
Q3. A table shows left-side values $1.9,\,1.99,\,1.999$ and right-side values $-1.9,\,-1.99,\,-1.999$ as $x\to a$. The limit:
Q4. What is $\displaystyle \lim_{x\to0}\frac{\sin x}{x}$ ($x$ in radians)?
Q5. For $f(x)=\left\{\begin{array}{cl} x+1 & \quad x<2 \\[2pt] x^2-4 & \quad x\ge2 \end{array}\right.$, the value of $\displaystyle \lim_{x\to2^+}f(x)$ is: