Chapter 1 · Limits and Continuity

1.2 The Limit Laws

Computing limits by algebra instead of tables — and the three standard rescues when direct substitution gives $\tfrac00$.

In the previous section we evaluated limits by looking at graphs or by constructing a table of values. Both are slow, and neither is exact. In this section we establish laws for calculating limits and learn how to apply them. We begin by restating two useful results from the previous section; together with the limit laws they are the foundation for calculating many limits.

📐 Theorem 1.2 — Basic Limit Results
For any real number $a$ and any constant $c$: $$\text{(a)}\quad \lim_{x\to a}x=a \qquad\qquad \text{(b)}\quad \lim_{x\to a}c=c$$
📐 Theorem 1.3 — The Limit Laws
Let $f(x)$ and $g(x)$ be defined for all $x\neq a$ over some open interval containing $a$. Assume $L$ and $M$ are real numbers with $\displaystyle\lim_{x\to a}f(x)=L$ and $\displaystyle\lim_{x\to a}g(x)=M$, and let $c$ be a constant. Then:
(a) Sum law$\displaystyle\lim_{x\to a}\big(f(x)+g(x)\big)=L+M$
(b) Difference law$\displaystyle\lim_{x\to a}\big(f(x)-g(x)\big)=L-M$
(c) Constant multiple law$\displaystyle\lim_{x\to a}\big(cf(x)\big)=c\cdot L$
(d) Product law$\displaystyle\lim_{x\to a}\big(f(x)\cdot g(x)\big)=L\cdot M$
(e) Quotient law$\displaystyle\lim_{x\to a}\frac{f(x)}{g(x)}=\frac{L}{M}$, for $M\neq0$
(f) Power law$\displaystyle\lim_{x\to a}\big(f(x)\big)^n=L^n$, $n$ a positive integer
(g) Root law$\displaystyle\lim_{x\to a}\sqrt[n]{f(x)}=\sqrt[n]{L}$, for all $L$ if $n$ is odd, and for $L\ge0$ if $n$ is even
📐 Theorem 1.4 — Limits of Polynomial and Rational Functions
Let $p(x)$ and $q(x)$ be polynomial functions and let $a$ be a real number. Then $$\lim_{x\to a}p(x)=p(a) \qquad\text{and}\qquad \lim_{x\to a}\frac{p(x)}{q(x)}=\frac{p(a)}{q(a)}\ \text{ when } q(a)\neq0.$$ In other words, for polynomials and for rational functions with a non-zero denominator at $a$, just substitute.
Interactive 1 — The Limit Law Explorer

Fix $f(x)=2x$ and $g(x)=x^2-1$ at $a=2$, so $L=4$ and $M=3$. Pick a law and watch the combined function approach the value the law predicts.

$x$combined value
from the left—
from the right—
what the law predicts—

Applying the Limit Laws

✏️ Example 1 — Evaluating a limit using the limit laws
Use the limit laws to evaluate $\displaystyle\lim_{x\to2}\frac{2x^2-3x+1}{x^3+4}$.
Step 1 Numerator and denominator are polynomials, so Theorem 1.4 applies provided the denominator is not zero at $x=2$. Here $2^3+4=12\neq0$, so we may substitute.
Step 2 Substitute: $$\lim_{x\to2}\frac{2x^2-3x+1}{x^3+4}=\frac{2(2)^2-3(2)+1}{(2)^3+4}=\frac{8-6+1}{12}=\frac{3}{12}=\frac14.$$
🧭 Strategy — when $\dfrac{f(x)}{g(x)}$ has the indeterminate form $\dfrac00$
  1. First check that the function really is in this form and cannot be evaluated immediately with the limit laws.
  2. Find a function equal to $h(x)=\dfrac{f(x)}{g(x)}$ for all $x\neq a$ on some interval containing $a$. One of these three moves usually does it:
    • If $f$ and $g$ are polynomials, factor each and cancel the common factor.
    • If a square root appears in a difference, multiply numerator and denominator by the conjugate.
    • If the expression is a complex fraction, simplify it first.
  3. Apply the limit laws to the simplified function.

The cancellation is legal because a limit never uses the value at $x=a$ — only the values nearby, where $x-a\neq0$.

Interactive 2 — The $\tfrac00$ Workshop
$x$value
value at the pointundefined
from the left—
from the right—
limit—
✏️ Example 2 — Evaluating a limit by factoring and cancelling
Evaluate $\displaystyle\lim_{x\to3}\frac{x^2-3x}{2x^2-5x-3}$.
Step 1 The function is undefined at $x=3$; substituting gives $\tfrac00$. Factoring and cancelling is a good strategy: $$\lim_{x\to3}\frac{x^2-3x}{2x^2-5x-3}=\lim_{x\to3}\frac{x(x-3)}{(x-3)(2x+1)}.$$
Step 2 For all $x\neq3$, $\dfrac{x^2-3x}{2x^2-5x-3}=\dfrac{x}{2x+1}$. Therefore $$\lim_{x\to3}\frac{x(x-3)}{(x-3)(2x+1)}=\lim_{x\to3}\frac{x}{2x+1}.$$
Step 3 Evaluate using the limit laws: $$\lim_{x\to3}\frac{x}{2x+1}=\frac{3}{7}.$$
✏️ Example 3 — Evaluating a limit by multiplying by a conjugate
Evaluate $\displaystyle\lim_{x\to-1}\frac{\sqrt{x+2}-1}{x+1}$.
Step 1 The expression has the form $\tfrac00$ at $-1$. Multiply numerator and denominator by $\sqrt{x+2}+1$, the conjugate of $\sqrt{x+2}-1$: $$\lim_{x\to-1}\frac{\sqrt{x+2}-1}{x+1}=\lim_{x\to-1}\frac{\sqrt{x+2}-1}{x+1}\cdot\frac{\sqrt{x+2}+1}{\sqrt{x+2}+1}.$$
Step 2 Multiply out the numerator only — we are hoping the $(x+1)$ in the denominator will cancel: $$\lim_{x\to-1}\frac{(x+2)-1}{(x+1)\left(\sqrt{x+2}+1\right)}=\lim_{x\to-1}\frac{x+1}{(x+1)\left(\sqrt{x+2}+1\right)}.$$
Step 3 Cancel: $$\lim_{x\to-1}\frac{1}{\sqrt{x+2}+1}.$$
Step 4 Apply the limit laws: $$\lim_{x\to-1}\frac{1}{\sqrt{x+2}+1}=\frac{1}{\sqrt{1}+1}=\frac12.$$
✏️ Example 4 — Evaluating a limit by simplifying a complex fraction
Evaluate $\displaystyle\lim_{x\to1}\frac{\dfrac{1}{x+1}-\dfrac12}{x-1}$.
Step 1 The expression has the form $\tfrac00$ at $1$. Combine the two fractions on top over the common denominator $2(x+1)$: $$\lim_{x\to1}\frac{\dfrac{1}{x+1}-\dfrac12}{x-1}=\lim_{x\to1}\frac{\dfrac{2-(x+1)}{2(x+1)}}{x-1}=\lim_{x\to1}\frac{-x+1}{2(x+1)(x-1)}=\lim_{x\to1}\frac{-(x-1)}{2(x+1)(x-1)}.$$
Step 2 Cancel the common factor $(x-1)$: $$=\lim_{x\to1}\frac{-1}{2(x+1)}.$$
Step 3 Evaluate using the limit laws: $$\lim_{x\to1}\frac{-1}{2(x+1)}=-\frac14.$$
✏️ Example 5 — A two-sided limit of a piecewise function
For $f(x)=\left\{\begin{array}{cl} 4x-3 & \quad x<2 \\[2pt] (x-3)^2 & \quad x\ge2 \end{array}\right.$ evaluate (a) $\displaystyle\lim_{x\to2^-}f(x)$, (b) $\displaystyle\lim_{x\to2^+}f(x)$, (c) $\displaystyle\lim_{x\to2}f(x)$.
(a) Since $f(x)=4x-3$ for all $x<2$, replace $f(x)$ by $4x-3$ and apply the limit laws: $$\lim_{x\to2^-}f(x)=\lim_{x\to2^-}(4x-3)=5.$$
(b) Since $f(x)=(x-3)^2$ for all $x\ge2$, replace $f(x)$ by $(x-3)^2$: $$\lim_{x\to2^+}f(x)=\lim_{x\to2^+}(x-3)^2=1.$$
(c) The one-sided limits are $5$ and $1$. They differ, so $\displaystyle\lim_{x\to2}f(x)$ does not exist.
Why this works On each side only one formula is ever used, so each one-sided limit is an ordinary polynomial limit — just substitute into the right branch.
Example 5 — the two branches arrive at different heights at $x=2$, so the two-sided limit fails.
✏️ Example 6 — A limit of the form $\dfrac{K}{0}$ with $K\neq0$
Evaluate $\displaystyle\lim_{x\to2^-}\frac{x-3}{x^2-2x}$.
Step 1 Substituting $x=2$ gives the form $\dfrac{-1}{0}$: the numerator approaches $-1$ while the denominator approaches $0$. Consequently the magnitude of the quotient becomes infinite. Factor the denominator to see the structure: $$\lim_{x\to2^-}\frac{x-3}{x^2-2x}=\lim_{x\to2^-}\frac{x-3}{x(x-2)}.$$
Step 2 Only $x-2$ vanishes at $2$, so separate that factor from the rest: $$=\lim_{x\to2^-}\frac{x-3}{x}\cdot\frac{1}{x-2}.$$
Step 3 Now $\displaystyle\lim_{x\to2^-}\frac{x-3}{x}=\frac{-1}{2}$ and $\displaystyle\lim_{x\to2^-}\frac{1}{x-2}=-\infty$. A negative quantity times $-\infty$ is $+\infty$: $$\lim_{x\to2^-}\frac{x-3}{x^2-2x}=+\infty.$$
Warning $\dfrac{K}{0}$ with $K\neq0$ is not indeterminate — there is nothing to cancel. The work is to decide the sign, and that depends on which side you approach from. From the right, $x-2>0$, and the same reasoning gives $-\infty$.
⚠️ Two forms, two completely different jobs

$\tfrac00$ is indeterminate. It tells you nothing yet. Factor, use a conjugate or simplify the complex fraction, then substitute.

$\tfrac{K}{0}$ with $K\neq0$ is not indeterminate. The magnitude blows up; your only task is the sign of the result on each side, so the answer is $+\infty$, $-\infty$, or "does not exist" when the two sides disagree.

Exercises

The Section 1.2 exercises come first (I–V), followed by extra practice sets VI–VIII. Work each one before opening its solution; the mini-labs let you check your answer numerically.

I. Direct substitution

I
Use direct substitution to evaluate each limit.
  • 1 $\displaystyle\lim_{x\to-2}\left(4x^2-1\right)$
  • 2 $\displaystyle\lim_{x\to3}\frac{2x^2-3x+1}{5x+4}$
  • 3 $\displaystyle\lim_{x\to0}\frac{1}{1+\sin x}$
  • 4 $\displaystyle\lim_{x\to1}\frac{2-7x}{x+6}$
  • 5 $\displaystyle\lim_{x\to2}e^{\,2x-x^2}$
  • 6 $\displaystyle\lim_{x\to3}\ln e^{3x}$
Substitution checker — watch the values close in
$x$value
value at the point—
from the left—
from the right—
1 A polynomial, so substitute: $4(-2)^2-1=16-1=\boxed{15}$.
2 Rational with $5(3)+4=19\neq0$: $\dfrac{2(9)-9+1}{19}=\dfrac{10}{19}$.
3 $\sin 0=0$ and the denominator $1+0=1\neq0$, so the value is $\dfrac{1}{1}=1$.
4 $\dfrac{2-7(1)}{1+6}=\dfrac{-5}{7}=-\dfrac57$.
5 The exponent is a polynomial: $2(2)-2^2=0$, so the limit is $e^0=1$.
6 $\ln e^{3x}=3x$ for every $x$, so the limit is $3(3)=9$.
Why substitution is legal Each function here is built from polynomials, $\sin$, $e^x$ and $\ln$ by sums, products and quotients with a non-zero denominator — exactly the situations Theorems 1.3 and 1.4 cover.

II. The indeterminate form $\dfrac00$

II
Use direct substitution to show that each limit leads to the indeterminate form $\dfrac00$. Then evaluate the limit.
  • 1 $\displaystyle\lim_{x\to4}\frac{x^2-16}{x-4}$
  • 2 $\displaystyle\lim_{x\to2}\frac{x-2}{x^2-2x}$
  • 3 $\displaystyle\lim_{x\to6}\frac{3x-18}{2x-12}$
  • 4 $\displaystyle\lim_{h\to0}\frac{(1+h)^2-1}{h}$
  • 5 $\displaystyle\lim_{x\to-3}\frac{\sqrt{x+4}-1}{x+3}$
Numerical check — values on both sides
$x$value
value at the pointundefined
from the left—
from the right—
limit—
1 At $x=4$: $\dfrac{16-16}{0}=\dfrac00$. Factor the difference of squares and cancel: $$\frac{(x-4)(x+4)}{x-4}=x+4\ (x\neq4)\ \Longrightarrow\ \lim_{x\to4}(x+4)=8.$$
2 At $x=2$: $\dfrac00$. Factor the denominator: $$\frac{x-2}{x(x-2)}=\frac1x\ (x\neq2)\ \Longrightarrow\ \lim_{x\to2}\frac1x=\frac12.$$
3 At $x=6$: $\dfrac00$. Take the common factor out of each line: $$\frac{3(x-6)}{2(x-6)}=\frac32\ (x\neq6)\ \Longrightarrow\ \text{the limit is }\frac32.$$
4 At $h=0$: $\dfrac00$. Expand the numerator: $$\frac{1+2h+h^2-1}{h}=\frac{h(2+h)}{h}=2+h\ (h\neq0)\ \Longrightarrow\ \lim_{h\to0}(2+h)=2.$$ This is exactly the difference quotient of $f(x)=x^2$ at $x=1$, so the answer $2$ is the slope of $y=x^2$ there.
5 At $x=-3$: $\dfrac{\sqrt1-1}{0}=\dfrac00$. Multiply by the conjugate $\sqrt{x+4}+1$: $$\frac{(x+4)-1}{(x+3)\left(\sqrt{x+4}+1\right)}=\frac{x+3}{(x+3)\left(\sqrt{x+4}+1\right)}=\frac{1}{\sqrt{x+4}+1},$$ so the limit is $\dfrac{1}{\sqrt{1}+1}=\dfrac12$.
II
Continued — same instruction: show the form is $\dfrac00$, then evaluate.
  • 6 $\displaystyle\lim_{x\to5}\frac{\sqrt{x-1}-2}{x-5}$
  • 7 $\displaystyle\lim_{t\to9}\frac{t-9}{\sqrt{t}-3}$
  • 8 $\displaystyle\lim_{x\to1}\frac{x^3-1}{x^2-1}$
  • 9 $\displaystyle\lim_{x\to\frac12}\frac{2x^2+3x-2}{2x-1}$
Numerical check — values on both sides
$x$value
value at the pointundefined
from the left—
from the right—
limit—
6 Form $\dfrac00$ at $5$. Multiply by the conjugate $\sqrt{x-1}+2$: $$\frac{(x-1)-4}{(x-5)\left(\sqrt{x-1}+2\right)}=\frac{x-5}{(x-5)\left(\sqrt{x-1}+2\right)}=\frac{1}{\sqrt{x-1}+2}\ \Longrightarrow\ \frac{1}{2+2}=\frac14.$$
7 Form $\dfrac00$ at $9$. Here the root is in the denominator, so multiply by $\sqrt t+3$: $$\frac{(t-9)\left(\sqrt t+3\right)}{(\sqrt t-3)(\sqrt t+3)}=\frac{(t-9)\left(\sqrt t+3\right)}{t-9}=\sqrt t+3\ \Longrightarrow\ \sqrt9+3=6.$$ Alternatively, factor $t-9=(\sqrt t-3)(\sqrt t+3)$ and cancel directly.
8 Form $\dfrac00$ at $1$. Use the difference-of-cubes and difference-of-squares factorisations: $$\frac{(x-1)(x^2+x+1)}{(x-1)(x+1)}=\frac{x^2+x+1}{x+1}\ \Longrightarrow\ \frac{3}{2}.$$
9 Form $\dfrac00$ at $\tfrac12$. Factor the numerator as $(2x-1)(x+2)$: $$\frac{(2x-1)(x+2)}{2x-1}=x+2\ \Longrightarrow\ \tfrac12+2=\tfrac52.$$
II
Continued — the last four.
  • 10 $\displaystyle\lim_{x\to-3}\frac{\dfrac{1}{x+2}+1}{x+3}$
  • 11 $\displaystyle\lim_{h\to0}\frac{\dfrac{1}{a+h}-\dfrac1a}{h}$, where $a$ is a real constant
  • 12 $\displaystyle\lim_{\theta\to\pi}\frac{\sin\theta}{\tan\theta}$
  • 13 $\displaystyle\lim_{x\to\frac{\pi}{2}}\frac{\cot x}{\cos x}$
Numerical check — values on both sides
$x$value
value at the pointundefined
from the left—
from the right—
limit—
10 At $x=-3$ the top is $\dfrac{1}{-1}+1=0$ and the bottom is $0$, so the form is $\dfrac00$. Combine the top over $x+2$: $$\frac{\dfrac{1+(x+2)}{x+2}}{x+3}=\frac{x+3}{(x+2)(x+3)}=\frac{1}{x+2}\ \Longrightarrow\ \frac{1}{-3+2}=-1.$$
11 Form $\dfrac00$ at $h=0$. Combine the top over $a(a+h)$: $$\frac{\dfrac{a-(a+h)}{a(a+h)}}{h}=\frac{-h}{h\,a(a+h)}=\frac{-1}{a(a+h)}\ \Longrightarrow\ \lim_{h\to0}\frac{-1}{a(a+h)}=-\frac{1}{a^2}.$$ This is the difference quotient of $f(x)=\dfrac1x$ at $x=a$, so the answer is the slope of $y=1/x$ there. With $a=2$ in the mini-lab it is $-\tfrac14$.
12 At $\theta=\pi$: $\sin\pi=0$ and $\tan\pi=0$, so the form is $\dfrac00$. Rewrite the tangent: $$\frac{\sin\theta}{\dfrac{\sin\theta}{\cos\theta}}=\cos\theta\ (\sin\theta\neq0)\ \Longrightarrow\ \lim_{\theta\to\pi}\cos\theta=-1.$$
13 At $x=\tfrac{\pi}{2}$: $\cot\tfrac\pi2=0$ and $\cos\tfrac\pi2=0$, so the form is $\dfrac00$. Rewrite the cotangent: $$\frac{\dfrac{\cos x}{\sin x}}{\cos x}=\frac{1}{\sin x}\ (\cos x\neq0)\ \Longrightarrow\ \lim_{x\to\frac\pi2}\frac{1}{\sin x}=1.$$

III. The form $\dfrac{K}{0}$ with $K\neq0$

III
Use direct substitution to show that each limit leads to the form $\dfrac{K}{0}$, $K\neq0$. Then evaluate the limit.
  • 1 $\displaystyle\lim_{x\to-5}\frac{x^2+25}{x+5}$
  • 2 $\displaystyle\lim_{x\to1^+}\frac{x+2}{(x-1)^2}$
  • 3 $\displaystyle\lim_{x\to-2}\frac{2x^2+7x-4}{x^2+x-2}$
  • 4 $\displaystyle\lim_{x\to1}\frac{2x^2+7x-4}{x^2+x-2}$
Watch the blow-up from each side
from the left—
from the right—
1 At $x=-5$ the top is $50$ and the bottom is $0$, form $\dfrac{50}{0}$. For $x\to-5^-$ the denominator is a small negative number, so the quotient $\to-\infty$; for $x\to-5^+$ it is small positive, so the quotient $\to+\infty$. The two sides disagree, so the two-sided limit does not exist (the line $x=-5$ is a vertical asymptote).
2 At $x=1$ the top is $3$ and the bottom is $0$, form $\dfrac{3}{0}$. Here the denominator $(x-1)^2$ is positive on both sides, so $$\lim_{x\to1^+}\frac{x+2}{(x-1)^2}=+\infty.$$ The two-sided limit is also $+\infty$ in this case, because the square never changes sign.
3 Factor both lines: $\dfrac{(2x-1)(x+4)}{(x+2)(x-1)}$. At $x=-2$ the top is $(-5)(2)=-10$ and the bottom is $0$, form $\dfrac{-10}{0}$. As $x\to-2^-$, $(x+2)\to0^-$ and $(x-1)\to-3$, so the denominator is $0^+$ and the quotient $\to-\infty$. As $x\to-2^+$ the denominator is $0^-$ and the quotient $\to+\infty$. The two-sided limit does not exist.
4 Same factored form. At $x=1$ the top is $(1)(5)=5$ and the bottom is $0$, form $\dfrac{5}{0}$. As $x\to1^-$, $(x-1)\to0^-$ and $(x+2)\to3$, so the quotient $\to-\infty$; as $x\to1^+$ it $\to+\infty$. The two-sided limit does not exist.
Pattern A factor that vanishes to an odd power flips sign across the point, so the two sides disagree. A factor vanishing to an even power, like $(x-1)^2$ in part 2, keeps its sign and the two sides agree.

IV. Limits from given limits

IV
Assume $\displaystyle\lim_{x\to6}f(x)=4$, $\displaystyle\lim_{x\to6}g(x)=9$ and $\displaystyle\lim_{x\to6}h(x)=6$. Use these three facts and the limit laws to evaluate each limit.
  • 1 $\displaystyle\lim_{x\to6}2f(x)g(x)$
  • 2 $\displaystyle\lim_{x\to6}\left(f(x)+\tfrac13 g(x)\right)$
  • 3 $\displaystyle\lim_{x\to6}\frac{g(x)-1}{f(x)}$
  • 4 $\displaystyle\lim_{x\to6}\frac{(h(x))^3}{2}$
  • 5 $\displaystyle\lim_{x\to6}\sqrt{g(x)-f(x)}$
  • 6 $\displaystyle\lim_{x\to6}\big(x\cdot h(x)\big)$
  • 7 $\displaystyle\lim_{x\to6}\big[(x+1)f(x)\big]$
  • 8 $\displaystyle\lim_{x\to6}\big(f(x)\cdot g(x)-h(x)\big)$
Which laws did you use?
1 Constant multiple then product: $2\cdot\left(\lim f\right)\left(\lim g\right)=2(4)(9)=72$.
2 Sum then constant multiple: $4+\tfrac13(9)=4+3=7$.
3 Difference on top, then quotient (legal since $\lim f=4\neq0$): $\dfrac{9-1}{4}=2$.
4 Power then constant multiple: $\dfrac{6^3}{2}=\dfrac{216}{2}=108$.
5 Difference then root (legal since $9-4=5\ge0$): $\sqrt{9-4}=\sqrt5$.
6 Product, using $\displaystyle\lim_{x\to6}x=6$ from Theorem 1.2: $6\cdot6=36$.
7 Sum then product: $(6+1)(4)=28$.
8 Product then difference: $(4)(9)-6=36-6=30$.
Watch the conditions Part 3 needs $\lim f\neq0$ for the quotient law, and part 5 needs the radicand's limit to be non-negative for the (even) root law. When a condition fails, the law simply does not apply and you must analyse the limit directly.

V. Piecewise functions

V
Evaluate the given limits.
  • 1 $f(x)=\left\{\begin{array}{cl} x^2 & \quad x\le3 \\[2pt] x+4 & \quad x>3 \end{array}\right.$   (a) $\displaystyle\lim_{x\to3^-}f(x)$   (b) $\displaystyle\lim_{x\to3^+}f(x)$   (c) $\displaystyle\lim_{x\to3}f(x)$
  • 2 $g(x)=\left\{\begin{array}{cl} x^3-1 & \quad x\le0 \\[2pt] 1 & \quad x>0 \end{array}\right.$   (a) $\displaystyle\lim_{x\to0^-}g(x)$   (b) $\displaystyle\lim_{x\to0^+}g(x)$   (c) $\displaystyle\lim_{x\to0}g(x)$
  • 3 $h(x)=\left\{\begin{array}{cl} x^2-2x+1 & \quad x<2 \\[2pt] 3-x & \quad x\ge2 \end{array}\right.$   (a) $\displaystyle\lim_{x\to2^-}h(x)$   (b) $\displaystyle\lim_{x\to2^+}h(x)$   (c) $\displaystyle\lim_{x\to2}h(x)$
One-sided approach on the graph
from the left—
from the right—
1 (a) For $x<3$ the rule is $x^2$, so $\displaystyle\lim_{x\to3^-}f(x)=9$. (b) For $x>3$ the rule is $x+4$, so $\displaystyle\lim_{x\to3^+}f(x)=7$. (c) The one-sided limits differ, so $\displaystyle\lim_{x\to3}f(x)$ does not exist.
2 (a) For $x<0$ the rule is $x^3-1$, so $\displaystyle\lim_{x\to0^-}g(x)=-1$. (b) For $x>0$ the rule is the constant $1$, so $\displaystyle\lim_{x\to0^+}g(x)=1$. (c) They differ, so $\displaystyle\lim_{x\to0}g(x)$ does not exist.
3 (a) For $x<2$, $x^2-2x+1=(x-1)^2\to1$. (b) For $x\ge2$, $3-x\to1$. (c) Both sides give $1$, so $\displaystyle\lim_{x\to2}h(x)=1$. Here the two branches meet, and since $h(2)=1$ as well, $h$ is continuous at $2$ — a preview of Section 1.3.
Method Never substitute into "the function"; substitute into the branch that governs the side you are approaching from. The value at the break point is irrelevant to the limit.

VI. More practice with $\dfrac00$: factoring, conjugates, complex fractions

VI
Show that each limit has the form $\dfrac00$, then choose a strategy (factor, conjugate, or combine fractions) and evaluate it.
  • 1 $\displaystyle\lim_{x\to-1}\frac{x^2+4x+3}{x^2-1}$
  • 2 $\displaystyle\lim_{x\to2}\frac{x^3-8}{x^2-4}$
  • 3 $\displaystyle\lim_{x\to0}\frac{\sqrt{9+x}-3}{x}$
  • 4 $\displaystyle\lim_{x\to1}\frac{\sqrt{x+3}-2}{\sqrt{x+8}-3}$
  • 5 $\displaystyle\lim_{x\to3}\frac{\dfrac1x-\dfrac13}{x-3}$
  • 6 $\displaystyle\lim_{h\to0}\frac{(2+h)^3-8}{h}$
Numerical check — values on both sides
$x$value
value at the pointundefined
from the left—
from the right—
limit—
1 At $x=-1$: $\dfrac{1-4+3}{1-1}=\dfrac00$. Factor both lines and cancel $x+1$: $$\frac{(x+1)(x+3)}{(x+1)(x-1)}=\frac{x+3}{x-1}\ (x\neq-1)\ \Longrightarrow\ \frac{-1+3}{-1-1}=-1.$$
2 At $x=2$: $\dfrac{8-8}{4-4}=\dfrac00$. Difference of cubes over difference of squares: $$\frac{(x-2)(x^2+2x+4)}{(x-2)(x+2)}=\frac{x^2+2x+4}{x+2}\ \Longrightarrow\ \frac{4+4+4}{4}=3.$$
3 At $x=0$: $\dfrac{3-3}{0}=\dfrac00$. Multiply by the conjugate $\sqrt{9+x}+3$: $$\frac{(9+x)-9}{x\left(\sqrt{9+x}+3\right)}=\frac{1}{\sqrt{9+x}+3}\ \Longrightarrow\ \frac{1}{3+3}=\frac16.$$
4 At $x=1$: $\dfrac{2-2}{3-3}=\dfrac00$. Roots appear on both lines, so multiply by both conjugates: $$\frac{\sqrt{x+3}-2}{\sqrt{x+8}-3}\cdot\frac{\sqrt{x+3}+2}{\sqrt{x+3}+2}\cdot\frac{\sqrt{x+8}+3}{\sqrt{x+8}+3}=\frac{(x-1)\left(\sqrt{x+8}+3\right)}{(x-1)\left(\sqrt{x+3}+2\right)}=\frac{\sqrt{x+8}+3}{\sqrt{x+3}+2},$$ so the limit is $\dfrac{3+3}{2+2}=\dfrac32$.
5 At $x=3$: $\dfrac{\frac13-\frac13}{0}=\dfrac00$. Combine the top over $3x$: $$\frac{\dfrac{3-x}{3x}}{x-3}=\frac{-(x-3)}{3x(x-3)}=-\frac{1}{3x}\ \Longrightarrow\ -\frac{1}{9}.$$
6 At $h=0$: $\dfrac{8-8}{0}=\dfrac00$. Expand $(2+h)^3=8+12h+6h^2+h^3$: $$\frac{12h+6h^2+h^3}{h}=12+6h+h^2\ (h\neq0)\ \Longrightarrow\ 12.$$ This is the slope of $y=x^3$ at $x=2$.

VII. One-sided limits, piecewise functions and an application

VII
  • 1 Let $f(x)=\left\{\begin{array}{cl} 2x+k & \quad x<1 \\[2pt] x^2+3 & \quad x\ge1. \end{array}\right.$ Find the value of the constant $k$ for which $\displaystyle\lim_{x\to1}f(x)$ exists, and state the limit. What happens when $k=0$?
  • 2 Evaluate $\displaystyle\lim_{x\to3^-}\frac{|x-3|}{x^2-9}$ and $\displaystyle\lim_{x\to3^+}\frac{|x-3|}{x^2-9}$. Does $\displaystyle\lim_{x\to3}\frac{|x-3|}{x^2-9}$ exist?
  • 3 Applied. A ball thrown upward has height $s(t)=20t-5t^2$ metres after $t$ seconds. Its instantaneous velocity at $t=1$ is $\displaystyle v(1)=\lim_{t\to1}\frac{s(t)-s(1)}{t-1}$. Evaluate this limit and interpret it.
One-sided approach on the graph
from the left—
from the right—
1 Left branch: $\displaystyle\lim_{x\to1^-}(2x+k)=2+k$. Right branch: $\displaystyle\lim_{x\to1^+}(x^2+3)=4$. The two-sided limit exists exactly when $2+k=4$, that is $k=2$, and then $\displaystyle\lim_{x\to1}f(x)=4$. With $k=0$ the left limit is $2\neq4$, so the limit does not exist (a jump of height $2$).
2 Factor $x^2-9=(x-3)(x+3)$. For $x<3$, $|x-3|=-(x-3)$, so $\dfrac{-(x-3)}{(x-3)(x+3)}=-\dfrac{1}{x+3}\to-\dfrac16$. For $x>3$, $|x-3|=x-3$, so the quotient is $\dfrac{1}{x+3}\to\dfrac16$. The one-sided limits differ ($-\tfrac16\neq\tfrac16$), so the two-sided limit does not exist.
3 $s(1)=20-5=15$. Substituting $t=1$ gives $\dfrac{15-15}{0}=\dfrac00$, so simplify: $$\frac{20t-5t^2-15}{t-1}=\frac{-5(t^2-4t+3)}{t-1}=\frac{-5(t-1)(t-3)}{t-1}=-5(t-3)\ (t\neq1)\ \Longrightarrow\ -5(1-3)=10.$$ So $v(1)=10$ m/s: one second after the throw the ball is still rising at $10$ metres per second.

VIII. The squeeze theorem and $\dfrac{\sin x}{x}$

VIII
Use the two tools below.
Squeeze theorem: if $g(x)\le f(x)\le h(x)$ for all $x\neq a$ near $a$ and $\displaystyle\lim_{x\to a}g(x)=\lim_{x\to a}h(x)=L$, then $\displaystyle\lim_{x\to a}f(x)=L$.   Key trig limit: $\displaystyle\lim_{x\to0}\frac{\sin x}{x}=1$ ($x$ in radians).
  • 1 $\displaystyle\lim_{x\to0}x^2\cos\frac1x$
  • 2 $\displaystyle\lim_{x\to0}\frac{\sin 5x}{x}$
  • 3 $\displaystyle\lim_{x\to0}\frac{\sin 3x}{\sin 7x}$
  • 4 $\displaystyle\lim_{\theta\to0}\frac{\tan\theta}{\theta}$
  • 5 $\displaystyle\lim_{x\to0}\frac{1-\cos x}{x}$
  • 6 Conceptual. Suppose $4x-9\le f(x)\le x^2-4x+7$ for all $x\ge0$. Find $\displaystyle\lim_{x\to4}f(x)$ and explain why the squeeze theorem applies.
Numerical check — values on both sides
$x$value
value at the pointundefined
from the left—
from the right—
limit—
1 $\cos\frac1x$ has no limit at $0$, so the product law cannot be used. Instead, $-1\le\cos\frac1x\le1$ gives, after multiplying by $x^2\ge0$, $$-x^2\le x^2\cos\frac1x\le x^2\quad(x\neq0).$$ Both $-x^2$ and $x^2$ tend to $0$, so by the squeeze theorem the limit is $0$.
2 Multiply and divide by $5$ and put $u=5x$ (so $u\to0$): $$\frac{\sin5x}{x}=5\cdot\frac{\sin5x}{5x}\ \Longrightarrow\ 5\cdot1=5.$$
3 Build two copies of the key limit: $$\frac{\sin3x}{\sin7x}=\frac{\dfrac{\sin3x}{3x}\cdot3x}{\dfrac{\sin7x}{7x}\cdot7x}=\frac37\cdot\frac{\dfrac{\sin3x}{3x}}{\dfrac{\sin7x}{7x}}\ \Longrightarrow\ \frac37\cdot\frac11=\frac37.$$
4 Write $\tan\theta=\dfrac{\sin\theta}{\cos\theta}$: $$\frac{\tan\theta}{\theta}=\frac{\sin\theta}{\theta}\cdot\frac{1}{\cos\theta}\ \Longrightarrow\ 1\cdot\frac{1}{\cos0}=1.$$
5 Multiply by $\dfrac{1+\cos x}{1+\cos x}$ and use $1-\cos^2x=\sin^2x$: $$\frac{1-\cos x}{x}=\frac{\sin^2x}{x(1+\cos x)}=\frac{\sin x}{x}\cdot\frac{\sin x}{1+\cos x}\ \Longrightarrow\ 1\cdot\frac{0}{2}=0.$$
6 The bounds are polynomials, so substitute: $\displaystyle\lim_{x\to4}(4x-9)=16-9=7$ and $\displaystyle\lim_{x\to4}(x^2-4x+7)=16-16+7=7$. The inequality holds for all $x$ near $4$ and both bounds have the same limit, so by the squeeze theorem $\displaystyle\lim_{x\to4}f(x)=7$. (Note $x^2-4x+7-(4x-9)=(x-4)^2\ge0$: the two bounds touch exactly at $x=4$.)
🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes

Research question. When does “just substitute” give a wrong or useless answer, and which technique fixes each case?

Why it matters. Computer algebra systems (such as those used in engineering design software) must detect $\frac00$ forms automatically and choose a method; this is an active area of symbolic computation research.

What to do.

  1. Collect 8 limits from this section and the exercises where substitution gives $\frac00$, e.g. $\lim_{x\to2}\frac{x^2-4}{x-2}$, $\lim_{x\to0}\frac{\sqrt{x+1}-1}{x}$.
  2. For each, name the technique: factor and cancel, rationalise, common denominator, or a special limit.
  3. Find one limit where substitution works but the function is not defined nearby on one side (e.g. $\sqrt{x}$ at $0$: only the one-sided limit $\lim_{x\to0^+}\sqrt x=0$ exists).
  4. Build a small table: limit, form, technique, answer. Check each answer in Desmos.

Tools: Pen and paper, Desmos

Hand in: A one-page classification table with a short sentence on which technique appears most often.

Going further. Anywhere: make a flowchart that a classmate could follow to choose a technique.

Instructor note: Students should notice that $\frac00$ says nothing about the answer; the value depends on the structure of the function (e.g. $\lim_{x\to0}\frac{\sqrt{x+1}-1}{x}=\frac12$).

🧪 Self-check quiz

0/5 correct

Q1. For which of these is direct substitution immediately valid?

Q2. Substituting gives $\dfrac00$. What does that tell you about the limit?

Q3. $\displaystyle\lim_{x\to9}\frac{x-9}{\sqrt{x}-3}$ equals:

Q4. Why does $\displaystyle\lim_{x\to1}\frac{x+2}{(x-1)^2}$ behave differently from $\displaystyle\lim_{x\to1}\frac{x+2}{x-1}$?

Q5. Given $\displaystyle\lim_{x\to6}f(x)=4$ and $\displaystyle\lim_{x\to6}g(x)=9$, the value of $\displaystyle\lim_{x\to6}\sqrt{g(x)-f(x)}$ is:

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