Differential Calculus
Limit Laws and Algebraic Techniques
Once you know what a limit is, the next job is to compute one. The limit laws let you break a complicated limit into simple pieces, direct substitution disposes of every function that is continuous where you are looking, and a small toolkit of algebraic moves rescues the cases where substitution returns the indeterminate form . This matters beyond limit questions: the derivative is defined as a limit, so every technique here is a technique you will use in the next topic.
The limit laws
Suppose is a constant and that both limits
exist and are finite. Then:
| Law | Statement | Hypothesis to check |
|---|---|---|
| Sum / Difference | both and exist | |
| Constant multiple | exists | |
| Product | both and exist | |
| Quotient | ||
| Power | a positive integer | |
| Root | odd: no extra condition. even: need , or with near , in which case the limit is | |
| Composition | and is continuous at |
Every law above is equally valid for one-sided limits (, ).
The laws run one way only. Knowing that exists tells you nothing about and . With and , the sum has limit at although neither piece has a limit there. So you must verify the pieces exist before you split.
Continuity and direct substitution
Definition. is continuous at when all three of these hold: (i) is defined; (ii) exists; (iii) .
Direct Substitution Property. If is continuous at , then — you may just plug in. This is always the first thing to try.
| Function | Continuous at |
|---|---|
| Polynomials | every real number |
| Rational | every with |
| , | every real number |
| , | every with |
| every ; only right-continuous at | |
| every | |
| and other exponentials | every real number |
Sums, differences, products, quotients (nonzero denominator) and compositions of continuous functions are continuous throughout their domains.
The indeterminate form 0/0
If and , the Quotient Law fails its hypothesis () and gives you nothing. The symbol is not a number and not an answer; it is a note to yourself that more algebra is required. Any outcome is still possible:
Why algebra is legal. A limit as never inspects . So for the factor is a genuine nonzero number and may be cancelled. Two functions that agree for all have the same limit at , even if one of them is undefined at . That single fact licenses every technique below.
Technique 1 — factor and cancel
Use it when top and bottom are polynomials. If both vanish at , the Factor Theorem guarantees divides both.
Technique 2 — rationalise
Use it when a square root causes the cancellation to hide. Multiply top and bottom by the conjugate, exploiting . For cube roots use instead. You may rationalise the numerator, the denominator, or both.
Technique 3 — clear a complex fraction
Combine the small fractions over a common denominator first, then cancel.
\frac{\frac{1}{3+h}-\frac{1}{3}}{h} &= \frac{1}{h}\cdot\frac{3-(3+h)}{3(3+h)}\\ &= \frac{1}{h}\cdot\frac{-h}{3(3+h)}\\ &= \frac{-1}{3(3+h)}\quad (h\ne 0) \end{aligned}$$ so the limit as $h\to 0$ is $-\frac{1}{9}$. ### Technique 4 — substitution (change of variable) **Statement.** If $\lim_{x\to a}g(x)=b$, if $g(x)\ne b$ for all $x\ne a$ in some open interval around $a$, and if $\lim_{u\to b}F(u)=L$, then $\lim_{x\to a}F\left(g(x)\right)=L$. The middle hypothesis is not decoration: without it the composite can land on the value $F(b)$ and give a different answer. Example: $\lim_{x\to 0}\dfrac{\sin 5x}{x}$. Put $u=5x$, so $x=\frac{u}{5}$ and $u\to 0$ with $u\ne 0$: $$\lim_{u\to 0}\frac{\sin u}{u/5}=5\lim_{u\to 0}\frac{\sin u}{u}=5.$$ ### The Squeeze Theorem **Theorem.** Suppose $g(x)\le f(x)\le h(x)$ for every $x$ in some open interval containing $a$, *except possibly at $x=a$ itself*, and suppose $$\lim_{x\to a}g(x)=\lim_{x\to a}h(x)=L.$$ Then $\lim_{x\to a}f(x)$ exists and equals $L$. Three things to notice. $f$ need not be continuous at $a$ and need not even be defined there. The two outer limits must be **equal** — if they differ, the theorem tells you nothing. The same statement holds for one-sided limits (use a one-sided interval) and for $x\to\pm\infty$. **Worked use.** Evaluate $\lim_{x\to 0}x^{2}\sin\frac{1}{x}$. Substitution is illegal (the function is undefined at $0$ and $\sin\frac1x$ has no limit there). But for every $x\ne 0$, $$-1\le \sin\frac{1}{x}\le 1.$$ Since $x^{2}\ge 0$, multiplying by $x^{2}$ preserves the inequalities: $$-x^{2}\le x^{2}\sin\frac{1}{x}\le x^{2}.$$ Both outer functions are polynomials with limit $0$ at $0$, so the Squeeze Theorem gives $\lim_{x\to 0}x^{2}\sin\frac{1}{x}=0$. ### The two standard trigonometric limits **Angles must be in radians** — the derivations use the sector-area formula $\frac12 r^{2}\theta$, which is a radian formula. $$\lim_{x\to 0}\frac{\sin x}{x}=1,\qquad \lim_{x\to 0}\frac{1-\cos x}{x}=0$$ **Deriving the first.** On the unit circle take $0 < x < \frac{\pi}{2}$, with $O=(0,0)$, $A=(1,0)$, $B=(\cos x,\sin x)$ and $T=(1,\tan x)$. Triangle $OAB$ sits inside sector $OAB$, which sits inside triangle $OAT$, so comparing areas: $$\frac{1}{2}\sin x<\frac{1}{2}x<\frac{1}{2}\tan x.$$ Multiply by $2$ and divide by $\sin x>0$: $$1<\frac{x}{\sin x}<\frac{1}{\cos x}\quad\implies\quad \cos x<\frac{\sin x}{x}<1.$$ As $x\to 0^{+}$ both $\cos x$ and $1$ tend to $1$, so the Squeeze Theorem gives $\lim_{x\to 0^{+}}\frac{\sin x}{x}=1$. The function $\frac{\sin x}{x}$ is even, so the left-hand limit is $1$ too, and the two-sided limit is $1$. **Deriving the second**, from the first: $$\begin{aligned} \frac{1-\cos x}{x} &= \frac{1-\cos x}{x}\cdot\frac{1+\cos x}{1+\cos x}\\ &= \frac{1-\cos^{2}x}{x(1+\cos x)}\\ &= \frac{\sin^{2}x}{x(1+\cos x)}\\ &= \frac{\sin x}{x}\cdot\frac{\sin x}{1+\cos x}\\ &\to 1\cdot\frac{0}{2}=0 \end{aligned}$$ The same algebra with one more $x$ downstairs gives the very useful $\frac{1-\cos x}{x^{2}}=\left(\frac{\sin x}{x}\right)^{2}\cdot\frac{1}{1+\cos x}\to\frac{1}{2}$. | Limit as $x\to 0$ | Value | |---|---| | $\dfrac{\sin x}{x}$ | $1$ | | $\dfrac{\tan x}{x}$ | $1$ | | $\dfrac{\sin kx}{x}$ | $k$ | | $\dfrac{\sin ax}{\sin bx}$ | $\dfrac{a}{b}$, for $b\ne 0$ | | $\dfrac{1-\cos x}{x}$ | $0$ | | $\dfrac{1-\cos x}{x^{2}}$ | $\dfrac{1}{2}$ | | $\dfrac{1-\cos kx}{x^{2}}$ | $\dfrac{k^{2}}{2}$ | ### Worked example A — rationalise, then factor Evaluate $\lim_{x\to 2}\dfrac{\sqrt{x+7}-3}{x^{2}-4}$. Substitution gives $\frac{3-3}{4-4}=\frac00$, so rationalise the numerator: $$\begin{aligned} \frac{\sqrt{x+7}-3}{x^{2}-4} &= \frac{\sqrt{x+7}-3}{x^{2}-4}\cdot\frac{\sqrt{x+7}+3}{\sqrt{x+7}+3}\\ &= \frac{(x+7)-9}{\left(x^{2}-4\right)\left(\sqrt{x+7}+3\right)}\\ &= \frac{x-2}{(x-2)(x+2)\left(\sqrt{x+7}+3\right)}\\ &= \frac{1}{(x+2)\left(\sqrt{x+7}+3\right)}\quad (x\ne 2) \end{aligned}$$ The last expression is continuous at $x=2$ with nonzero denominator, so substitute: $$\lim_{x\to 2}\frac{\sqrt{x+7}-3}{x^{2}-4}=\frac{1}{4\cdot 6}=\frac{1}{24}.$$ ### Worked example B — manufacturing the standard ratios Evaluate $\lim_{x\to 0}\dfrac{\tan 3x}{\sin 5x}$. Substitution gives $\frac00$. Rewrite $\tan 3x=\frac{\sin 3x}{\cos 3x}$ and insert the matching arguments: $$\begin{aligned} \frac{\tan 3x}{\sin 5x} &= \frac{\sin 3x}{\cos 3x\sin 5x}\\ &= \frac{1}{\cos 3x}\cdot\frac{\sin 3x}{3x}\cdot\frac{5x}{\sin 5x}\cdot\frac{3x}{5x}\\ &= \frac{3}{5}\cdot\frac{1}{\cos 3x}\cdot\frac{\sin 3x}{3x}\cdot\frac{5x}{\sin 5x} \end{aligned}$$ As $x\to 0$: $\cos 3x\to 1$, $\frac{\sin 3x}{3x}\to 1$ and $\frac{5x}{\sin 5x}\to 1$ (substitute $u=3x$ and $u=5x$). All three limits exist, so the Product Law applies and $$\lim_{x\to 0}\frac{\tan 3x}{\sin 5x}=\frac{3}{5}.$$ ### Common mistakes | Wrong | Right | |---|---| | $\frac{0}{0}=1$, or $\frac00=0$ | $\frac00$ is indeterminate — it is an instruction to do more algebra, never a value | | "$\lim_{x\to 1}\frac{x^{2}-1}{x-1}$ is undefined because you cannot divide by zero" | the function is undefined at $1$, the *limit* is fine: cancel to get $x+1\to 2$ | | $\frac{x^{2}-9}{x-3}=x+3$ | these agree only for $x\ne 3$ — different domains, but the same limit at $3$, which is all the limit needs | | $\lim\frac{f}{g}=\frac{\lim f}{\lim g}$ always | only when $\lim g\ne 0$ | | $\lim_{x\to 0}\left(\frac{1}{x}-\frac{1}{x}\right)=\lim\frac1x-\lim\frac1x$ | neither piece has a limit, so the Difference Law does not apply; simplify to $0$ first, then the limit is $0$ | | $\lim_{x\to 0}\frac{\sin 3x}{x}=1$ | $=3$: the standard limit needs the argument and the denominator to be identical | | $\lim_{x\to \pi}\frac{\sin x}{x}=1$ | the standard limit lives only at $x\to 0$; here substitution works and gives $\frac{0}{\pi}=0$ | | Using degrees in $\frac{\sin x}{x}$ | in degrees the limit is $\frac{\pi}{180}$, not $1$ — always radians | | Squeezing $x\sin\frac1x$ with $-x\le x\sin\frac1x\le x$ | multiplying an inequality by $x$ reverses it when $x<0$; use a nonnegative multiplier such as $x^{2}$, or squeeze with $-\sqrt{x^{2}}\le x\sin\frac1x\le\sqrt{x^{2}}$ | | "$\lim_{x\to 0}\frac{1}{x^{2}}=\infty$, so the limit exists" | it does not exist; $\infty$ only records *how* it fails | | Squeezing when $\lim g=2$ and $\lim h=5$ | the outer limits must be equal, otherwise the theorem gives no conclusion | | $\lim\left(f\right)^{1/2}=L^{1/2}$ with $L<0$ | if $L<0$ then $\sqrt{f(x)}$ is not even defined for $x$ near $a$; the even-root law needs $L>0$, or $L=0$ together with $f(x)\ge 0$ near $a$, and then the limit is $0$ |Key terms
- Limit laws
- Direct substitution
- Continuity at a point
- Indeterminate form 0/0
- Factor and cancel
- Conjugate / rationalising
- Complex fraction
- Change of variable (substitution)
- Squeeze Theorem
- One-sided limit
- Standard trigonometric limits
- Radian measure
Practice Problems
Suppose and . Evaluate
and state which hypothesis you must check before applying the Quotient Law.
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Work out the limit of the numerator and the limit of the denominator separately, then look hard at the denominator's value.
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Both given limits exist and are finite, so the Constant Multiple, Difference and Sum Laws apply to the top and bottom separately.
Numerator.
Denominator.
Check the hypothesis. The Quotient Law requires the limit of the denominator to be nonzero. Here it is , so the law applies:
Answer: . The hypothesis that must be checked is . Had it been , the Quotient Law would say nothing and the limit could be finite, infinite, or nonexistent.
Evaluate .
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Try substitution first and see what form you get; then factor both quadratics.
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Step 1 — try substitution. At the numerator is and the denominator is . The form is , so the Quotient Law does not apply and more work is needed.
Step 2 — factor. Both quadratics vanish at , so is a factor of each:
Step 3 — cancel. For the factor is a nonzero number, so it may be cancelled:
Step 4 — substitute into the simplified form. is a rational function whose denominator at is , so it is continuous there and direct substitution is valid:
Answer: .
Evaluate .
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Substitution gives 0/0 and there is nothing to factor, so get rid of the root by multiplying by something clever that equals 1.
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Step 1 — try substitution. Numerator , denominator . Indeterminate .
Step 2 — rationalise the numerator. Multiply by the conjugate over itself (this is legal since for near ):
\frac{\sqrt{9+x}-3}{x} &= \frac{\sqrt{9+x}-3}{x}\cdot\frac{\sqrt{9+x}+3}{\sqrt{9+x}+3}\\ &= \frac{\left(\sqrt{9+x}\right)^{2}-3^{2}}{x\left(\sqrt{9+x}+3\right)}\\ &= \frac{(9+x)-9}{x\left(\sqrt{9+x}+3\right)}\\ &= \frac{x}{x\left(\sqrt{9+x}+3\right)} \end{aligned}$$ **Step 3 — cancel.** For $x\ne 0$: $$\frac{x}{x\left(\sqrt{9+x}+3\right)}=\frac{1}{\sqrt{9+x}+3}$$ **Step 4 — substitute.** The function $\frac{1}{\sqrt{9+x}+3}$ is continuous at $x=0$ (the root is defined and the denominator is $6\ne 0$): $$\lim_{x\to 0}\frac{1}{\sqrt{9+x}+3}=\frac{1}{\sqrt{9}+3}=\frac{1}{6}$$ **Answer: $\dfrac{1}{6}$.** (Sanity check: at $x=0.001$ the original quotient is $0.166662$..., close to $\frac16\approx 0.1667$.)Evaluate .
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You know , so engineer a under the top sine and a under the bottom one without changing the expression.
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Step 1 — check the form. As both and , so this is ; the Quotient Law does not apply directly.
Step 2 — insert the matching arguments. For small enough that , multiply and divide to build the standard ratios:
\frac{\sin 7x}{\sin 3x} &= \frac{\sin 7x}{7x}\cdot\frac{7x}{3x}\cdot\frac{3x}{\sin 3x}\\ &= \frac{7}{3}\cdot\frac{\sin 7x}{7x}\cdot\frac{3x}{\sin 3x} \end{aligned}$$ **Step 3 — apply the standard limit via substitution.** Put $u=7x$: as $x\to 0$ with $x\ne 0$ we get $u\to 0$ with $u\ne 0$, so $\lim_{x\to 0}\frac{\sin 7x}{7x}=\lim_{u\to 0}\frac{\sin u}{u}=1$. Likewise with $v=3x$, $\lim_{x\to 0}\frac{\sin 3x}{3x}=1$, and since that limit is nonzero the Quotient Law gives $\lim_{x\to 0}\frac{3x}{\sin 3x}=\frac{1}{1}=1$. **Step 4 — combine.** All three factors have limits, so the Product Law applies: $$\lim_{x\to 0}\frac{\sin 7x}{\sin 3x}=\frac{7}{3}\cdot 1\cdot 1=\frac{7}{3}$$ **Answer: $\dfrac{7}{3}$.** (Numerical check: at $x=0.001$ the quotient is $2.33332$, heading to $\frac{7}{3}=2.3333$...)Evaluate .
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This is an situation, so combine the two terms over a single denominator before you do anything else.
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Step 1 — recognise the form. Each term blows up as , so the Difference Law does not apply (neither piece has a limit). Combine first.
Step 2 — common denominator. For and :
This is now .
Step 3 — rationalise the numerator with the conjugate :
\frac{1-\sqrt{1+t}}{t\sqrt{1+t}} &= \frac{1-\sqrt{1+t}}{t\sqrt{1+t}}\cdot\frac{1+\sqrt{1+t}}{1+\sqrt{1+t}}\\ &= \frac{1-(1+t)}{t\sqrt{1+t}\left(1+\sqrt{1+t}\right)}\\ &= \frac{-t}{t\sqrt{1+t}\left(1+\sqrt{1+t}\right)} \end{aligned}$$ **Step 4 — cancel** the factor $t\ne 0$: $$\frac{-t}{t\sqrt{1+t}\left(1+\sqrt{1+t}\right)}=\frac{-1}{\sqrt{1+t}\left(1+\sqrt{1+t}\right)}$$ **Step 5 — substitute.** This is continuous at $t=0$ and its denominator there is $1\cdot(1+1)=2\ne 0$: $$\lim_{t\to 0}\frac{-1}{\sqrt{1+t}\left(1+\sqrt{1+t}\right)}=\frac{-1}{1\cdot 2}=-\frac{1}{2}$$ **Answer: $-\dfrac{1}{2}$.** (Check at $t=0.001$: $999.5004-1000=-0.4996$, consistent.)It is known that
for all . Find , justifying every hypothesis of the theorem you use.
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Compare the two bounding functions at the point in question before you conclude anything.
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Step 1 — name the bounds. Let and , so .
Step 2 — check the interval hypothesis. The Squeeze Theorem needs the inequality to hold on an open interval containing , except possibly at . The inequality is given for all , and the open interval is contained in and contains . Hypothesis satisfied.
Step 3 — compute the outer limits. Both and are polynomials, hence continuous everywhere, so direct substitution applies:
Step 4 — the outer limits agree. Both equal , which is the remaining hypothesis.
Step 5 — conclude. By the Squeeze Theorem, exists and
Answer: . Note that we never needed a formula for , and need not be continuous at — the squeeze forces the limit regardless. Note also that the argument works only at : at , for instance, and differ, so the theorem would give no conclusion there.
Evaluate .
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The conjugate trick that produced works here too.
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Step 1 — check the form. and , so this is .
Step 2 — multiply by the conjugate . For near with we have and , so:
\frac{1-\cos x}{x\sin x} &= \frac{1-\cos x}{x\sin x}\cdot\frac{1+\cos x}{1+\cos x}\\ &= \frac{1-\cos^{2}x}{x\sin x\left(1+\cos x\right)}\\ &= \frac{\sin^{2}x}{x\sin x\left(1+\cos x\right)}\\ &= \frac{\sin x}{x}\cdot\frac{1}{1+\cos x} \end{aligned}$$ using the Pythagorean identity $1-\cos^{2}x=\sin^{2}x$ and cancelling one factor of $\sin x\ne 0$. **Step 3 — take limits factor by factor.** $\lim_{x\to 0}\frac{\sin x}{x}=1$ (standard limit), and $\frac{1}{1+\cos x}$ is continuous at $0$ with value $\frac{1}{1+1}=\frac12$. Both limits exist, so the Product Law applies: $$\lim_{x\to 0}\frac{1-\cos x}{x\sin x}=1\cdot\frac{1}{2}=\frac{1}{2}$$ **Answer: $\dfrac{1}{2}$.** *Alternative route:* write $\frac{1-\cos x}{x\sin x}=\frac{1-\cos x}{x^{2}}\cdot\frac{x}{\sin x}\to\frac12\cdot 1=\frac12$, using the standard result $\lim_{x\to 0}\frac{1-\cos x}{x^{2}}=\frac12$. Same answer, which is a good independent check.Evaluate .
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Fractional exponents with denominators 3 and 2 both become whole numbers if you rename as a suitable power of a new variable.
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Step 1 — check the form. At both and equal : indeterminate .
Step 2 — change variable. Let , so with , and then , .
Hypotheses for the substitution: is continuous on with as , and whenever (the sixth-root function is strictly increasing on , hence one-to-one). So the change of variable is valid and
Step 3 — factor. Using and :
Step 4 — substitute. The simplified expression is a rational function whose denominator at is , so it is continuous there:
Answer: . (Check at : the original quotient is ...)
Evaluate without using l'Hopital's Rule.
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The conjugate trick generalises: there is a three-term factor that turns a difference of cube roots into a difference of the radicands.
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Step 1 — check the form. As , , so the numerator and the denominator : indeterminate .
Step 2 — set up the cube-root conjugate. Write , so , and . The identity is
so multiply top and bottom by :
\frac{(1+x)^{1/3}-1}{x} &= \frac{a-1}{x}\cdot\frac{a^{2}+a+1}{a^{2}+a+1}\\ &= \frac{a^{3}-1}{x\left(a^{2}+a+1\right)}\\ &= \frac{(1+x)-1}{x\left(a^{2}+a+1\right)}\\ &= \frac{x}{x\left(a^{2}+a+1\right)} \end{aligned}$$ This multiplication is legitimate because $a^{2}+a+1\ne 0$ for real $a$ (its discriminant is $1-4=-3<0$), and in fact $a^{2}+a+1\to 3$ near $x=0$. **Step 3 — cancel** the factor $x\ne 0$: $$\frac{x}{x\left(a^{2}+a+1\right)}=\frac{1}{a^{2}+a+1}=\frac{1}{(1+x)^{2/3}+(1+x)^{1/3}+1}$$ **Step 4 — substitute.** The cube root is continuous everywhere, so the denominator is continuous at $x=0$ with value $1+1+1=3\ne 0$: $$\lim_{x\to 0}\frac{(1+x)^{1/3}-1}{x}=\frac{1}{3}$$ **Answer: $\dfrac{1}{3}$.** (Check at $x=0.001$: $\frac{1.000333222-1}{0.001}=0.33322$..., heading to $\frac13$. This limit is also the derivative of $(1+x)^{1/3}$ at $x=0$, and $\frac{d}{dx}(1+x)^{1/3}=\frac13(1+x)^{-2/3}$ equals $\frac13$ there — an independent confirmation.)Evaluate .
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Convert the tangent to sine over cosine, then arrange every sine so that its argument also appears underneath it.
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Step 1 — check the form. and : indeterminate .
Step 2 — remove the tangent and the cosine. For small (so , , ):
\frac{1-\cos 4x}{x\tan 2x} &= \frac{1-\cos 4x}{x}\cdot\frac{\cos 2x}{\sin 2x}\\ &= \frac{1-\cos 4x}{x}\cdot\frac{\cos 2x}{\sin 2x}\cdot\frac{1+\cos 4x}{1+\cos 4x}\\ &= \frac{1-\cos^{2}4x}{x\sin 2x}\cdot\frac{\cos 2x}{1+\cos 4x}\\ &= \frac{\sin^{2}4x}{x\sin 2x}\cdot\frac{\cos 2x}{1+\cos 4x} \end{aligned}$$ **Step 3 — force the standard ratios.** Replace $\sin 4x=\frac{\sin 4x}{4x}\cdot 4x$ and $\sin 2x=\frac{\sin 2x}{2x}\cdot 2x$: $$\frac{\sin^{2}4x}{x\sin 2x}=\frac{\left(\frac{\sin 4x}{4x}\right)^{2}\left(4x\right)^{2}}{x\cdot\frac{\sin 2x}{2x}\cdot 2x}=\frac{16x^{2}}{2x^{2}}\cdot\frac{\left(\frac{\sin 4x}{4x}\right)^{2}}{\frac{\sin 2x}{2x}}=8\cdot\frac{\left(\frac{\sin 4x}{4x}\right)^{2}}{\frac{\sin 2x}{2x}}$$ so altogether $$\frac{1-\cos 4x}{x\tan 2x}=8\cdot\frac{\left(\frac{\sin 4x}{4x}\right)^{2}}{\frac{\sin 2x}{2x}}\cdot\frac{\cos 2x}{1+\cos 4x}$$ **Step 4 — take limits.** By substitution $u=4x$ and $u=2x$, both $\frac{\sin 4x}{4x}\to 1$ and $\frac{\sin 2x}{2x}\to 1$ (nonzero, so the Quotient Law is allowed); $\cos 2x\to 1$; $1+\cos 4x\to 2\ne 0$. Hence $$\lim_{x\to 0}\frac{1-\cos 4x}{x\tan 2x}=8\cdot\frac{1^{2}}{1}\cdot\frac{1}{2}=4$$ **Answer: $4$.** *Independent check with the standard results.* $\frac{1-\cos 4x}{x^{2}}\to\frac{4^{2}}{2}=8$ and $\frac{x\tan 2x}{x^{2}}=\frac{\tan 2x}{x}\to 2$, so the ratio tends to $\frac{8}{2}=4$. Numerically at $x=0.01$ the quotient is $3.99893$, and at $x=0.001$ it is $3.99999$.Use the Squeeze Theorem to evaluate
stating the inequality you use and why it is valid.
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Bound the exponent first, then use the fact that the exponential function is increasing.
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Step 1 — why substitution fails. The function is undefined at , and oscillates without limit as , so no limit law reaches this directly. But the oscillating factor is bounded, which is exactly the squeeze situation.
Step 2 — bound the exponent. For every ,
Step 3 — apply the exponential. The exponential function is increasing on all of the reals, so applying it preserves the inequalities:
Step 4 — multiply by . Since , multiplying an inequality by preserves its direction (this is the step that would fail if we multiplied by , which is negative on one side of ):
valid for every in, say, the open interval except — precisely the hypothesis the Squeeze Theorem asks for.
Step 5 — the outer limits. By the Constant Multiple Law and continuity of :
They are equal, so .
Step 6 — conclude. By the Squeeze Theorem the middle limit exists and
Answer: . (Numerically at the value is about , and at about .)
Find constants and such that
Justify why is forced to take the value you claim.
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If the bottom tends to 0 and the whole quotient tends to a finite number, what is forced on the top? Write the numerator as (quotient) times (denominator).
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Step 1 — force the numerator to vanish. Write and , and suppose the stated limit holds, so . Since and both of these limits exist, the Product Law gives
So the numerator must tend to . (If instead , the quotient would be a nonzero number divided by something tending to , which is unbounded — it could never converge to .)
Step 2 — solve for . We need for the root to be defined, and the square-root function is continuous there, so . Setting :
Step 3 — now find by rationalising. With , note for all close enough to , so the expression is defined on a punctured interval around . For :
\frac{\sqrt{9+bx}-3}{x} &= \frac{\sqrt{9+bx}-3}{x}\cdot\frac{\sqrt{9+bx}+3}{\sqrt{9+bx}+3}\\ &= \frac{(9+bx)-9}{x\left(\sqrt{9+bx}+3\right)}\\ &= \frac{bx}{x\left(\sqrt{9+bx}+3\right)}\\ &= \frac{b}{\sqrt{9+bx}+3} \end{aligned}$$ **Step 4 — take the limit and match.** The last expression is continuous at $x=0$ with denominator $3+3=6\ne 0$, so $$\lim_{x\to 0}\frac{\sqrt{9+bx}-3}{x}=\frac{b}{6}$$ Setting $\frac{b}{6}=2$ gives $b=12$. **Answer: $a=9$ and $b=12$.** **Verification.** With these values, $\frac{\sqrt{9+12x}-3}{x}=\frac{12}{\sqrt{9+12x}+3}\to\frac{12}{6}=2$. Numerically at $x=0.001$: $\frac{\sqrt{9.012}-3}{0.001}=1.99933$..., as required.