5 Problems
The error term, limits, and $0/0$
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Before every limit, substitute. Most of the work in this lecture is deciding which step of Method 5.24 you are in.
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The price of an approximation
Read the theory: 1. The price of an approximation →Let $f(x)=\ln(1+x)$ and $a=0.$
(a) Find the Taylor polynomial $p_{2}$ and use it to approximate $\ln1.1.$
(b) Use Fact 5.4 to bound the error. Which number $M$ do you use, and why does it work on the whole interval?
(c) Show that the approximation in (a) is too small, so that $\ln1.1$ lies between $0.095$ and $0.095$ plus your bound.
Hint
You need $f',$ $f''$ and $f'''.$ In (b), the remainder of $p_{2}$ contains $f'''(z)$ with $z$ between $0$ and $0.1{:}$ where on that interval is $|f'''|$ largest? In (c), look at the sign of the remainder in (5.1) rather than its size.
(a) The derivatives at $0.$ $$ \begin{aligned} f'(x)&=\frac{1}{1+x}\,,\\ f''(x)&=-\frac{1}{(1+x)^{2}}\,,\\ f'''(x)&=\frac{2}{(1+x)^{3}}\,. \end{aligned} $$ So $f(0)=0,$ $f'(0)=1$ and $f''(0)=-1,$ and $$ p_{2}(x)=x-\frac{x^{2}}{2}\,. $$ At $x=0.1{:}$ $p_{2}(0.1)=0.1-0.005=0.095.$
(b) The bound $M.$ For $t$ between $0$ and $0.1$ the denominator $(1+t)^{3}$ is at least $1,$ so $|f'''(t)|=\dfrac{2}{(1+t)^{3}}\le2.$ Take $M=2$ (the value at the left-hand end, where $|f'''|$ is largest).
(b) The bound. By (5.2) with $n=2,$ $$ \begin{aligned} |R_{3}(0.1)|&\le\frac{2}{3!}\,(0.1)^{3}\\ &=\frac{1}{3}\cdot\frac{1}{1000}=\frac{1}{3000}\,, \end{aligned} $$ which is about $0.00033.$
(c) The sign of the remainder. By (5.1), $R_{3}(0.1)=\dfrac{f'''(z)}{3!}\,(0.1)^{3}$ for some $z$ between $0$ and $0.1,$ and $f'''(z)=2/(1+z)^{3}>0.$ So $R_{3}(0.1)>0{:}$ $\ln1.1=p_{2}(0.1)+R_{3}(0.1)$ is larger than $0.095.$ Together with (b), $$ 0.095<\ln1.1<0.095+\tfrac{1}{3000}\,. $$ (A calculator gives $\ln1.1=0.09531\ldots,$ inside the interval.)
Answer
(a) $p_{2}(x)=x-\tfrac12x^{2},$ and $\ln1.1\approx0.095.$
(b) $M=2,$ and the error is at most $\tfrac{1}{3000}\approx0.00033.$
(c) The remainder is positive, so $0.095<\ln1.1<0.095+\tfrac1{3000}.$
How did it go?
Stuck? Read 1. The price of an approximation →
The number $e$ is $f(1)$ for $f(x)=e^{x}.$ We want to compute it from the Taylor polynomial $p_{n}$ about $a=0$ with an error guaranteed to be below $10^{-3}.$
(a) Explain why $M=3$ is a valid bound for $|f^{(n+1)}(t)|$ for every $t$ between $0$ and $1,$ whatever $n$ is.
(b) Find the smallest $n$ for which (5.2) with $M=3$ guarantees an error below $10^{-3}.$
(c) Compute that $p_{n}(1)$ as a fraction and as a decimal.
Hint
Every derivative of $e^{x}$ is $e^{x}.$ In (b), with $x-a=1$ the power in (5.2) is $1,$ so only the factorial is left to do the work.
(a) Every derivative is $e^{t}.$ So $|f^{(n+1)}(t)|=e^{t},$ which is increasing, and for $0\le t\le1$ it is at most $e^{1}=e<3.$ The bound does not depend on $n.$
(b) The bound as a function of $n.$ With $a=0,$ $x=1$ and $M=3,$ $$ \begin{aligned} |R_{n+1}(1)|&\le\frac{3}{(n+1)!}\cdot1^{n+1}\\ &=\frac{3}{(n+1)!}\,. \end{aligned} $$ This is below $10^{-3}$ exactly when $(n+1)!>3000.$
(b) The smallest $n.$ $6!=720$ is too small and $7!=5040>3000.$ So $n+1=7,$ that is $n=6.$ (With $n=5$ the bound is $\tfrac{3}{720}\approx0.0042,$ which is not enough.)
(c) The polynomial at $1.$ $p_{6}(1)=\sum_{k=0}^{6}\dfrac{1}{k!}{:}$ $$ \begin{aligned} &1+1+\tfrac12+\tfrac16+\tfrac1{24}+\tfrac1{120}+\tfrac1{720}\\ &\quad=\frac{1957}{720}\approx2.71806\,. \end{aligned} $$ The true value $e=2.71828\ldots$ differs by about $0.0002,$ below the guaranteed $\tfrac{3}{5040}\approx0.0006.$
Answer
(a) $|f^{(n+1)}(t)|=e^{t}\le e<3$ on $[0,1].$
(b) $n=6,$ since $7!=5040>3000$ but $6!=720<3000.$
(c) $p_{6}(1)=\tfrac{1957}{720}\approx2.71806.$
How did it go?
Stuck? Read 1. The price of an approximation →
Limits, properly
Read the theory: 2. Limits, properly →Decide whether each limit exists. Give the reason, using the one-sided limits.
(a) $\displaystyle\lim_{x\to2}\frac{|x-2|}{x-2}$
(b) $\displaystyle\lim_{x\to1}\frac{1}{(x-1)^{2}}$
(c) $\displaystyle\lim_{x\to1}f(x),$ where $f(x)=x^{2}$ for $x<1,$ $f(x)=2-x$ for $x>1,$ and $f(1)=5.$
Hint
In (a), write $|x-2|$ without the bars, separately for $x>2$ and for $x<2.$ In (c), remember what a limit never looks at.
(a) The two sides. For $x>2,$ $|x-2|=x-2$ and the quotient is $1.$ For $x<2,$ $|x-2|=-(x-2)$ and the quotient is $-1.$ So the right-hand limit is $1$ and the left-hand limit is $-1.$ They differ, and by Fact 5.8 there is no limit.
(b) Both sides run away. For $x$ near $1,$ on either side, $(x-1)^{2}$ is a small positive number, so $\dfrac{1}{(x-1)^{2}}\to+\infty$ from both sides. Neither one-sided limit is a number, so there is no limit. One writes $\to\infty$ to say how it fails (Watch out 5.10).
(c) The value plays no part. From the left $f(x)=x^{2}\to1;$ from the right $f(x)=2-x\to1.$ Both one-sided limits are $1,$ so the limit exists and is $1$ (Fact 5.8). The value $f(1)=5$ changes nothing: a limit never looks at $f(a).$ (It does decide continuity: $f$ is not continuous at $1.$)
Answer
(a) No limit: $-1$ from the left, $1$ from the right.
(b) No limit: $+\infty$ from both sides, and $\infty$ is not a number.
(c) The limit is $1,$ although $f(1)=5.$
How did it go?
Stuck? Read 2. Limits, properly →
Continuity, and the licence to substitute
Read the theory: 3. Continuity, and the licence to substitute →Let $$ f(x)=\frac{x^{2}-1}{x^{2}-3x+2}\,. $$
(a) Where is $f$ defined, and where is it continuous?
(b) At each point where $f$ is not continuous, decide whether the discontinuity is removable or irremovable. Where it is removable, say which value repairs it.
Hint
Factor the denominator first. At each zero of the denominator, substitute into the numerator too: $\tfrac00$ and $c/0$ lead to different steps of Method 5.24.
(a) Domain and continuity. $x^{2}-3x+2=(x-1)(x-2),$ so $f$ is defined for all $x$ except $1$ and $2.$ A rational function is continuous wherever it is defined (Fact 5.17), so $f$ is continuous at every $x\ne1,2.$ At $1$ and $2$ it is not, since $f(1)$ and $f(2)$ do not exist.
(b) At $x=1{:}$ the form $\tfrac00.$ The numerator is $(x-1)(x+1),$ so for $x\ne1,2$ $$ \begin{aligned} f(x)&=\frac{(x-1)(x+1)}{(x-1)(x-2)}\\ &=\frac{x+1}{x-2}\,, \end{aligned} $$ and $\lim_{x\to1}f(x)=\dfrac{2}{-1}=-2.$ The limit exists, so the discontinuity is removable: setting $f(1)=-2$ repairs it.
(b) At $x=2{:}$ the form $c/0.$ The numerator is $3$ there and the denominator $0.$ For $x$ just above $2$ the reduced form $\dfrac{x+1}{x-2}$ is positive and tends to $+\infty;$ just below $2$ it tends to $-\infty.$ The limit does not exist, so the discontinuity is irremovable: no value of $f(2)$ can repair it.
Answer
(a) Defined and continuous for all $x$ except $1$ and $2.$
(b) At $1$ removable (limit $-2;$ set $f(1)=-2$). At $2$ irremovable ($-\infty$ from the left, $+\infty$ from the right).
How did it go?
Stuck? Read 3. Continuity, and the licence to substitute →
Reading the signal: l'Hôpital's rule
Read the theory: 4. Reading the signal: l'Hôpital's rule →Each attempt below goes wrong. Say in words what went wrong, and find the right limit.
(a) $\displaystyle\lim_{x\to0}\frac{\ln(1+3x)}{x}{:}$ “by l'Hôpital, the derivative of the quotient is $\dfrac{\frac{3x}{1+3x}-\ln(1+3x)}{x^{2}},$ which is $\tfrac00$ at $0$ again …”
(b) $\displaystyle\lim_{x\to1}\frac{x^{2}+1}{x+1}{:}$ “by l'Hôpital, $\dfrac{2x}{1}\to2.$”
(c) $\displaystyle\lim_{x\to0}\frac{\ln(1+x)-x}{x^{2}}{:}$ “by l'Hôpital, $\dfrac{\frac{1}{1+x}-1}{2x},$ and the numerator is $0$ at $x=0,$ so the limit is $0.$”
Hint
For each one, ask the questions of Method 5.24 in order: what does substituting give, and is the rule allowed? Then: was it a fraction of derivatives?
(a) The quotient rule, not l'Hôpital. The form is $\tfrac00,$ so the rule applies — but it differentiates numerator and denominator separately (Watch out 5.21): $$ \lim_{x\to0}\frac{\ln(1+3x)}{x}=\lim_{x\to0}\frac{\frac{3}{1+3x}}{1}=3\,. $$
(b) The rule never applied. At $x=1$ the quotient is $\dfrac{2}{2},$ not indeterminate (Watch out 5.22). Substitute: the limit is $1.$ The “answer” $2$ is the silent failure of Example 5.23.
(c) Stopped at a second $\tfrac00.$ After one application the numerator and the denominator $2x$ are $0$ at $x=0.$ That is $\tfrac00$ again, a signal, not the value $0.$ Apply the rule once more: $$ \begin{aligned} \lim_{x\to0}\frac{\frac{1}{1+x}-1}{2x} &=\lim_{x\to0}\frac{-\frac{1}{(1+x)^{2}}}{2}\\ &=-\frac12\,. \end{aligned} $$
Answer
(a) $3.$ (b) $1.$ (c) $-\tfrac12.$
How did it go?
Stuck? Read 4. Reading the signal: l'Hôpital's rule →
For a constant $a,$ consider $$ \lim_{x\to0}\frac{e^{ax}-1-3x}{x^{2}}\,. $$ Show that the limit exists for exactly one value of $a,$ and find the limit for that value.
Hint
The form is $\tfrac00$ for every $a.$ Write the quotient as $\dfrac{1}{x}\Bigl(\dfrac{e^{ax}-1}{x}-3\Bigr)$ and find the limit of the bracket first. What happens if it is not $0?$
The form. At $x=0$ the numerator is $1-1-0=0$ and so is the denominator, for every $a.$
A rewriting. For $x\ne0,$ $$ \begin{aligned} &\frac{e^{ax}-1-3x}{x^{2}}\\ &\quad=\frac{1}{x}\Bigl(\frac{e^{ax}-1}{x}-3\Bigr)\,. \end{aligned} $$ The quotient $\dfrac{e^{ax}-1}{x}$ is $\tfrac00$ at $0,$ and by l'Hôpital's rule it tends to $\dfrac{ae^{0}}{1}=a.$ So the bracket tends to $a-3.$
If $a\ne3.$ The bracket tends to the nonzero number $a-3$ while $x\to0.$ That is the situation $c/0$ of step 2 of Method 5.24: the quotient grows past every bound in size, with opposite signs on the two sides of $0.$ There is no limit.
If $a=3.$ Now the form is $\tfrac00$ and the rule may be applied twice: $$ \begin{aligned} &\lim_{x\to0}\frac{e^{3x}-1-3x}{x^{2}}\\ &\quad=\lim_{x\to0}\frac{3e^{3x}-3}{2x}\\ &\quad=\lim_{x\to0}\frac{9e^{3x}}{2}=\frac92\,. \end{aligned} $$ After the first application the form is $\tfrac00$ again, which is why the second is allowed.
A remark on the rule. For $a\ne3$ one application gives $\dfrac{ae^{ax}-3}{2x},$ which is $\dfrac{a-3}{0}$ at $0.$ That alone does not prove that the original limit fails: Fact 5.19 says the two limits agree when the right-hand one exists. The rewriting above is what proves it.
Answer
Only $a=3$ gives a limit, and then the limit is $\tfrac92.$ For $a\ne3$ the quotient is $c/0$ in form, with $c=a-3\ne0,$ and there is no limit.
How did it go?
Stuck? Read 4. Reading the signal: l'Hôpital's rule →
Check the form first
Read the theory: 5. Check the form first →For each limit, say which step of Method 5.24 settles it, and why. Then find the limit, or show that there is none.
(a) $\displaystyle\lim_{x\to2}\frac{x^{3}-8}{x^{2}-4}$
(b) $\displaystyle\lim_{x\to-1}\frac{x^{2}+1}{(x+1)^{2}}$
(c) $\displaystyle\lim_{x\to0}\frac{e^{x}+x}{1+\ln(1+x)}$
(d) $\displaystyle\lim_{x\to1}\frac{\ln x}{x^{2}-1}$
Hint
Substitute first, every time. Only $\tfrac00$ (or $\tfrac{\infty}{\infty}$) is a reason to factor or to use l'Hôpital's rule. In (a), $x^{3}-8$ has the factor $x-2.$
(a) Step 3: $\tfrac00.$ Both polynomials are $0$ at $x=2,$ so both have the factor $x-2{:}$ $$ \begin{aligned} \frac{x^{3}-8}{x^{2}-4}&=\frac{(x-2)(x^{2}+2x+4)}{(x-2)(x+2)}\\ &=\frac{x^{2}+2x+4}{x+2} \end{aligned} $$ for $x\ne2.$ Step 4: substitute, $\dfrac{4+4+4}{4}=3.$
(b) Step 2: $c/0.$ Substituting gives $\tfrac20.$ There is no limit. The denominator is positive on both sides, so the quotient tends to $+\infty$ from both sides.
(c) Step 1: substitute. Numerator and denominator are continuous at $0$ and the denominator is $1+\ln1=1\ne0.$ The limit is the value, $\dfrac{1+0}{1}=1.$ (Differentiating top and bottom here would give $\dfrac{e^{x}+1}{1/(1+x)}\to2,$ a wrong number: Watch out 5.22.)
(d) Step 3: $\tfrac00.$ At $x=1$ both $\ln1$ and $1-1$ are $0.$ By l'Hôpital's rule, $$ \lim_{x\to1}\frac{\ln x}{x^{2}-1}=\lim_{x\to1}\frac{1/x}{2x}=\frac12\,, $$ where the last quotient is continuous at $1,$ so step 1 finishes it.
Answer
(a) $3$ (factor and cancel). (b) No limit; $+\infty$ from both sides. (c) $1$ (substitute). (d) $\tfrac12$ (l'Hôpital once).
How did it go?
Stuck? Read 5. Check the form first →
Addition: limits as $x\to\infty$
Read the theory: 6. Addition: limits as $x\to\infty$ →In each of these, both terms tend to $\infty$ as $x\to\infty.$
(a) $(x+5)-x$
(b) $x^{2}-x$
(c) $\sqrt{x^{2}+1}-x$
Find what each difference does as $x\to\infty.$ What do the three answers say about the form $\infty-\infty?$
Hint
(a) and (b) need only algebra: simplify, or take out a factor $x.$ In (c), multiply and divide by the conjugate.
(a) $(x+5)-x=5$ for every $x,$ so the limit is $5.$
(b) $x^{2}-x=x(x-1),$ a product of two factors that both grow past every bound. It tends to $\infty{:}$ there is no limit.
(c) The conjugate. $$ \begin{aligned} \sqrt{x^{2}+1}-x&=\frac{(x^{2}+1)-x^{2}}{\sqrt{x^{2}+1}+x}\\ &=\frac{1}{\sqrt{x^{2}+1}+x}\,. \end{aligned} $$ The denominator grows past every bound, so the limit is $0.$
What it says. The same form $\infty-\infty$ gave $5,$ no limit and $0.$ So the form alone determines nothing: it is indeterminate, just as $\tfrac00$ is (Definition 5.18). Writing “$\infty-\infty=0$” treats $\infty$ as a number, which it is not (Watch out 5.10).
Answer
(a) $5.$ (b) $\infty$ (no limit). (c) $0.$ The form $\infty-\infty$ is indeterminate.
How did it go?
Stuck? Read 6. Addition: limits as $x\to\infty$ →
Mixed, as on the exam
Here the sections are mixed, and nothing says which step to use. Write the solution on paper, with reasons, before you look.
Fact 5.19 also covers the form $\tfrac{\infty}{\infty}.$ Use it to find
(a) $\displaystyle\lim_{x\to\infty}\frac{x^{2}}{e^{x}};$
(b) $\displaystyle\lim_{x\to0^{+}}x\ln x.$
In (b), first write the product as a quotient.
Hint
(a) needs the rule twice. In (b), $x\ln x=\dfrac{\ln x}{1/x}{:}$ what form is that as $x\to0^{+}?$
(a) The form $\tfrac{\infty}{\infty}.$ Both $x^{2}$ and $e^{x}$ grow past every bound. By the rule, $$ \lim_{x\to\infty}\frac{x^{2}}{e^{x}}=\lim_{x\to\infty}\frac{2x}{e^{x}}\,, $$ which is $\tfrac{\infty}{\infty}$ again.
(a) Again. A second application gives $\dfrac{2}{e^{x}},$ whose denominator grows past every bound, so the limit is $0.$ The exponential wins the race against $x^{2}.$
(b) A product becomes a quotient. For $x>0,$ $x\ln x=\dfrac{\ln x}{1/x}.$ As $x\to0^{+},$ $\ln x\to-\infty$ and $1/x\to+\infty{:}$ the form is $\tfrac{\infty}{\infty}$ (up to sign), so the rule applies.
(b) The rule. $$ \begin{aligned} \lim_{x\to0^{+}}\frac{\ln x}{1/x} &=\lim_{x\to0^{+}}\frac{1/x}{-1/x^{2}}\\ &=\lim_{x\to0^{+}}(-x)=0\,. \end{aligned} $$ Simplify the fraction of derivatives before substituting: it is just $-x.$
Answer
(a) $0.$ (b) $0.$
How did it go?
Stuck? Read 4. Reading the signal: l'Hôpital's rule →
As on the exam
3 problems · 25 pProblems in exam form, with points. Solve them on paper or in the box. Then open the solution one step at a time and mark yourself against the rubric.
A parameter from a 0/0 limit, and a root at infinity
Two limits.
- (a)5 p
Determine the value of the parameter $b$ such that $$ \lim_{x\to0}\frac{\ln(1+bx)}{e^{4x}-e^{x}}=2\,. $$
- (b)5 p
Calculate the limit $$ \lim_{x\to\infty}\Bigl(2x-\sqrt{4x^{2}-12x}\Bigr)\,. $$
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(a) The form. At $x=0$ the numerator is $\ln1=0$ and the denominator is $e^{0}-e^{0}=0.$ The quotient is of the indeterminate form $\tfrac00$ (Definition 5.18), whatever $b$ is, so l'Hôpital's rule may be used.
(a) Differentiate numerator and denominator separately. By the chain rule, $$ \begin{aligned} \frac{d}{dx}\ln(1+bx)&=\frac{b}{1+bx}\,,\\ \frac{d}{dx}\bigl(e^{4x}-e^{x}\bigr)&=4e^{4x}-e^{x}\,. \end{aligned} $$
(a) The limit as an expression in $b.$ The new quotient is continuous at $0$ and its denominator is $4-1=3\ne0$ there, so we substitute (Fact 5.17): $$ \begin{aligned} &\lim_{x\to0}\frac{\ln(1+bx)}{e^{4x}-e^{x}}\\ &\quad=\lim_{x\to0}\frac{\frac{b}{1+bx}}{4e^{4x}-e^{x}} =\frac{b}{3}\,. \end{aligned} $$
(a) The parameter. $\dfrac{b}{3}=2$ gives $b=6.$
(b) The form, and the conjugate. For $x\ge3$ the root is defined. Both terms grow past every bound, so the form is $\infty-\infty.$ Multiply and divide by the conjugate $2x+\sqrt{4x^{2}-12x}{:}$ $$ \begin{aligned} &2x-\sqrt{4x^{2}-12x}\\ &\quad=\frac{4x^{2}-(4x^{2}-12x)}{2x+\sqrt{4x^{2}-12x}}\\ &\quad=\frac{12x}{2x+\sqrt{4x^{2}-12x}}\,. \end{aligned} $$
(b) Divide by $x>0.$ As $x\to\infty$ we may take $x>0,$ and then $x=\sqrt{x^{2}},$ so $$ \frac{\sqrt{4x^{2}-12x}}{x}=\sqrt{4-\frac{12}{x}}\,, $$ and the expression equals $$ \frac{12}{2+\sqrt{4-\frac{12}{x}}}\,. $$
(b) Let $x\to\infty.$ $\dfrac{12}{x}\to0,$ and the root is continuous, so $\sqrt{4-12/x}\to2.$ The limit is $\dfrac{12}{2+2}=3.$
Answer
(a) $b=6.$
(b) $\displaystyle\lim_{x\to\infty}\Bigl(2x-\sqrt{4x^{2}-12x}\Bigr)=3.$
Mark your solution
Tick each criterion that is really in your solution. A criterion scores only if what it asks for is there.
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Common errors
- Using l'Hôpital's rule without checking the form. Here it happens to be $\tfrac00,$ but the exam's solutions name the form before the rule is used, and Watch out 5.22 says why: applied to a form that is not indeterminate, the rule returns a wrong number silently.
- Forgetting the inner derivative: $\dfrac{1}{1+bx}$ for the numerator gives the limit $\tfrac13$ with no $b$ in it, and the equation has no solution. Equally, $e^{x}$ for the derivative of $e^{4x}$ gives the denominator $0$ at $x=0$ and a quotient $b/0.$
- Differentiating the whole fraction with the quotient rule and calling it l'Hôpital (Watch out 5.21). In the live vote on that error about two answers in three were right at first, and nearly all after the discussion.
- In (b), “$\infty-\infty=0$”, or dividing by $x$ and writing $\sqrt{4x^{2}-12x}/x=\sqrt{4-12x},$ with $x$ instead of $x^{2}$ under the root. Divide inside the root by $x^{2}=\bigl(\sqrt{x^{2}}\bigr)^{2}.$
l'Hôpital twice with a parameter, and two differences at infinity
Two problems on limits.
- (a)5 p
Determine all real values of the parameter $a$ such that $$ \lim_{x\to0}\frac{\ln(1+ax)-ax}{2x^{2}}=-4\,. $$
- (b)5 p
Calculate the following limits, or show that they do not exist.
(i) $\displaystyle\lim_{x\to\infty}\Bigl(\sqrt{x^{2}+7x}-\sqrt{x^{2}+x}\Bigr)$
(ii) $\displaystyle\lim_{x\to\infty}\Bigl(x-\sqrt{4x^{2}+x}\Bigr)$
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(a) The form. At $x=0$ the numerator is $\ln1-0=0$ and the denominator is $0.$ The form is $\tfrac00$ for every $a,$ so l'Hôpital's rule may be used. (For $a=0$ the numerator is $0$ for every $x$ and the limit is $0\ne-4,$ so $a\ne0$ in what follows.)
(a) The first application. Differentiating numerator and denominator separately, $$ \begin{aligned} &\lim_{x\to0}\frac{\ln(1+ax)-ax}{2x^{2}}\\ &\quad=\lim_{x\to0}\frac{\frac{a}{1+ax}-a}{4x}\,. \end{aligned} $$ At $x=0$ the new numerator is $a-a=0$ and the denominator is $0{:}$ the form is $\tfrac00$ again.
(a) The second application. The derivative of $\dfrac{a}{1+ax}-a$ is $-\dfrac{a^{2}}{(1+ax)^{2}},$ and that of $4x$ is $4.$ The new quotient is continuous at $0,$ so $$ \lim_{x\to0}\frac{-\frac{a^{2}}{(1+ax)^{2}}}{4}=-\frac{a^{2}}{4}\,. $$ (Alternatively, simplify after the first step: $\dfrac{a}{1+ax}-a=-\dfrac{a^{2}x}{1+ax},$ so the quotient is $-\dfrac{a^{2}}{4(1+ax)}\to-\dfrac{a^{2}}{4}.$)
(a) Every value. $-\dfrac{a^{2}}{4}=-4$ gives $a^{2}=16,$ so $a=4$ or $a=-4.$ Both work: $\ln(1+ax)$ is defined for $x$ near $0$ in either case.
(b)(i) The conjugate. For $x>0$ both roots are defined. Multiply and divide by $\sqrt{x^{2}+7x}+\sqrt{x^{2}+x}{:}$ $$ \begin{aligned} &\sqrt{x^{2}+7x}-\sqrt{x^{2}+x}\\ &\quad=\frac{(x^{2}+7x)-(x^{2}+x)}{\sqrt{x^{2}+7x}+\sqrt{x^{2}+x}}\\ &\quad=\frac{6x}{\sqrt{x^{2}+7x}+\sqrt{x^{2}+x}}\,. \end{aligned} $$
(b)(i) Divide by $x>0$ (so that $x=\sqrt{x^{2}}$): $$ \frac{6}{\sqrt{1+\frac7x}+\sqrt{1+\frac1x}}\,. $$ As $x\to\infty$ both roots tend to $1,$ and the limit is $\dfrac{6}{2}=3.$
(b)(ii) Take $x$ out of the root. For $x>0,$ $\sqrt{4x^{2}+x}=\sqrt{x^{2}}\sqrt{4+\tfrac1x}=x\sqrt{4+\tfrac1x},$ so $$ \begin{aligned} &x-\sqrt{4x^{2}+x}\\ &\quad=x\Bigl(1-\sqrt{4+\tfrac1x}\Bigr)\,. \end{aligned} $$
(b)(ii) The verdict. The bracket tends to $1-\sqrt4=-1,$ and $x$ grows past every bound. So the expression tends to $-\infty,$ and the limit does not exist (Watch out 5.10). (The conjugate gives the same: $\dfrac{-3x^{2}-x}{x+\sqrt{4x^{2}+x}}=\dfrac{-3x-1}{1+\sqrt{4+1/x}}\to-\infty.$)
Answer
(a) $a=4$ or $a=-4.$
(b) (i) The limit is $3.$ (ii) The expression tends to $-\infty{:}$ the limit does not exist.
Mark your solution
Tick each criterion that is really in your solution. A criterion scores only if what it asks for is there.
Your points 0 of 10 p
Common errors
- Stopping after the first application: the numerator $\tfrac{a}{1+ax}-a$ is $0$ at $x=0,$ but so is the denominator. A second $\tfrac00$ is a signal to go on, not the answer $0.$
- Giving only $a=4.$ The equation is $a^{2}=16,$ and the exam's solutions require every value of the parameter.
- In (b)(i), “$x-x=0$”: replacing each root by $x$ because the leading terms agree. The terms of lower order decide a difference; that is what the conjugate brings out.
- In (b)(ii), answering “the limit is $-\infty$” as if $-\infty$ were a value (Watch out 5.10), or a sign error that gives $+\infty.$ The root is about $2x,$ larger than $x,$ so the difference is negative.
Quotients at infinity: the leading terms decide
Calculate the limits.
- (a)1 p$\displaystyle\lim_{x\to\infty}\frac{6x^{3}-x+4}{2x^{3}+5x^{2}}$
- (b)2 p$\displaystyle\lim_{x\to\infty}\frac{\sqrt{16x^{2}+3x}+2x}{3x-1}$
- (c)2 p$\displaystyle\lim_{x\to\infty}\frac{\bigl(8x^{3}+x\bigr)^{11}}{(1-2x)^{33}}$
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(a) Divide by $x^{3}.$ For $x\ne0,$ $$ \frac{6x^{3}-x+4}{2x^{3}+5x^{2}} =\frac{6-\frac{1}{x^{2}}+\frac{4}{x^{3}}}{2+\frac{5}{x}}\,, $$ and every fraction with $x$ in its denominator tends to $0.$ The limit is $\dfrac{6}{2}=3.$
(b) Divide by $x>0.$ As $x\to\infty$ we take $x>0,$ so $x=\sqrt{x^{2}}$ and $\dfrac{\sqrt{16x^{2}+3x}}{x}=\sqrt{16+\dfrac3x}.$ Hence $$ \begin{aligned} &\frac{\sqrt{16x^{2}+3x}+2x}{3x-1}\\ &\quad=\frac{\sqrt{16+\frac3x}+2}{3-\frac1x}\,. \end{aligned} $$
(b) The limit. The root tends to $\sqrt{16}=4,$ so the limit is $\dfrac{4+2}{3}=2.$
(c) Divide by $x^{33}=\bigl(x^{3}\bigr)^{11}.$ For $x>0,$ $$ \begin{aligned} \frac{\bigl(8x^{3}+x\bigr)^{11}}{x^{33}}&=\Bigl(8+\frac{1}{x^{2}}\Bigr)^{11},\\ \frac{(1-2x)^{33}}{x^{33}}&=\Bigl(\frac1x-2\Bigr)^{33}. \end{aligned} $$
(c) The limit. The first tends to $8^{11}=2^{33},$ the second to $(-2)^{33}=-2^{33},$ because $33$ is odd. The limit is $\dfrac{2^{33}}{-2^{33}}=-1.$
Answer
(a) $3.$ (b) $2.$ (c) $-1.$
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Common errors
- In (c), losing the sign: $(1-2x)^{33}$ has the leading term $(-2x)^{33}=-2^{33}x^{33},$ negative because the power is odd. The answer $1$ comes from writing $(2x)^{33}.$
- In (b), dividing by $x$ inside the root as if it were outside: $\sqrt{16x^{2}+3x}/x$ is $\sqrt{(16x^{2}+3x)/x^{2}},$ not $\sqrt{16x+3}.$ And forgetting that $x=\sqrt{x^{2}}$ needs $x>0.$
- In (a), writing the answer $3$ with no argument. Reading it off the leading terms is right, but the solution has to show why: divide through by $x^{3},$ or say that every other term is of lower order and tends to $0$ after the division.