Corrected exercises: limits and continuity (201-103)

Direct substitution: the denominator equals $2\times 3-1=5\neq 0$, so the limit is $\frac{9+3}{5}=\frac{12}{5}=2.4$. No transformation is needed when substitution answers.

Substitution gives $\frac{0}{0}$. Difference of squares: $\frac{x^{2}-25}{x+5}=\frac{(x-5)(x+5)}{x+5}=x-5$ for $x\neq -5$, so the limit is $-5-5=-10$. Pitfall: keeping the wrong factor, $x+5$, and answering $0$.

$\frac{0}{0}$ at $x=2$, so $x-2$ is a factor of both trinomials: $x^{2}+x-6=(x-2)(x+3)$ and $x^{2}-3x+2=(x-2)(x-1)$. What remains is $\frac{x+3}{x-1}$, which equals $\frac{5}{1}=5$ at $x=2$.

$\frac{0}{0}$ at $x=1$. On top, the difference of cubes $x^{3}-1=(x-1)(x^{2}+x+1)$; on the bottom, $x^{2}-1=(x-1)(x+1)$. What remains is $\frac{x^{2}+x+1}{x+1}$, hence the limit $\frac{3}{2}=1.5$. Pitfall: writing $x^{3}-1=(x-1)^{3}$, which is false as soon as $x=0$.

$\frac{0}{0}$ with a square root: multiply top and bottom by the conjugate $\sqrt{x+7}+3$. The numerator becomes $(x+7)-9=x-2$, which cancels: $\frac{\sqrt{x+7}-3}{x-2}=\frac{1}{\sqrt{x+7}+3}$, hence $\frac{1}{3+3}=\frac{1}{6}\approx 0.167$. Pitfall: multiplying the numerator alone, which changes the value of the fraction.

$\frac{0}{0}$ with stacked fractions: put the numerator over a common denominator, $\frac{1}{x}-\frac{1}{5}=\frac{5-x}{5x}=\frac{-(x-5)}{5x}$. Divided by $x-5$, what remains is $-\frac{1}{5x}$, hence the limit $-\frac{1}{25}=-0.04$. Pitfall: the minus sign hidden in $5-x=-(x-5)$.

Substitution gives $\frac{7}{0}$: an infinite limit, and everything depends on the sign. For $x<3$, $x-3<0$ and the numerator is close to $7>0$, so $\lim_{x\to 3^{-}}\frac{2x+1}{x-3}=-\infty$. For $x>3$, $x-3>0$, so $\lim_{x\to 3^{+}}\frac{2x+1}{x-3}=+\infty$. The two-sided limit does not exist, and the line $x=3$ is a vertical asymptote.

Substitution gives $\frac{-5}{0}$. The denominator $(x+1)^{2}$ is a SQUARE, positive on both sides of $-1$; the numerator is close to $-5$. The quotient is therefore negative on both sides, and $\lim_{x\to -1}\frac{x-4}{(x+1)^{2}}=-\infty$ from both sides. Pitfall: answering $+\infty$ because the denominator is positive, forgetting the sign of the numerator.

At infinity, keep the dominant terms: $\frac{6x^{3}-2x}{1-3x^{3}}\approx\frac{6x^{3}}{-3x^{3}}=-2$. Equal degrees: the limit is the ratio of the leading coefficients, sign included. Pitfall: reading $\frac{6}{1}$ by taking the first term written instead of the term of highest degree.

As $t\to+\infty$, $-0.3t\to-\infty$, so $e^{-0.3t}\to 0$. The denominator tends to $1+4\times 0=1$ and the limit is $600$: this is the ceiling of the model, its horizontal asymptote $y=600$.

Take $x^{2}$ out of the root: $\sqrt{9x^{2}+1}=|x|\sqrt{9+\frac{1}{x^{2}}}=x\sqrt{9+\frac{1}{x^{2}}}$ for $x>0$. So $\frac{\sqrt{9x^{2}+1}}{2x-5}=\frac{\sqrt{9+\frac{1}{x^{2}}}}{2-\frac{5}{x}}\to\frac{3}{2}=1.5$. Pitfall: comparing the “degrees” $2$ and $1$ and concluding $+\infty$, when the root brings the degree of the top down to $1$.

Form $\infty-\infty$: multiply and divide by the conjugate $\sqrt{x^{2}+5x+1}+x$. The numerator becomes $(x^{2}+5x+1)-x^{2}=5x+1$, so the expression equals $\frac{5x+1}{\sqrt{x^{2}+5x+1}+x}$. Dividing top and bottom by $x$: $\frac{5+\frac{1}{x}}{\sqrt{1+\frac{5}{x}+\frac{1}{x^{2}}}+1}\to\frac{5}{2}=2.5$. Pitfall: answering $0$ because “infinity minus infinity cancels”.

Approaching 2 from the LEFT, we follow the branch $x+3$, which tends to $2+3=5$. Approaching from the RIGHT, we follow the branch $7-x$, which tends to $7-2=5$. The two one-sided limits are equal, so $\lim_{x\to 2}f(x)=5$. But the filled dot is at $(2, 3)$: $f(2)=3$. The limit exists and equals 5, the value exists and equals 3, and they differ. This is the case that on its own justifies the invention of the limit: the function knows where it is going without being there.

To the left of 5, we follow $7-x$, which tends to $7-5=2$. To the right, the branch is constant at 4, so the right-hand limit is 4. $f(5)=2$, the filled dot. The two one-sided limits are DIFFERENT, so $\lim_{x\to 5}f(x)$ DOES NOT EXIST. Writing “the limit is 2 because $f(5)=2$” is the typical error: the value at a point has never been used to compute a limit.

At $x=2$, the limit exists but is not equal to $f(2)$: a HOLE (point discontinuity). At $x=5$, both one-sided limits exist and differ: a JUMP, of height $4-2=2$. Neither is infinite, since the function stays bounded.

The three conditions for continuity at $a$: first, $f(a)$ must exist; second, $\lim_{x\to a}f(x)$ must exist, so the two one-sided limits must be equal; third, the two must coincide, $\lim_{x\to a}f(x)=f(a)$. At $x=2$, the first two are met, $f(2)=3$ exists and the limit is 5: it is the THIRD that fails. Naming which of the three breaks down is what a grader looks for on the paper, not the word “discontinuous”.

Because a hole comes from ONE misplaced value: by setting $f(2)=5$ instead of 3, we get a function that is continuous at 2 without touching anything else. A jump comes from two different one-sided limits, and no value chosen at $x=5$ can equal both 2 and 4. We say that a hole is REMOVABLE. This is exactly what we will do in the next chapter when we simplify a fraction: the simplification fills the hole, and that is what makes the derivative computable.

Substitution gives $\dfrac{0}{0}$. Factor the numerator: $q^{2}-16=(q-4)(q+4)$, so for $q\ne 4$ the fraction equals $q+4$. The limit is $4+4=8$. The cancellation is legitimate because we are NOT computing the value at 4, we are computing what happens around it: $q-4$ is nonzero as long as we have not arrived.

Substitution again gives $\dfrac{0}{0}$. Factor both: $x^{2}-5x+6=(x-2)(x-3)$ and $x^{2}-9=(x-3)(x+3)$. The factor $(x-3)$ cancels and what remains is $\dfrac{x-2}{x+3}$, whose limit at 3 is $\dfrac{1}{6}$. That the common factor is $(x-3)$ is no accident: the form $\dfrac{0}{0}$ guaranteed it, since 3 makes both polynomials vanish.

Substitution gives $\dfrac{0}{0}$. Expand: $(5+h)^{2}-25=25+10h+h^{2}-25=10h+h^{2}=h(10+h)$. The fraction therefore equals $10+h$ for $h\ne 0$, and the limit is 10. This limit has a name that the next chapter will supply: it is the DERIVATIVE of $x^{2}$ at $x=5$. The whole construction of the derivative is a $\dfrac{0}{0}$ indeterminate form resolved by factoring.

Substitution gives $\dfrac{0}{0}$, and nothing factors at a glance. Multiply top and bottom by the CONJUGATE $\sqrt{x}+1$: the numerator becomes $(\sqrt{x})^{2}-1^{2}=x-1$, and the fraction equals $\dfrac{x-1}{(x-1)(\sqrt{x}+1)}=\dfrac{1}{\sqrt{x}+1}$. The limit is $\dfrac{1}{1+1}=\dfrac{1}{2}$. The conjugate is the third tool of the chapter, after factoring and expanding, and it is recognized by a square root in a difference.

Because $\dfrac{0}{0}$ is not a value, it is a REPORT OF FAILURE of substitution. The four previous questions show it: the same form gave 8, then $\dfrac{1}{6}$, then 10, then $\dfrac{1}{2}$. A form that can give any number cannot designate any one of them, hence the word INDETERMINATE. Nor does it announce that the limit is absent: all four limits exist. What it announces is more modest and more useful: numerator and denominator vanish together, so they share a factor, so there is something to cancel.

Both degrees equal 2. Divide top and bottom by $x^{2}$: $\dfrac{3+\dfrac{5}{x}}{2-\dfrac{7}{x^{2}}}$. Both fractions tend to 0, so the limit is $\dfrac{3}{2}=1.5$. The shortcut, once the method is understood, is the RATIO OF THE LEADING COEFFICIENTS, here $\dfrac{3}{2}$. The terms $5x$ and $-7$ do not vanish because they are small, but because they are negligible COMPARED TO $x^{2}$ as $x$ grows.

The degree of the top, 1, is lower than that of the bottom, 2. The denominator wins, so the limit is 0. Horizontal asymptote $y=0$. In other words, a quantity that grows linearly is always eventually crushed by a quantity that grows quadratically, however far you have to go.

The degree of the top, 3, exceeds that of the bottom, 2. The numerator wins and the limit is $+\infty$: there is NO horizontal asymptote. Writing $\dfrac{2}{5}$ out of habit is the most common error of the chapter; the ratio of the coefficients only applies when the degrees are EQUAL.

$M(12)=\dfrac{960}{24}=40$%, $M(36)=\dfrac{2{,}880}{48}=60$%. The degrees of the top and the bottom are both 1, so the limit is the ratio of the leading coefficients, $\dfrac{80}{1}=80$%. The horizontal asymptote is $y=80$. Reading of the model: market share climbs quickly at first, 40% after one year, then levels off at 80% without ever reaching it. The parameter 12 is the delay after which HALF of the ceiling is reached, which gives the right order of magnitude for a campaign.

Three cases, three readings. Degree of the top LOWER: limit 0, asymptote $y=0$, the model dies out. Degrees EQUAL: limit equal to the ratio of the leading coefficients, nonzero horizontal asymptote, the model levels off. Degree of the top HIGHER: infinite limit, no horizontal asymptote, the model blows up. The third case is an alarm signal in the social sciences: a market share, a rate, a proportion cannot exceed 100, so a model that blows up is a poorly chosen model, not a prediction.

The form is $\infty-\infty$. Multiply and divide by the conjugate $\sqrt{x^{2}+6x}+x$: the numerator becomes $(x^{2}+6x)-x^{2}=6x$, so the expression equals $\dfrac{6x}{\sqrt{x^{2}+6x}+x}$. Dividing top and bottom by $x$, which is positive at infinity: $\dfrac{6}{\sqrt{1+\frac{6}{x}}+1}$, whose limit is $\dfrac{6}{2}=3$. The shortcut to remember for what follows: $\sqrt{x^{2}+bx}-x$ tends to $\dfrac{b}{2}$.

Same method with the conjugate $x+\sqrt{x^{2}-4x}$: the numerator becomes $x^{2}-(x^{2}-4x)=4x$, and the expression equals $\dfrac{4x}{x+\sqrt{x^{2}-4x}}$, with limit $\dfrac{4}{2}=2$. We recover $\dfrac{b}{2}$ up to the sign, and we see that an $\infty-\infty$ form can perfectly well tend to a FINITE nonzero number.

$A(t)-B(t)=0.5t^{2}+300t-0.5t^{2}-280t-5{,}000=20t-5{,}000$. The quadratic terms cancel exactly, which was the indeterminacy. The limit is $+\infty$: the gap widens without bound. It vanishes at $t=\dfrac{5{,}000}{20}=250$ years: before year 250, $B$ has cost more; after, it is $A$. The fixed amount of 5,000 is absorbed in two and a half centuries by a gap of 20 per year.

$\dfrac{A(t)}{B(t)}=\dfrac{0.5t^{2}+300t}{0.5t^{2}+280t+5{,}000}$. Both degrees equal 2, so the limit is the ratio of the leading coefficients, $\dfrac{0.5}{0.5}=1$. The two programs therefore end up costing the SAME in proportion: their ratio tends to 1, that is, to a relative gap of 0%.

They are not contradictory, they do not measure the same thing. The ABSOLUTE gap $A-B$ grows like $20t$, the RELATIVE gap $\dfrac{A-B}{B}$ behaves like $\dfrac{20t}{0.5t^{2}}=\dfrac{40}{t}$, which tends to 0. A gap that grows in dollars can become negligible in percentage, simply because the two budgets grow faster than their difference. All of budget rhetoric lies in this choice: whoever wants to alarm quotes the millions of difference, whoever wants to reassure quotes the percentage, and both tell the truth. Knowing which of the two quotients was computed is exactly what the course teaches you to ask.

$S$ is continuous at $R_{0}$ if $S(R_{0})$ exists, if $\lim_{R\to R_{0}}S(R)$ exists, which requires the two one-sided limits to be equal, and if these two numbers coincide. On a piecewise function, this always comes down to the same calculation: the two formulas that meet at the junction must give the same value there.

At $R=70$, the left-hand limit is $4{,}500-a(70-25)=4{,}500-45a$ and the right-hand limit is 0. Setting them equal gives $45a=4{,}500$, so $a=100$. The continuous schedule is therefore the one that withdraws 100 dollars of benefit per thousand dollars of additional income, that is, 10 cents per dollar earned.

At $R=25$: the first formula gives $4{,}500$; the second gives $4{,}500-100\times(25-25)=4{,}500$. The two coincide, so $S$ is continuous at 25. Continuity there was guaranteed from the start, by the very form of the expression $R-25$: this is why schedules are always written with the distance to the lower bound of the bracket, and not with gross income.

$S(50)=4{,}500-100\times(50-25)=4{,}500-2{,}500=2{,}000$ dollars. The number $a=100$ is the RECOVERY RATE: each additional thousand dollars earned costs 100 dollars of benefit. Stated per dollar, the family keeps 90 cents of each additional dollar, the other 10 being taken back by the schedule.

With $a=120$, the second formula vanishes at $R=25+\dfrac{4{,}500}{120}=62.5$, then becomes NEGATIVE between 62.5 and 70, where it reaches $4{,}500-120\times 45=-900$ dollars. Two defects, and the second is the worse. First, a negative benefit makes no sense: the formula would have to be truncated at zero, so the schedule rewritten. Second, if we do not truncate, $S$ has at $R=70$ a JUMP from $-900$ to 0, which means that a household earning 70,000 dollars would be better treated than one earning 69,000. A discontinuity in a public schedule always creates a zone where earning more means receiving less, and this is exactly what the continuity condition is there to rule out.

If $f$ is CONTINUOUS on a CLOSED interval $[a, b]$, and if $k$ is a number between $f(a)$ and $f(b)$, then there is at least one $c$ in $[a, b]$ such that $f(c)=k$. Three hypotheses, and none is decorative: continuity, the closed interval, and $k$ lying between the two endpoint values. The conclusion is an EXISTENCE, never a value.

$P(8)=-0.2\times 512+6\times 64+160+4{,}000=-102.4+384+160+4{,}000=4{,}441.6$. $P(9)=-0.2\times 729+6\times 81+180+4{,}000=-145.8+486+180+4{,}000=4{,}520.2$. $P$ is a polynomial, so it is continuous on $\mathbb{R}$, in particular on $[8, 9]$. Since $4{,}441.6<4{,}500<4{,}520.2$, the theorem gives the existence of a $t$ in $[8, 9]$ such that $P(t)=4{,}500$: the town crossed the threshold between 2018 and 2019.

$P(8.5)=-0.2\times 614.125+6\times 72.25+170+4{,}000=-122.825+433.5+170+4{,}000=4{,}480.675$, still below the threshold. Apply the theorem on $[8.5, 9]$. Then $P(8.75)=-0.2\times 669.921875+6\times 76.5625+175+4{,}000=-133.984+459.375+175+4{,}000=4{,}500.39$, this time above. The moment sought is therefore in $[8.5, 8.75]$, and very close to 8.75. Two applications of the theorem divided the uncertainty by four, without ever solving the cubic equation.

No. The theorem gives “there is AT LEAST one $c$”: it says nothing about the number of solutions. Here the curve rises without falling back on $[8, 9]$, which ensures uniqueness, but that information comes from MONOTONICITY, not from the theorem. A model that crossed the threshold, fell back, then crossed it again would satisfy the same hypotheses with three solutions.

No, and it is continuity that is missing. The flat rate jumps from 255 to 250 dollars at $x=50$: no consumption level makes you pay 252 dollars, even though 252 lies between the values taken before and after. What is lost is exactly the conclusion of the theorem, the existence of a preimage for every intermediate value. This is the counterexample to keep in mind so as never to invoke the theorem without checking its least visible hypothesis: a piecewise-defined function is continuous only if you have checked that it is.