Corrected exercises: continuity at a point (201-SN2-RE)

A polynomial is continuous on all of $\mathbb{R}$: no denominator, no root, no change of formula, so no suspicious point. Substitute without a second thought: $f(2)=2\times 8-10+1=7$, and $\lim_{x\to 2}f(x)=7$ as well. CONTINUOUS. Computing both one-sided limits of a polynomial is the most expensive beginner's mistake on a test: it earns no marks and eats the minutes of the exercises that really do hide something.

The denominator vanishes at 5, so $g(5)$ DOES NOT EXIST: the first condition already fails and the function is discontinuous, whatever happens next. What remains is to name the discontinuity, and for that the limit is needed: $\dfrac{x^{2}-25}{x-5}=\dfrac{(x-5)(x+5)}{x-5}=x+5$ for $x\neq 5$, so $\lim_{x\to 5}g(x)=10$. The limit exists, the value does not: a REMOVABLE discontinuity, a hole, the only one of the three that can be repaired.

Here too $h(3)$ does not exist, but the numerator does not vanish at 3: nothing cancels. To the left of 3, $x-3$ is a small NEGATIVE number, so $h(x)\to-\infty$; to the right, it is small and POSITIVE, so $h(x)\to+\infty$. Both limits are infinite, with opposite signs: the line $x=3$ is a vertical asymptote and the discontinuity is INFINITE. No value can repair it, since no real number joins up with $\pm\infty$.

The formula changes at 2, so 2 is a suspicious point, and the only one. $k(2)=2+1=3$, because the piece $x\le 2$ is the one that contains the point. From the left, $\lim_{x\to 2^{-}}(x+1)=3$. From the right, $\lim_{x\to 2^{+}}(3x-3)=6-3=3$. Both one-sided limits equal 3, so the limit exists and equals 3, and $k(2)=3$: all three conditions hold. CONTINUOUS. Two different formulas do not create a discontinuity, they create a point to CHECK.

Same structure, opposite answer. $m(1)=2\times 1=2$, because the piece $x\ge 1$ is the one that contains the point. From the left, $\lim_{x\to 1^{-}}x^{2}=1$. From the right, $\lim_{x\to 1^{+}}2x=2$. The two one-sided limits are DIFFERENT, so $\lim_{x\to 1}m(x)$ does not exist: a JUMP discontinuity, of height $2-1=1$. No redefinition of $m(1)$ will change anything, since it is the limit that is missing, not the value.

$n(4)=|0|=0$. To the left of 4, $|x-4|=4-x\to 0$; to the right, $|x-4|=x-4\to 0$. Both limits equal 0 and $n(4)=0$: CONTINUOUS. Yet the graph has a SHARP POINT, a corner. This is the central pitfall of the chapter: continuity forbids lifting the pencil, it does not forbid turning sharply. A corner is visible, but it does not count as a discontinuity, and exercise 9 shows what it costs when the function describes a real object.

$f(0)=1$ by definition, so the first condition holds. The limit is the fundamental trigonometric limit: $\lim_{x\to 0}\dfrac{\sin x}{x}=1$, with angles in RADIANS. It equals 1 from both sides, so it exists, and it equals $f(0)=1$: CONTINUOUS. Note that the function $\dfrac{\sin x}{x}$ on its own is NOT continuous at 0, since it is not even defined there; it is the value $f(0)=1$, chosen on purpose, that repairs the hole.

$g(0)=0$ exists. For the limit, multiply by the conjugate: $\dfrac{\sqrt{x+4}-2}{x}\times\dfrac{\sqrt{x+4}+2}{\sqrt{x+4}+2}=\dfrac{x+4-4}{x(\sqrt{x+4}+2)}=\dfrac{1}{\sqrt{x+4}+2}$, whose limit at 0 is $\dfrac{1}{4}$. The limit exists, the value exists, but $\dfrac{1}{4}\neq 0$: the THIRD condition fails. A REMOVABLE discontinuity, repaired by setting $g(0)=\dfrac14$. This is the sneakiest case: the function is defined everywhere and yet discontinuous.

The domain of $p$ is $[3, +\infty)$: to the left of 3 the function does not exist, so $\lim_{x\to 3^{-}}p(x)$ makes no sense, and speaking of a discontinuity would be a misreading. What you check is continuity FROM THE RIGHT: $p(3)=0$ and $\lim_{x\to 3^{+}}\sqrt{x-3}=0$. They agree, so $p$ is continuous from the right at 3, and continuous on its whole domain $[3, +\infty)$. Answering “discontinuous because the left-hand limit is missing” is the classic error: a function is not blamed for failing to exist where its domain stops.

Two suspicious points, two answers. Factor: $\dfrac{x^{2}-x-6}{x^{2}-9}=\dfrac{(x-3)(x+2)}{(x-3)(x+3)}=\dfrac{x+2}{x+3}$ for $x\neq 3$. At $a=3$: the function is not defined, but $\lim_{x\to 3}q(x)=\dfrac{5}{6}$, so a hole, a REMOVABLE discontinuity. At $a=-3$: the factor $x+3$ did not cancel, the numerator $x+2$ tends to $-1\neq 0$, so the quotient blows up: an INFINITE discontinuity. One function carries both types, and it is the simplification that tells them apart.

$r(0)=0$ exists. From the left, $\dfrac{1}{x}\to-\infty$, so $e^{1/x}\to e^{-\infty}=0$: the left-hand limit is 0 and even agrees with $r(0)$. From the right, $\dfrac{1}{x}\to+\infty$, so $e^{1/x}\to+\infty$. The one-sided limits differ, and one of them is infinite: $\lim_{x\to 0}r(x)$ does not exist and the discontinuity is INFINITE, on the right side only. The function is continuous from the left at 0, which is never enough: continuity requires BOTH sides when both exist.

$s(0)=0$. The factor $\sin\dfrac{1}{x}$ has no limit at 0, it oscillates endlessly, so no product rule applies: the Squeeze Theorem is needed. For every $x\neq 0$, $-1\le\sin\dfrac{1}{x}\le 1$, so $-x^{2}\le s(x)\le x^{2}$. Both bounds tend to 0 as $x\to 0$, so $\lim_{x\to 0}s(x)=0=s(0)$: CONTINUOUS. The oscillation is crushed by the factor $x^{2}$, and that is the whole mechanism of the squeeze: trap a function with no limit between two functions that do have one, and the SAME one.

The complete statement is given above, and that is what you copy, with “for $x$ near $a$” and the “if... and...” about the two limits: the grader takes off a quarter mark for each missing piece. On the right, $x>0$: multiplying $-1\le\cos\frac{1}{x}\le 1$ by $x$ keeps the direction, $-x\le u(x)\le x$. On the left, $x<0$: the multiplication REVERSES the inequalities, $-x\ge u(x)\ge x$, that is, $x\le u(x)\le -x$. The inequalities “$-x\le u(x)\le x$” on the left side are false, and they show it on their own: at $x=-0.1$ they claim $0.1\le\dots\le -0.1$. On both sides the bounds tend to 0, so $\lim_{x\to 0}u(x)=0=u(0)$: $u$ is continuous at 0. The result does not change, so the error is invisible in the answer; it still costs the justification, which is what gets marked.

“For $x$ near 0” lets you work in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, where the sign of $\tan x$ is known. For $0<x<\frac{\pi}{2}$, $\tan x>0$: $-\tan x\le t(x)\le\tan x$. For $-\frac{\pi}{2}<x<0$, $\tan x<0$ and the inequalities reverse: $\tan x\le t(x)\le -\tan x$. The tangent is continuous at 0, so $\tan x\to\tan 0=0$: cite continuity, since it is what allows you to replace $x$ by 0. The squeeze gives a limit of zero on each side, equal to $t(0)$: $t$ is continuous at 0. The form changes because the multiplier changes sign at 0; the result does not.

Only one source of suspicion: the denominator. $x^{2}-4=0$ gives $x=-2$ and $x=2$, TWO suspicious points. The numerator $x+1$ vanishes neither at $-2$ nor at 2, so nothing cancels and both discontinuities are INFINITE. $f$ is continuous on $(-\infty, -2)\cup(-2, 2)\cup(2, +\infty)$, that is, on its whole domain. A rational function is always continuous on its domain: the only real question is where that domain is.

Two simultaneous conditions, and it is their INTERSECTION that counts. The root requires $2x-6\ge 0$, so $x\ge 3$: the inequality is NON-STRICT, because $\sqrt{0}=0$ exists. The logarithm requires $x-4>0$, so $x>4$: the inequality is STRICT, because $\ln 0$ does not exist. The intersection is $(4, +\infty)$, and $g$ is continuous there as the sum of two continuous functions. The point 3 is not a point of discontinuity: it is a point where the function does not exist at all, which is not the same thing.

The root requires $x\ge 0$ and the denominator excludes $x=4$. The domain is therefore $[0, 4)\cup(4, +\infty)$ and $h$ is continuous there. Two points deserve a word: at $x=4$, the numerator equals $2\neq 0$, so an INFINITE discontinuity; at $x=0$, the function exists, $h(0)=0$, and it is continuous from the right, since the domain stops there. So you count ONE point of discontinuity, not two.

A piecewise function has two families of suspicious points: the joins, and the trouble spots INSIDE a piece. Joins: $x=0$ and $x=3$. Inside trouble: the middle piece has a denominator that vanishes at $x=1$, and 1 does lie in $[0, 3)$. So three suspicious points. At 0: from the left $x^{2}\to 0$, from the right $\dfrac{1}{x-1}\to -1$, and $k(0)=-1$: a JUMP. At 1: an INFINITE discontinuity. At 3: from the left $\dfrac{1}{x-1}\to\dfrac12$, from the right 5, and $k(3)=5$: a JUMP. Forgetting the point 1, which is not a join, is the error this exercise is looking for.

$x^{3}-x=x(x-1)(x+1)$ vanishes at $-1$, 0 and 1: three suspicious points. But the numerator $x-1$ cancels with the factor $x-1$: for $x\neq 1$, $m(x)=\dfrac{1}{x(x+1)}$. At $x=1$, the limit is therefore $\dfrac{1}{2}$: a REMOVABLE discontinuity. At $x=0$ and $x=-1$, nothing cancels: INFINITE discontinuities. Three suspicious points, three discontinuities, but not of the same type, and only the factoring tells you which.

An identity to know: $x^{3}-8=(x-2)(x^{2}+2x+4)$. So for $x\neq 2$, $f(x)=x^{2}+2x+4$, and $\lim_{x\to 2}f(x)=4+4+4=12$. Set $f(2)=12$: the extension is the whole parabola $x^{2}+2x+4$. A numerical check, always useful: $f(2.001)=12.006$, so we are indeed close to 12.

Conjugate: $\dfrac{\sqrt{x+1}-1}{x}\times\dfrac{\sqrt{x+1}+1}{\sqrt{x+1}+1}=\dfrac{x}{x(\sqrt{x+1}+1)}=\dfrac{1}{\sqrt{x+1}+1}$, whose limit at 0 is $\dfrac{1}{2}$. Set $g(0)=\dfrac12$. The reflex to keep: a root in the numerator and a zero over zero call for the conjugate before any other idea, because it turns a difference of roots into a difference of squares, and so into a factor that cancels.

Bring it back to the fundamental limit by making the same argument appear above and below: $\dfrac{\sin 5x}{2x}=\dfrac{5}{2}\times\dfrac{\sin 5x}{5x}$. As $x\to 0$, $5x\to 0$ too, so the second factor tends to 1 and the limit is $\dfrac{5}{2}$. Set $h(0)=\dfrac52$. The usual mistake is to write $\dfrac52$ straight away without the change of argument, which gives the right answer for the wrong reason and collapses as soon as the question puts $\sin 5x$ over $\sin 3x$.

No extension is possible. To the left of 0, $|x|=-x$, so $k(x)=-1$; to the right, $|x|=x$, so $k(x)=1$. The one-sided limits are $-1$ and 1: the limit does not exist, and assigning a value at 0 will never create the missing equality, whatever that value is. This is a JUMP discontinuity, and a jump cannot be repaired: you do not fill a ditch with a single point.

Standard limit: $\lim_{h\to 0}\dfrac{e^{h}-1}{h}=1$. Force the argument: $\dfrac{e^{2x}-1}{x}=2\times\dfrac{e^{2x}-1}{2x}$, whose limit is $2\times 1=2$. Set $m(0)=2$. Those who have already seen the derivative will recognize the rate of change of $e^{2x}$ at 0: limits of this kind are exactly derivatives in disguise, and computing them is the reason the chapter on limits exists.

The substitution $u=\sqrt{x}$ tidies everything up: $x=u^{2}$, and as $x\to 1$, $u\to 1$. Then $\dfrac{x^{2}-1}{\sqrt{x}-1}=\dfrac{u^{4}-1}{u-1}=\dfrac{(u-1)(u+1)(u^{2}+1)}{u-1}=(u+1)(u^{2}+1)$, whose limit at $u=1$ is $2\times 2=4$. Set $n(1)=4$. The conjugate would also have worked, but in two stages; when a root is the only obstacle, setting $u=\sqrt{x}$ brings the problem back to a polynomial.

The reflex “the numerator must vanish” gives nothing here: at 1, it equals $1+b-b-1=0$ for EVERY $b$. The condition is always true, so it fixes nothing. Factor: $x^{3}+bx-b-1=(x^{3}-1)+b(x-1)=(x-1)(x^{2}+x+1+b)$. For $x\neq 1$, $p(x)=\frac{x^{2}+x+1+b}{x-1}$: the denominator STILL contains $x-1$, and the new numerator must vanish at 1, that is, $3+b=0$, $b=-3$. Then $x^{2}+x-2=(x-1)(x+2)$ and $p(x)=x+2\to 3$: set $p(1)=3$. Altogether, $x^{3}-3x+2=(x-1)^{2}(x+2)$, the factor appears twice in the numerator, just as in the denominator. For $b\neq -3$, the form $\left[\frac{3+b}{0}\right]$ gives opposite infinities on either side of 1.

First cancellation: $9+3p+c=0$, so $c=-9-3p$ and the numerator is $(x-3)(x+3+p)$. For $x\neq 3$, $q(x)=\frac{x+3+p}{x-3}$, and the second cancellation requires $6+p=0$: $p=-6$, then $c=9$. The numerator is then $(x-3)^{2}$, $q(x)=1$ for $x\neq 3$, and you set $q(3)=1$. Shorter: a degree-2 polynomial with leading coefficient 1 that contains $(x-3)$ twice IS $(x-3)^{2}$. With a single cancellation, the limit is infinite with opposite signs; with none, the square, positive on both sides, gives $\pm\infty$: never a number.

None. The cancellation requires $4m-8=0$, so $m=2$ is the only candidate; but $2x^{2}-8=2(x-2)(x+2)$ contains $x-2$ only once, and $r(x)=\frac{2(x+2)}{x-2}$ for $x\neq 2$, form $\left[\frac{8}{0}\right]$: $-\infty$ on the left, $+\infty$ on the right. For $m\neq 2$, the form $\left[\frac{4m-8}{0^{+}}\right]$ gives $+\infty$ or $-\infty$ on both sides. This could be predicted: $mx^{2}-8$ takes the same value at $x$ and at $-x$, so if it vanishes at 2 it also vanishes at $-2$, and 2 cannot be a double root.