Corrected practice exam 1 in differential calculus (201-SN2-RE)

A quotient: the only restriction is a zero denominator. Factor $2x^{2}+3x+1$. The discriminant is $9-8=1$, the roots are $x=\frac{-3\pm 1}{4}$, that is $-\frac{1}{2}$ and $-1$, so $2x^{2}+3x+1=(2x+1)(x+1)$. The denominator vanishes at $x=2$, $x=-\frac{1}{2}$ and $x=-1$. Domain: $(-\infty,-1)\cup(-1,-\tfrac{1}{2})\cup(-\tfrac{1}{2},2)\cup(2,+\infty)$. The pitfall: many students also exclude $x=-2$ because the numerator vanishes there. A zero numerator creates no restriction, it creates a zero of the function. The domain question never looks at the top of the fraction.

A square root, so an even root: we require $4-x\ge 0$, that is $x\le 4$. Domain: $(-\infty,4]$. The bracket is closed at 4 because $\sqrt{0}=0$ exists. Writing $(-\infty,4)$ costs a mark for nothing.

A quotient again, but here the denominator never vanishes: the discriminant of $3x^{2}-4x+7$ is $16-4(3)(7)=16-84=-68<0$. A quadratic with a negative discriminant keeps the sign of its leading coefficient, here $3>0$, so $3x^{2}-4x+7>0$ for every real number. Domain: $\mathbb{R}$. This is the false restriction of the title: the presence of a fraction is not enough to remove anything, you must check that the denominator really vanishes.

Two restrictions combine, and they combine into a single STRICT inequality. The root requires $x^{2}-5x+4\ge 0$, the fraction requires $\sqrt{x^{2}-5x+4}\ne 0$, so $x^{2}-5x+4\ne 0$. Altogether, $x^{2}-5x+4>0$. Now $x^{2}-5x+4=(x-1)(x-4)$, a product that is positive outside the roots. Domain: $(-\infty,1)\cup(4,+\infty)$. A root in the denominator is not handled like a root in the numerator: it turns a $\ge$ into a $>$.

Handle the two constraints separately. Root: $x^{2}-3x+1\ge 0$. The discriminant is $9-4=5$, the roots are $\frac{3\pm\sqrt{5}}{2}$, approximately $0.382$ and $2.618$. The quadratic, with a positive leading coefficient, is positive outside its roots: $x\le\frac{3-\sqrt{5}}{2}$ or $x\ge\frac{3+\sqrt{5}}{2}$. Denominator: $3-x\ne 0$, so $x\ne 3$. Intersect: $\left(-\infty,\frac{3-\sqrt{5}}{2}\right]\cup\left[\frac{3+\sqrt{5}}{2},3\right)\cup(3,+\infty)$. Note that $3$ does lie in the region allowed by the root ($3>2.618$), so it must be removed explicitly: this is exactly the kind of point you forget when you handle both conditions in your head instead of writing them one under the other.

Without simplifying: $(f\circ g)(x)=\left(\sqrt{x^{2}-5}+1\right)^{2}-\left(\sqrt{x^{2}-5}+1\right)+3$. Expand the square: $\left(\sqrt{x^{2}-5}\right)^{2}+2\sqrt{x^{2}-5}+1=x^{2}-5+2\sqrt{x^{2}-5}+1$. Then subtract $\sqrt{x^{2}-5}+1$ and add 3: $x^{2}-5+2\sqrt{x^{2}-5}+1-\sqrt{x^{2}-5}-1+3=x^{2}-2+\sqrt{x^{2}-5}$. Numerical check at $x=3$: $g(3)=\sqrt{4}+1=3$, then $f(3)=9-3+3=9$; the formula gives $9-2+\sqrt{4}=9$. The two agree.

$(g\circ f)(x)=\sqrt{(x^{2}-x+3)^{2}-5}+1$. Domain: we need $(x^{2}-x+3)^{2}\ge 5$. The quadratic $x^{2}-x+3$ has discriminant $1-12=-11<0$, so it stays strictly positive, and its minimum, reached at $x=\frac{1}{2}$, is $\frac{1}{4}-\frac{1}{2}+3=\frac{11}{4}=2.75$. Hence $(x^{2}-x+3)^{2}\ge 2.75^{2}=7.5625>5$ for every real number. Domain: $\mathbb{R}$. Compare with a): $f\circ g$ and $g\circ f$ have neither the same expression nor the same domain; composition is not commutative.

Domain of $f\circ g$: first, $x$ must be in the domain of $g$, so $x^{2}-5\ge 0$, that is $|x|\ge\sqrt{5}$; then $g(x)$ must be in the domain of $f$, which is $\mathbb{R}$, and that adds nothing. Domain: $(-\infty,-\sqrt{5}]\cup[\sqrt{5},+\infty)$. The simplified expression $x^{2}-2+\sqrt{x^{2}-5}$ still carries the root, so here it gives the same restriction. But the general rule goes the other way: the domain of a composite is read from the composite BEFORE simplifying, never after. Simplifying can hide a restriction without removing it.

Apply $h$ first, then $g$, then $f$: $(f\circ g\circ h)(x)=f\!\left(g(x+3)\right)=\left(\sqrt{(x+3)^{2}-5}+1\right)^{2}-\left(\sqrt{(x+3)^{2}-5}+1\right)+3$. No simplification is asked for: on an exam, a chain of compositions is written by substituting the variable mechanically, and every forgotten parenthesis costs the result.

NO. Their rules agree wherever both are defined, since $\frac{x^{2}-4}{x-2}=\frac{(x-2)(x+2)}{x-2}=x+2$ as soon as $x\ne 2$. But $\operatorname{dom}(p)=\mathbb{R}\setminus\{2\}$, whereas $\operatorname{dom}(q)=\mathbb{R}$. Two functions are equal if and only if they have the SAME domain and the same rule. The graph of $p$ is the graph of $q$ with a hole punched at $(2,4)$. This distinction is not a matter of vocabulary: it is exactly what separates a vertical asymptote from a hole, and it comes back unchanged in exercise 7.

A polynomial is continuous everywhere, so the limit is the value: $8-24+4=-12$. No work at all, and it is the first reflex to have: you only look for a technique if substitution fails.

$3^{-x}=\left(\frac{1}{3}\right)^{x}$ with a base strictly between 0 and 1: the powers shrink towards zero. Limit $=0$. This is not an indeterminate form, it is the standard behaviour of a decreasing exponential.

$x^{2}+6x+9=(x+3)^{2}$. Substitution gives $\frac{5}{0}$, with $k=5\ne 0$: this is NOT an indeterminate form. The denominator is a square, so it is positive on both sides of $-3$, and the numerator stays $5>0$. The quotient blows up on both sides: $\lim_{x\to -3}\frac{5}{(x+3)^{2}}=+\infty$. The line $x=-3$ is a vertical asymptote.

Substitution: $\frac{0}{0}$, so a common factor is guaranteed. $25-x^{2}=(5-x)(5+x)$, hence $\frac{x+5}{(5-x)(5+x)}=\frac{1}{5-x}$ for $x\ne -5$. At $-5$: $\frac{1}{10}$. The pitfall is believing that $\frac{0}{0}$ always gives infinity or always gives zero: it gives nothing at all until you have simplified.

$\frac{0}{0}$: $4-4-1+7-6=0$ on top, $3-8+6-1=0$ at the bottom. Both polynomials therefore have $(x-1)$ as a factor. Synthetic division of the numerator with the coefficients $4,-4,-1,7,-6$: bring down $4$, then $4-4=0$, then $0-1=-1$, then $-1+7=6$, then $6-6=0$. The quotient is $4x^{3}+0x^{2}-x+6$, which equals $4-1+6=9$ at $x=1$. In the denominator, coefficients $3,-8,6,-1$: bring down $3$, then $3-8=-5$, then $-5+6=1$, then $1-1=0$. The quotient is $3x^{2}-5x+1$, which equals $3-5+1=-1$. Limit $=\frac{9}{-1}=-9$. Here synthetic division is faster and safer than L'Hôpital's rule, and it requires no theorem.

A compound fraction: first combine the numerator over a common denominator. $\frac{1}{x+1}-\frac{1}{4}=\frac{4-(x+1)}{4(x+1)}=\frac{3-x}{4(x+1)}$. The denominator factors: $x^{2}-2x-3=(x-3)(x+1)$. The expression becomes $\frac{3-x}{4(x+1)}\cdot\frac{1}{(x-3)(x+1)}=\frac{-(x-3)}{4(x+1)^{2}(x-3)}=\frac{-1}{4(x+1)^{2}}$. At $x=3$: $\frac{-1}{4\cdot 16}=-\frac{1}{64}$. Spotting $3-x=-(x-3)$ is the whole exercise.

$\frac{0}{0}$ with a root in the DENOMINATOR. The numerator factors: $2x^{2}-10x-12=2(x^{2}-5x-6)=2(x-6)(x+1)$. Multiply top and bottom by the conjugate $\sqrt{6}+\sqrt{x}$: the denominator becomes $\left(\sqrt{6}\right)^{2}-\left(\sqrt{x}\right)^{2}=6-x$. You get $\frac{2(x-6)(x+1)\left(\sqrt{6}+\sqrt{x}\right)}{6-x}=-2(x+1)\left(\sqrt{6}+\sqrt{x}\right)$ since $\frac{x-6}{6-x}=-1$. At $x=6$: $-2(7)(2\sqrt{6})=-28\sqrt{6}\approx -68.59$. The minus sign comes from the reversal $6-x=-(x-6)$: it is the most frequent loss in the whole exercise.

The limit at infinity of a quotient of polynomials of the SAME degree: the behaviour is dictated by the leading terms, and all the others become negligible. Limit $=\frac{3}{2}$. A clean write-up: divide top and bottom by $x^{4}$, which gives $\frac{3-5/x^{2}+7/x^{4}}{2-4/x+1/x^{3}-1/x^{4}}\longrightarrow\frac{3}{2}$. The quotient $\frac{3}{2}$ is the value of the horizontal asymptote $y=\frac{3}{2}$.

Form $\frac{0}{0}$. Combine the numerator: $\frac{1}{x}+\frac{1}{5}=\frac{5+x}{5x}$. The denominator factors: $x^{2}-25=(x-5)(x+5)$. The expression becomes $\frac{x+5}{5x}\cdot\frac{1}{(x-5)(x+5)}=\frac{1}{5x(x-5)}$. At $x=-5$: $\frac{1}{5(-5)(-10)}=\frac{1}{250}=0.004$. The factor $(x+5)$ was there all along, but hidden in a compound fraction: that is why you combine BEFORE trying to simplify.

Form $\frac{0}{0}$: on top $-729+729+6-6=0$, at the bottom $108-126+39-21=0$. The numerator factors by grouping, without any division: $-x^{6}+3x^{5}+2x-6=-x^{5}(x-3)+2(x-3)=(x-3)\left(2-x^{5}\right)$, which equals $2-243=-241$ at $x=3$. In the denominator, synthetic division by $(x-3)$ with the coefficients $4,-14,13,-21$: bring down $4$, then $12-14=-2$, then $-6+13=7$, then $21-21=0$. The quotient is $4x^{2}-2x+7$, which equals $36-6+7=37$. Limit $=-\frac{241}{37}\approx -6.51$. Looking for a grouping before launching into a division of degree 6 saves five minutes of exam time.

Form $\frac{0}{0}$ with a root on TOP. Conjugate: multiply by $\frac{2+\sqrt{x+2}}{2+\sqrt{x+2}}$, which gives $\frac{4-(x+2)}{(x-2)\left(2+\sqrt{x+2}\right)}=\frac{2-x}{(x-2)\left(2+\sqrt{x+2}\right)}=\frac{-1}{2+\sqrt{x+2}}$. At $x=2$: $\frac{-1}{2+2}=-\frac{1}{4}$. This is exactly what the conjugate is for: turning a difference of roots into a difference of squares, hence into a polynomial in which the common factor becomes visible.

Substitution: $\frac{8}{0}$, with $8\ne 0$. This is NOT an indeterminate form, so there is nothing to factor, but both sides must be studied. To the left of 2, $x-2<0$ and the numerator is about $8>0$: the quotient tends to $-\infty$. To the right, $x-2>0$: the quotient tends to $+\infty$. The two one-sided limits are different, so $\lim_{x\to 2}\frac{x^{2}+4}{x-2}$ DOES NOT EXIST. Writing $\infty$ on its own is wrong here: $\infty$ without a sign describes neither branch, and the line $x=2$ is still a vertical asymptote even though the limit does not exist.

A quotient of polynomials of DIFFERENT degrees: degree 5 on top, degree 3 at the bottom. Dividing top and bottom by $x^{3}$ gives $\frac{4x^{2}+3/x+9/x^{3}}{2-7/x+8/x^{3}}$, whose numerator blows up like $4x^{2}$ while the denominator tends to 2. The function behaves like $2x^{2}$, and $x^{2}\to+\infty$ even when $x\to-\infty$. Limit $=+\infty$. The classic pitfall is to conclude $-\infty$ because $x\to-\infty$: here it is the EVEN remaining degree, together with the positive ratio of the leading coefficients, that decides the sign, not the direction you are heading in.

The value first: $-2$ falls in the second piece, since that piece is valid for $-2\le x<0$. So $f(-2)=3(-2)-7=-13$, which exists. Left-hand limit: this is the first piece, $\lim_{x\to -2^{-}}\left(-x^{2}+5\right)=-4+5=1$. Right-hand limit: second piece, $\lim_{x\to -2^{+}}(3x-7)=-13$. The two one-sided limits are finite but DIFFERENT, so $\lim_{x\to -2}f(x)$ does not exist: the second condition fails. $f$ is discontinuous at $-2$. Watch the inequality trap: the $\le$ of the second piece alone decides the value $f(-2)$, and many students read the first piece out of habit.

Value: the third line imposes it, $f(0)=-7$. Left-hand limit: $\lim_{x\to 0^{-}}(3x-7)=-7$. Right-hand limit: the fourth piece must be simplified first, $\frac{x^{2}+14x}{-2x}=\frac{x(x+14)}{-2x}=\frac{x+14}{-2}$ for $x\ne 0$, which tends to $\frac{14}{-2}=-7$. Both one-sided limits equal $-7$, so $\lim_{x\to 0}f(x)=-7=f(0)$. All three conditions hold: $f$ is CONTINUOUS at $0$. A function defined in five pieces is not doomed to break at every join, and the expression $\frac{x^{2}+14x}{-2x}$, undefined at 0, does not prevent it: it is the line $x=0$ that gives the value.

Value: $1$ falls in the fifth piece. Substituting there gives $\frac{30-30}{4-4}=\frac{0}{0}$, which is not a number. So $f(1)$ DOES NOT EXIST, and the first condition already fails. The limit, however, does exist. On the left: $\frac{x+14}{-2}\to\frac{15}{-2}=-\frac{15}{2}$. On the right: $\frac{30-30x}{4x^{3}-4x^{2}}=\frac{30(1-x)}{4x^{2}(x-1)}=\frac{-30}{4x^{2}}\to\frac{-30}{4}=-\frac{15}{2}$. The two agree, so $\lim_{x\to 1}f(x)=-\frac{15}{2}$. $f$ is discontinuous at $1$, not because the limit is missing, but because the function is not defined there.

At $x=-2$: a JUMP discontinuity, since the two one-sided limits are finite and distinct. It is NOT removable: no single value can equal both 1 and $-13$, the curve really tears. At $x=1$: a HOLE, called a REMOVABLE discontinuity, because setting $f(1)=-\frac{15}{2}$ is enough to make all three conditions true. That is the fundamental difference between the two accidents: the hole is a defect of DEFINITION, the function was badly written at one point; the jump is a defect of BEHAVIOUR, the function itself tears, and no touch-up at a single point can mend a tear.

The denominator vanishes at $x=-7$, $x=\frac{1}{2}$ and $x=1$, so $\operatorname{dom}(f)=\mathbb{R}\setminus\left\{-7,\tfrac{1}{2},1\right\}$. The key point is that $3x-3=3(x-1)$: the factor $(x-1)$ appears ON TOP and AT THE BOTTOM. For $x\ne 1$ you can therefore simplify, $f(x)=\frac{x^{2}+4}{3(x+7)(2x-1)}$. This expression is valid on the whole domain, but it does not define $f$ at $x=1$: the domain is read from the original expression, never from the simplified one.

After simplification, only two factors remain in the denominator, hence two vertical asymptotes: $x=-7$ and $x=\frac{1}{2}$. The numerator $x^{2}+4$ is always strictly positive, so it never changes the sign. At $x=-7$: $(2x-1)$ is about $-15<0$ nearby, and $(x+7)$ changes sign. On the left, $(x+7)<0$, the denominator $3(x+7)(2x-1)$ is positive, so $\lim_{x\to -7^{-}}f(x)=+\infty$. On the right, $(x+7)>0$, the denominator is negative, so $\lim_{x\to -7^{+}}f(x)=-\infty$. At $x=\frac{1}{2}$: $(x+7)$ is about $7.5>0$, and $(2x-1)$ changes sign. On the left, $\lim_{x\to \frac{1}{2}^{-}}f(x)=-\infty$; on the right, $\lim_{x\to \frac{1}{2}^{+}}f(x)=+\infty$. A vertical asymptote is described by its two one-sided limits, not by its equation alone.