02Continuity, asymptotes, and limits at infinity

both sides finite, and the function is visibly broken at $0$

both sides finite and equal, and still no continuity, because the third condition has no value to test

two independent ways to satisfy the hypothesis and fail the conclusion

the cancellation is legal exactly off $x=2$, so the restriction travels with the formula instead of being dropped once the page turns

this is true whatever real number $f(2)$ is, so the identity does not name the value; any claim about $f(2)$ has to come from somewhere else in the question

an explicit witness rather than a doubt: it satisfies every instance of the identity and lands nowhere near $4$

one counterexample settles an implication, and no amount of algebra can rescue it

continuity is the only thing that converts a limit into a value, so with that word added the claim becomes correct and the number $4$ is earned

$\sin$ stays inside $[-1,1]$ however large $x$ gets, so dividing by $x$ traps the quotient

both bounds tend to $0$, so the Squeeze Theorem applies and $y=0$ is an asymptote

the quotient vanishes exactly where the numerator does, and there are infinitely many such points to the right

one function both settles on the line and meets it endlessly

the interval of admissible values is empty, so the theorem produces no $c$ at all

continuous, matches both given values, has no zero and takes no value other than $4$

any conclusion that this example contradicts cannot follow from the hypotheses

the common factor is the whole difference between the two candidates

the reduced formula is continuous at $2$, so substitution gives the limit

finite limit, no value: the first branch of the classification

nonzero over zero, so the values explode and only the sign is left to decide

left of $-2$ the denominator is a small negative, so a negative over a negative is a large positive; on the right it flips

the other candidate cancelled, so it never had a chance

the three part test fixes $k$ before any computation, so the problem is really a limit problem

the conjugate is chosen so the numerator becomes a difference of squares, which has no root left in it

the factor $x-2$ that blocked substitution is now visible on both floors

the reduced formula is continuous at $2$, so substitution is finally legal

the same factor $x-3$ sits on both floors, which is what makes the $\tfrac00$ removable

substitution is legal now: the denominator is $3$, not $0$

cosine is continuous on all of $\mathbb{R}$, so the check costs one line and never fails

the limit moves inside the cosine because the outer function is continuous at the inner limit

dividing by $x^{2}$, the dominant power on both floors

the leading coefficients are what survive; the constants play no part

on the far left $x$ is negative, so $\vert x\vert$ resolves to $-x$; this is the only place a sign enters

dividing by $x$, not by $x^{2}$: both floors are of size $x$

the small terms die and the minus stays

the conjugate is what makes the two giants cancel on paper instead of in your head

on the far right the bars cost nothing, since $\vert x\vert=x$ there

the root contributes a $2$ of its own, which is why the answer is not $\tfrac52$

only the sign matters: the theorem compares $N=0$ with the two endpoint values, and size is irrelevant

$f$ is continuous on the subinterval because it is continuous on all of $[0,5]$

the same argument twice more

three flips, three guarantees, and no guarantee anywhere else

the two candidates for vertical asymptotes

the numerator vanishes nowhere at all, so nothing cancels and both candidates survive

near $3$ the factor $x+2$ is about $5>0$, so the sign of the quotient follows $x-3$

near $-2$ the factor $x-3$ is about $-5<0$, which flips both directions

equal degrees, so the leading coefficients decide, and the same computation serves both ends

three lines, each earned by a limit

if this is not $0$, the quotient is a nonzero number over something shrinking to zero, and no finite limit is possible

one condition, one unknown

with $a=1$ the numerator carries the factor $x-2$, which is what makes the break removable

continuity means the value equals the limit, and the limit is now a substitution

the signature of an infinite break

and no choice of $b$ can repair it, since no finite value competes with an explosion

$\tfrac00$ with roots in it: the conjugate is the standard repair

the roots disappear from the numerator, which is the whole purpose of the move

the factor $x$ that blocked substitution is gone

the reduced formula is continuous at $0$, so this last step is legal

dividing an inequality by a positive number keeps its direction; this is why the restriction $x>0$ was stated

a constant over a growing power dies at both ends

the Squeeze Theorem, with the two bounds agreeing on $0$

that is precisely what a limit of $0$ at $\infty$ means

a fraction is zero exactly when its numerator is, and the denominator never vanishes for $x>0$

there are infinitely many positive multiples of $\pi$

left of the junction the function is a polynomial, so substitution is legal

just right of $1$ the denominator is a small positive number, so the quotient grows past every bound

the third branch of the classification needs only one runaway side

the definition asks for at least one infinite one sided limit, not for both

the piece with $x\le1$ contains the junction, so it supplies the value

the left hand test passes even though the two sided one cannot

the radicand must be strictly positive: where it is zero the denominator vanishes too, so no point of $[0,4]$ can be in the domain

two finite boundary points and two infinite ends, so at most two vertical asymptotes and exactly two end behaviours to describe

the numerator tends to $10\neq0$ while the denominator tends to $0$ through positive values; approaching $4$ from the left is impossible, so this is the only limit there is

same argument with a negative numerator; the square root is positive wherever it is defined, so the sign of $f$ is the sign of $x^{2}-6$

$\sqrt{x^{2}}=|x|$ and never $x$; this single line is where the two ends part company

for $x>0$ the bars give $|x|=x$, so the denominator is $x^{2}\sqrt{1-4/x}$ and dividing top and bottom by $x^{2}$ finishes it

for $x<0$ the bars give $|x|=-x$; the sign is the entire content of this step, and it is the step the degree rule cannot see

$b$ is by definition $\lim\left(f(x)-mx\right)$; with $m=1$ this is an $\infty-\infty$ difference and has to be repaired before any limit is read off

multiplying by the conjugate turns the difference into a quotient; the two quartic terms cancel and a cubic survives on top

divide top and bottom by $x^{3}$, using $\sqrt{x^{2}-4x}/x\to1$ at the right end

the same conjugate but with $m=-1$; now $\sqrt{x^{2}-4x}/x\to-1$, which flips the sign of the whole denominator and hence of $b$

the instruction to prove the claims means the two limits per end are the answer, not the lines

the complete map: two one sided vertical asymptotes and two different slants, all of them supported

a solution of the equation is precisely a zero of $h$, and nothing has been added or lost by moving the $2$

a polynomial is continuous everywhere, a quotient of polynomials is continuous wherever the denominator is nonzero, and a difference of continuous functions is continuous; since $x=1$ is outside the domain, no interval used below may straddle it

cubing is strictly increasing on all of $\mathbb{R}$

on a single branch $x-1$ keeps one sign, so $1/(x-1)$ is strictly decreasing there and its negative is strictly increasing; this reasoning collapses across $x=1$, where the term swings from $+\infty$ to $-\infty$

a sum of strictly increasing functions is strictly increasing, and a strictly increasing function takes the value $0$ at most once, so each branch holds at most one solution

both test points lie in $(-\infty,1)$, so $h$ is continuous on the closed interval $\left[0,\tfrac12\right]$ joining them

Intermediate Value Theorem with $N=0$, which lies strictly between $h(0)$ and $h(\tfrac12)$

both test points lie in $(1,\infty)$, so $\left[\tfrac32,2\right]$ again avoids the break

Intermediate Value Theorem a second time, on an interval where the hypothesis of continuity is genuinely satisfied

at least one and at most one solution on each branch, and every admissible $x$ lies on one of the two branches, so no third solution can exist anywhere

both terms grow like $x$; subtracting and adding $x$ leaves two differences that each settle on a finite number, and one conjugate could never serve a cube root and a square root at the same time

the cube root conjugate $a^{3}-b^{3}=(a-b)(a^{2}+ab+b^{2})$; the cubes cancel and only $x$ survives on top

$A/x=\sqrt[3]{1+1/x^{2}}\to1$, so the denominator behaves like $3x^{2}$ and beats the numerator

the ordinary conjugate; divide top and bottom by $x$ and use $B/x=\sqrt{1+1/x}\to1$, which is legitimate because $x>0$ at this end

the limit of a difference is the difference of the limits, now that both of them exist

$\sin(\pi n)=0$ at every integer, so the oscillating term is switched off along this sequence

$\sin\!\left(2\pi n+\tfrac{\pi}{2}\right)=1$, so the oscillating term is at full strength along this sequence

if the limit existed, every sequence tending to infinity would have to reproduce it; two sequences with two values is the proof of failure, and it is the part of the answer that carries the marks

$|\sin(\pi x)|\le1$ gives $2x\le x\sin(\pi x)+3x\le4x$, so $Q$ is trapped between $2x/(x+2)$ and $4x/(x+2)$, and its values accumulate on the whole of $[2,4]$: bounded, and still without a limit, hence without a horizontal asymptote

the Sandwich Theorem concludes only when the two bounds close on the same number; here they close on a band, so the argument delivers boundedness and nothing more

the usual squeeze slogan is bounded times something tending to zero; here the bounded factor is multiplied by $x$, which runs away, so the slogan does not apply

the graphs meet at $c$ exactly when $f(c)=1/c$, that is when $g(c)=0$; the theorem talks about one function reaching a level, not about two curves crossing

$f$ is continuous by hypothesis, $1/x$ is continuous at every $x\neq0$ and $[1,3]$ avoids $0$, and a difference of continuous functions is continuous; this sentence is the hypothesis the theorem will be applied with, so it has to be written

expand and clear the fraction; the aim is to make the given inequality appear on the page unchanged

the hypothesis says precisely that $g(1)$ and $g(3)$ are not both positive and not both negative; the factor $3$ is positive and does not affect the sign

$N=0$ lies strictly between $g(1)$ and $g(3)$ and $g$ is continuous on $[1,3]$: this is the Intermediate Value Theorem, applied on an interval with no break in it

the hypothesis allows equality, and then one endpoint is already a meeting point; this is why the conclusion is claimed on the closed interval $[1,3]$ and not on $(1,3)$

both cases end at the same conclusion, so the proof is complete without extra assumptions on $f$

a constant function is continuous everywhere, so nothing at all is wrong with $f$; the defect will have to be somewhere else

the sign information the theorem asks for is present, so sign alone cannot be what is missing

$1/x$ is never $0$, so the graphs never meet; the only hypothesis left to blame is continuity, and indeed $g$ is undefined at $x=0$, so it is not continuous on $[-1,1]$

these are the only points where the identity cannot be divided; everywhere else it is an ordinary formula and there is nothing to discuss

the restriction travels with the formula, because the division that produced it is only legal off $\pm1$

every real number satisfies this, so the identity cannot be the thing that determines $h(\pm1)$; something else in the question has to

shifting to $t\to0$ turns the numerator into a small angle sine, which is the only shape whose limit we are allowed to quote

$\sin(\pi t)/(\pi t)\to1$ supplies the factor $\pi$, and $t+2\to2$ is plain substitution in a continuous factor

the equality between value and limit is the continuity hypothesis and nothing else; at $-1$ the shift $t=x+1$ gives denominator $t(t-2)$, whose sign flips the answer

the identity is a separate statement for each $x$, and at $x=1$ it only ever demanded $0\cdot H(1)=0$

so $H$ satisfies every instance of the identity and is still not continuous at $1$: continuity is an extra hypothesis, never a consequence of the identity

the half integers are where $\sin(\pi x)$ is $\pm1$ or $0$, so they are the only endpoints whose value can be written without a calculator

the second exact point; it sits below the level $1$ while the first sits above it, which is the sign change the theorem needs

$\left[-\tfrac12,0\right]$ avoids $\pm1$, so $h$ is continuous there for the cheapest possible reason, and the level $1$ lies strictly between the two endpoint values