Exercises — Intuitive concept of a limit — table of values, graphical

WHAT: Read the left column (as $x$ rises to $2$) and the right column (as $x$ falls to $2$). WHY: The definition says both sides must aim at the same number. Left outputs: $4.9\to 4.99\to 4.999$ — climbing toward $5$. Right outputs: $5.001\to 5.01\to 5.1$... wait, read them toward $2$: $5.001\to 5.01\to 5.1$ is moving away; toward $2$ they are $5.001,5.0\dots$ i.e. shrinking toward $5$. Both sides squeeze on $5$. ${\lim }_{x\to 2}f(x)=5$

WHAT: Compare the two one-sided limits, ignore $f(3)$. WHY: The limit watches the approach, never the point itself. Both one-sided values equal $7$, so they agree. The stray value $f(3)=100$ is a lone dot that the trend never touches. ${\lim }_{x\to 3}f(x)=7$

WHAT: Recall the existence condition. WHY: $f(a)$ is irrelevant (holes and stray dots have limits). The two one-sided limits must each exist and be equal. Correct: (b), (c), (d). Not (a).

WHAT: Plug in values near $1$ (never $1$ itself).

Both sides aim at $2$. WHY confirm algebraically: a table only suggests. Factor to be sure — legal because $x\to 1$ means $x\ne 1$, so $x-1\ne 0$: $\frac{x^2-1}{x-1}=\frac{(x-1)(x+1)}{x-1}=x+1(x\ne 1).$ As $x\to 1$, $x+1\to 2$. ${\lim }_{x\to 1}\frac{x^2-1}{x-1}=2.$

WHAT: Try plugging in: $\frac{16-16}{0}=\frac{0}{0}$ — a $0/0$ trap, so substitution fails. WHY factor: the numerator is a difference of squares $x^2-16=(x-4)(x+4)$; the $(x-4)$ cancels (valid since $x\ne 4$): $\frac{(x-4)(x+4)}{x-4}=x+4(x\ne 4).$ As $x\to 4$: $x+4\to 8$. ${\lim }_{x\to 4}\frac{x^2-16}{x-4}=8.$

WHAT: Substitution gives $\frac{3-3}{0}=\frac{0}{0}$. Table:

Both sides squeeze toward $\approx 0.1667$. WHY the algebra: multiply by the conjugate $\frac{\sqrt{x+9}+3}{\sqrt{x+9}+3}$ to clear the root: $\frac{(\sqrt{x+9}-3)(\sqrt{x+9}+3)}{x(\sqrt{x+9}+3)}=\frac{(x+9)-9}{x(\sqrt{x+9}+3)}=\frac{x}{x(\sqrt{x+9}+3)}=\frac{1}{\sqrt{x+9}+3}.$ As $x\to 0$: $\frac{1}{\sqrt{9}+3}=\frac{1}{6}$. ${\lim }_{x\to 0}\frac{\sqrt{x+9}-3}{x}=\frac{1}{6}\approx \mathrm{0.1667.}$

WHAT / WHAT IT LOOKS LIKE: In the figure, the left piece (lavender) is the line $y=x+1$; as $x\to 2^{-}$ its height climbs to $2+1=3$. The right piece (coral) is $y=2x-1$; as $x\to 2^{+}$ its height falls to $2(2)-1=3$. The lone butter dot at $(2,5)$ is $g(2)$.

• ${\lim }_{x\to 2^{-}}g=3$
• ${\lim }_{x\to 2^{+}}g=3$ WHY the two-sided limit exists: both sides agree at $3$, so ${\lim }_{x\to 2}g=3.$ But $g(2)=5$ — the stray dot. The limit ($3$) and the value ($5$) differ. This is exactly a removable-discontinuity picture; see Continuity at a point.

WHAT: For $x>0$, $|x|=x$ so $h(x)=\frac{x}{x}=1$. For $x<0$, $|x|=-x$ so $h(x)=\frac{-x}{x}=-1$. WHAT IT LOOKS LIKE: two flat rays — the mint ray at height $+1$ on the right, the coral ray at height $-1$ on the left, with a jump across $x=0$.

• ${\lim }_{x\to 0^{-}}h=-1$
• ${\lim }_{x\to 0^{+}}h=+1$ WHY no limit: left $(-1)\ne$ right $(+1)$. The two paths aim at different heights, so ${\lim }_{x\to 0}h\text{ does not exist.}$

WHAT: Feed values close to $3$. From the right ($x=3.1,3.01,3.001$): $x-3$ is a tiny positive number, so $\frac{1}{x-3}$ is huge and positive — it blows up to $+\infty$. From the left ($x=2.9,2.99,2.999$): $x-3$ is a tiny negative number, so $\frac{1}{x-3}$ is huge and negative — it dives to $-\infty$. WHY no finite limit: the outputs don't settle on any number; they run off to infinity, and to opposite infinities on the two sides. ${\lim }_{x\to 3}p\text{ does not exist (finite); }{\lim }_{x\to 3^{+}}p=+\infty ,{\lim }_{x\to 3^{-}}p=-\infty .$ This "blow-up" behaviour is the topic of One-sided limits and limits at infinity.

WHAT: Compute both one-sided limits. Left: $x^2+1\to 1^2+1=2$. Right: $3x-1\to 3(1)-1=2$. WHY the limit exists regardless of $k$: both sides already agree at $2$, and the limit ignores $q(1)=k$ entirely. So: ${\lim }_{x\to 1}q=2\text{for **any** }k.$ Choosing $k=2$ additionally makes $q$ continuous; but the limit is $2$ no matter what $k$ is. (Contrast: continuity would demand $k=2$; the limit does not — see Continuity at a point.)

WHAT: Substitution gives $\frac{\sin 0}{0}=\frac{0}{0}$ — indeterminate. There is no algebraic factor to cancel ($\sin x$ is not a polynomial with an $x$-factor you can lift out cleanly), so we table it:

WHY both sides matter: left and right both climb toward $1$ as $x\to 0$. ${\lim }_{x\to 0}\frac{\sin x}{x}=1.$ WHY substitution/cancelling fail: substitution hits $0/0$; and $\sin x$ can't be factored like $x^2-1$, so we cannot cancel our way out. This is precisely the case the Squeeze (Sandwich) Theorem was built to nail rigorously, and it is the seed of The derivative as a limit of difference quotients.

WHAT / WHAT IT LOOKS LIKE: In the figure, the coral line is a secant — it cuts the parabola at $(3,9)$ and $(3+h,(3+h{)}^2)$. As $h$ shrinks, the second point slides toward the first and the secant pivots toward the lavender tangent. The difference quotient is the secant's slope. Substitution gives $\frac{0}{0}$, so table it:

Both sides aim at $6$. WHY / algebra to confirm: expand (valid because $h\ne 0$ in the limit, so we may cancel $h$): $\frac{(3+h{)}^2-9}{h}=\frac{9+6h+h^2-9}{h}=\frac{6h+h^2}{h}=6+h(h\ne 0).$ As $h\to 0$: $6+h\to 6$. ${\lim }_{h\to 0}\frac{f(3+h)-f(3)}{h}=6.$ MEANING: $6$ is the exact steepness (slope) of $y=x^2$ at the single point $x=3$ — the slope of the lavender tangent line. This is the derivative; see The derivative as a limit of difference quotients.

WHAT: As $x\to 0$, the inside $\frac{1}{x}$ races off to $\pm \infty$, so $\cos (1/x)$ swings through its entire range $[-1,1]$ over and over — faster and faster.

WHY the claim fails: the outputs never settle — they alternate between $-1$ and $1$ no matter how close $x$ gets to $0$. There is no single approached value. ${\lim }_{x\to 0}\cos \left(\frac{1}{x}\right)\text{ does not exist.}$ (Contrast: ${\lim }_{x\to 0}x\cos (1/x)=0$, because the $x$ factor crushes the wobble — a job for the Squeeze (Sandwich) Theorem.)

(1) Jump — left and right one-sided limits differ (Ex 3.2). (2) Blow-up — outputs run to $\pm \infty$ (Ex 3.3). (3) Oscillation — outputs never settle on one value (Ex 5.2).

$6$ a slope and not just a number? Because the difference quotient is the slope of the secant line through two points on $y=x^2$; as $h\to 0$ the secant becomes the tangent at $x=3$, whose slope is $6$.

$0/0$ NOT mean "no limit" in Ex 2.1–2.3 but a blow-up DID mean no finite limit in Ex 3.3? $0/0$ is indeterminate — a shared vanishing factor can cancel to reveal a finite value. In Ex 3.3 the denominator alone went to $0$ (numerator was $1$), so the output genuinely exploded.