Exercises — One-sided limits — left-hand, right-hand

Tool 1 — Read the direction. $x\to a^{+}$ means $x$ slides toward $a$ staying bigger than $a$ (from the right). $x\to a^{-}$ means $x$ stays smaller (from the left). The little sign is a direction, never a "minus number". Tool 2 — Pick the right piece of the function. For a piecewise rule, the side of the arrow tells you which formula is active. Coming from the right at $a$? Use the rule that holds for $x>a$. Tool 3 — The handshake. The two-sided limit ${\lim }_{x\to a}f(x)$ exists only when both one-sided limits exist AND are equal. ${\lim }_{x\to a}f(x)=L⟺{\lim }_{x\to a^{-}}f(x)={\lim }_{x\to a^{+}}f(x)=L$

WHAT the arrow says: $5^{-}$ = approach $5$ from below (values less than $5$). $5^{+}$ = approach from above (values greater than $5$).

• $x\to 5^{-}$: e.g. $4.9,4.99,4.999$ — all creeping up toward $5$.
• $x\to 5^{+}$: e.g. $5.1,5.01,5.001$ — all creeping down toward $5$. None are negative. The minus sign describes which side of $5$, not the sign of $x$. All six numbers are positive.

Apply the handshake. Both sides exist and both equal $4$, so they shake hands: ${\lim }_{x\to 2}f(x)=4.$ Why $f(2)=9$ is ignored: a limit describes where $f$ is heading near $2$, never the single value sitting at $2$. The point $(2,9)$ is a lone dot floating above the curve; the approach still aims at height $4$.

• $x<a$ (from below) $\Rightarrow$ the minus sign: ${\lim }_{x\to a^{-}}$.
• $x>a$ (from above) $\Rightarrow$ the plus sign: ${\lim }_{x\to a^{+}}$. Mnemonic from the parent note: PLUS pushes from the upper side, MINUS from the lower.

Left ($x<1$): the active rule is $2x+1$. Substitute the target $1$: ${\lim }_{x\to 1^{-}}(2x+1)=2(1)+1=3.$ Right ($x>1$): the active rule is $5x-2$: ${\lim }_{x\to 1^{+}}(5x-2)=5(1)-2=3.$ Handshake: both equal $3$, so ${\lim }_{x\to 1}f(x)=3$. The value $f(1)=10$ plays no role.

Right ($x>0$): here $|x|=x$ (that is the definition of absolute value for positives), so $f(x)=\frac{x}{x}=1$. Thus ${\lim }_{x\to 0^{+}}f=1$. Left ($x<0$): here $|x|=-x$ (flipping the negative to positive), so $f(x)=\frac{-x}{x}=-1$. Thus ${\lim }_{x\to 0^{-}}f=-1$. Handshake fails: $1\ne -1$, so ${\lim }_{x\to 0}f(x)$ does not exist.

What to see in the figure below: the black curve sits flat at height $+1$ for all positive $x$ and flat at height $-1$ for all negative $x$. The two open circles at $x=0$ mark that neither height is reached at $0$; the red dotted segment between them is the jump of size $2$ — visual proof the two sides cannot shake hands.

Right ($x$ just above $4$, e.g. $4.001$): the greatest integer $\le 4.001$ is $4$. So ${\lim }_{x\to 4^{+}}\lfloor x\rfloor =4$. Left ($x$ just below $4$, e.g. $3.999$): the greatest integer $\le 3.999$ is $3$. So ${\lim }_{x\to 4^{-}}\lfloor x\rfloor =3$. Since $3\ne 4$, no two-sided limit at $x=4$. This mismatch happens at every integer.

What to see in the figure below: the graph is a staircase of flat black steps, each one unit wide. Every step ends in a solid dot on the left (value included) and an open circle on the right (value excluded). The red dotted segment at $x=4$ shows the step below it sitting at height $3$ while the next step jumps up to height $4$ — the left limit reads the lower step, the right limit reads the upper one.

Let $u=x-3$, so $h=\frac{1}{u}$ and $x\to 3$ becomes $u\to 0$. Right ($x>3\Rightarrow u>0$, tiny positive): a small positive denominator gives a huge positive output: ${\lim }_{x\to 3^{+}}h=+\infty$. Left ($x<3\Rightarrow u<0$, tiny negative): a small negative denominator gives a huge negative output: ${\lim }_{x\to 3^{-}}h=-\infty$. Verdict: the sides disagree and neither is finite, so the two-sided limit does not exist. There is a vertical asymptote at $x=3$.

What to see in the figure below: the red dashed vertical line marks the asymptote $x=3$. Follow the black curve: just to the right of the line it rockets upward toward $+\infty$; just to the left it plunges downward toward $-\infty$. The two arms escape in opposite directions — that opposition is exactly why the sides disagree.

Domain first: $\sqrt{x-2}$ needs $x-2\ge 0$, i.e. $x\ge 2$. There is no real domain for $x<2$. Left ($x<2$): impossible — no points to approach along. ${\lim }_{x\to 2^{-}}f$ does not exist (no domain). Right ($x>2$): allowed, and as $x\to 2^{+}$, $x-2\to 0^{+}$, so $\sqrt{x-2}\to 0$. Thus ${\lim }_{x\to 2^{+}}f=0$. Two-sided: fails, because one required side (left) has no domain. At a domain edge, only one side exists.

Naive substitution first (and why it fails): putting $x=3$ into $\frac{x^2-9}{x-3}$ gives $\frac{0}{0}$ — an indeterminate form, so substitution alone is useless here. We must reshape the expression. For $x\ne 3$ factor the numerator: $x^2-9=(x-3)(x+3)$, so $\frac{x^2-9}{x-3}=\frac{(x-3)(x+3)}{x-3}=x+3(x\ne 3).$ Why cancel is legal: near $x=3$ but not at $3$, the factor $x-3$ is nonzero, so dividing by it is fine. Right: ${\lim }_{x\to 3^{+}}(x+3)=6$. Left: ${\lim }_{x\to 3^{-}}(x+3)=6$. Handshake: both $=6$, so ${\lim }_{x\to 3}g(x)=6$. The stray value $g(3)=5$ is irrelevant — this is a removable discontinuity.

Left ($x\le 2$, use $x^2+k$): ${\lim }_{x\to 2^{-}}(x^2+k)=4+k$. Right ($x>2$, use $4x-3$): ${\lim }_{x\to 2^{+}}(4x-3)=8-3=5$. Force the handshake: set left $=$ right: $4+k=5⟹k=1.$ With $k=1$ both sides equal $5$, so ${\lim }_{x\to 2}f(x)=5$ exists.

Right ($x>1$, use $x^2+2$): ${\lim }_{x\to 1^{+}}(x^2+2)=1+2=3$. Left ($x<1$, use $ax+b$): ${\lim }_{x\to 1^{-}}(ax+b)=a+b$. Condition 1 — handshake: the two sides must match the target $3$: $a+b=3.$ That is one equation with two unknowns, so infinitely many $(a,b)$ work — e.g. $a=1,b=2$ or $a=0,b=3$. Why continuity is already satisfied: the right limit is $3=f(1)$, so once $a+b=3$ the full limit is $3=f(1)$. Continuity needs limit $=$ value, and both are $3$. Answer: any $(a,b)$ with $a+b=3$; a clean choice is $a=1,b=2$.

We need different constants on each side and a lone value at $0$: $f(x)=\{\begin{array}{ll}-3& x<0\\ 100& x=0\\ +3& x>0\end{array}$ Check: left branch is constant $-3\Rightarrow {\lim }_{x\to 0^{-}}=-3$; right branch constant $+3\Rightarrow {\lim }_{x\to 0^{+}}=+3$; and $f(0)=100$ by construction. The sides disagree ($-3\ne 3$), so this is a genuine jump and the two-sided limit does not exist — exactly as designed.

Setup: for any real $t$, the floor obeys $t-1<\lfloor t\rfloor \le t$. Put $t=\frac{1}{x}$: $\frac{1}{x}-1<\lfloor \frac{1}{x}\rfloor \le \frac{1}{x}.$ Multiply by $x$. Since $x\to 0^{+}$ we have $x>0$, so the inequalities keep their direction: $x\left(\frac{1}{x}-1\right)<x\lfloor \frac{1}{x}\rfloor \le x\cdot \frac{1}{x},$ $1-x<x\lfloor \frac{1}{x}\rfloor \le 1.$ Squeeze: as $x\to 0^{+}$, the left bound $1-x\to 1$ and the right bound is constant $1$. Both squeeze the middle to $1$: ${\lim }_{x\to 0^{+}}x\lfloor \frac{1}{x}\rfloor =1.$ Why one-sided matters: we needed $x>0$ to multiply through an inequality without flipping it; on the left the sign flip changes the bounds.

Case A: $r(x)=\frac{x+1}{(x-2{)}^2}$. Near $x=2$ the numerator $x+1\to 3>0$. The denominator $(x-2{)}^2$ is a square, always $>0$, shrinking to $0$. So a positive-over-tiny-positive gives $+\infty$ from both sides: ${\lim }_{x\to 2^{-}}r=+\infty ,{\lim }_{x\to 2^{+}}r=+\infty .$ Both sides agree in behaviour ($+\infty$), yet the two-sided limit still does not exist (infinite is not a number) — though we say it diverges to $+\infty$. Case B: $\frac{x+1}{x-2}$. Now the denominator changes sign. For $x\to 2^{+}$, $x-2\to 0^{+}$: $\frac{3}{0^{+}}=+\infty$. For $x\to 2^{-}$, $x-2\to 0^{-}$: $\frac{3}{0^{-}}=-\infty$. Opposite infinities. Lesson: an even power in the denominator keeps the sign the same on both sides; an odd power flips it. See Vertical asymptotes.

Left ($x<0$, use $\frac{\sin x}{x}$): this expression tends to $1$ regardless of side, so ${\lim }_{x\to 0^{-}}f=1$. Right ($x>0$, use $\frac{1-\cos x}{x}$): this standard limit is $0$, so ${\lim }_{x\to 0^{+}}f=0$. Handshake fails: $1\ne 0$, so the two-sided limit does not exist — and therefore no value of $c$ can make $f$ continuous at $0$. Continuity needs a two-sided limit to exist first; here it never does. This is a jump discontinuity with a floating point at height $c$.