Exercises — Continuity — definition, types of discontinuity (removable, jump, infinite)

Both one-sided limits are equal and finite ($=3$), so the two-sided limit exists. The only thing wrong is the missing value. Condition 2 ✓ (limit exists), condition 1 ✗ (value undefined) → removable. Patch: define $f(2)=3$.

At least one side is infinite (here both), so the limit does not exist as a finite number. → Infinite discontinuity (a vertical asymptote at $x=5$). See Vertical asymptotes and rational functions.

Both one-sided limits are finite but unequal ($1\ne 4$) → jump discontinuity. Jump size $=|L^{+}-L^{-}|=|4-1|=3$. (The value of $f(0)$ is irrelevant to the type here — no single value can close a gap between two different heights.)

Step 1: $f(3)=\frac{9-9}{0}=\frac{0}{0}$ — undefined. Step 2: Factor the numerator: $x^2-9=(x-3)(x+3)$. For $x\ne 3$, $\frac{(x-3)(x+3)}{x-3}=x+3.$ Why factor? The limit ignores the point $x=3$ itself; cancelling the common factor reveals the height the graph approaches. So ${\lim }_{x\to 3}f(x)=3+3=6$. Verdict: limit exists ($=6$), value undefined → removable. Patch: $f(3)=6$.

Step 1: $f(4)=\frac{6}{0}$ — undefined (numerator $\ne 0$, denominator $=0$: no cancellation possible). Step 2: Just below $4$, $x-4$ is a tiny negative, numerator $\approx 6>0$, so $\frac{6}{{\text{tiny}}^{-}}\to -\infty$. Just above, $x-4$ is a tiny positive, so $\to +\infty$. $L^{-}=-\infty ,L^{+}=+\infty .$ Verdict: infinite discontinuity (vertical asymptote at $x=4$).

Step 1: $x=1$ satisfies $x\ge 1$, so use the second piece: $f(1)=1^2+3=4$. (Defined.) Step 2: $L^{-}={\lim }_{x\to 1^{-}}(2x+1)=2(1)+1=3,L^{+}={\lim }_{x\to 1^{+}}(x^2+3)=4.$ Verdict: both finite, $3\ne 4$ → jump, size $|4-3|=1$.

Step 1: $g(0)=2$ (defined). Step 2: Use the standard limit $\underset{u\to 0}{\lim }\frac{\sin u}{u}=1$. Why this tool? Direct substitution gives $\frac{0}{0}$; the standard limit is the machine built exactly to resolve $\frac{\sin (\cdot )}{\cdot }$. Rewrite so the argument matches the denominator: $\frac{\sin 3x}{x}=3\cdot \frac{\sin 3x}{3x}\underset{x\to 0}{\to }3\cdot 1=3.$ Verdict: limit $=3$ exists and is finite, but $g(0)=2\ne 3$ → removable (redefine $g(0)=3$). Why removable, not jump? Left and right both approach the same value $3$; only the plotted dot is misplaced.

The denominator factors: $x^2-4=(x-2)(x+2)$, zero at $x=2$ and $x=-2$. Both are suspect points. $f(x)=\frac{x-2}{(x-2)(x+2)}=\frac{1}{x+2}(x\ne 2).$ At $x=2$: the $(x-2)$ cancels, so ${\lim }_{x\to 2}f(x)=\frac{1}{2+2}=\frac{1}{4}$ (finite), but $f(2)=\frac{0}{0}$ undefined → removable, patch $f(2)=\frac{1}{4}$. At $x=-2$: the factor $(x+2)$ does not cancel. ${\lim }_{x\to -2^{-}}\frac{1}{x+2}=-\infty$, ${\lim }_{x\to -2^{+}}\frac{1}{x+2}=+\infty$ → infinite (asymptote at $x=-2$). Summary: removable at $x=2$, infinite at $x=-2$.

Continuity at $x=2$ means $L^{-}=L^{+}=f(2)$.

• $L^{-}={\lim }_{x\to 2^{-}}(3x-1)=5$.
• $f(2)=L^{+}=k(2)+4=2k+4$ (the second piece also gives the value since it's $x\ge 2$). Set them equal: $2k+4=5\Rightarrow 2k=1\Rightarrow \boxed{k=\frac{1}{2}}$. Check: with $k=\frac{1}{2}$, $L^{+}=f(2)=5=L^{-}$. All three conditions hold — pen never lifts.

Only the two seams $x=1$ and $x=3$ can break; each middle/edge piece is a line (continuous on its own). At $x=1$: left value $=1+1=2$; right approach $=a(1)+b=a+b$. Continuity ⇒ a + b = 2. \tag{i} At $x=3$: left approach $=a(3)+b=3a+b$; right value $=2(3)-3=3$. Continuity ⇒ 3a + b = 3. \tag{ii} Subtract (i) from (ii): $2a=1\Rightarrow a=\frac{1}{2}$. Then $b=2-\frac{1}{2}=\frac{3}{2}$. $\boxed{a=\frac{1}{2},b=\frac{3}{2}.}$ Check: at $x=1$, $\frac{1}{2}+\frac{3}{2}=2$ ✓; at $x=3$, $\frac{3}{2}+\frac{3}{2}=3$ ✓.

The absolute value splits by sign of $x-3$:

• For $x>3$: $|x-3|=x-3$, so $f(x)=\frac{x-3}{x-3}=+1$.
• For $x<3$: $|x-3|=-(x-3)$, so $f(x)=\frac{-(x-3)}{x-3}=-1$. Hence $L^{+}=+1$, $L^{-}=-1$, both finite but unequal. And $f(3)=\frac{0}{0}$ is undefined. Verdict: jump, size $|1-(-1)|=2$. Why not removable, despite the $\frac{0}{0}$? A $\frac{0}{0}$ form does not guarantee the limit exists — here the two sides settle on different constants, so no single patch works.

$p$ is a polynomial, hence continuous on the closed interval $[0,1]$ — including the two endpoints, each of which we check with the relevant one-sided limit (no holes, jumps, or poles anywhere, so this is automatic). Evaluate the endpoints: $p(0)=0-0+1=1>0,p(1)=1-4+1=-2<0.$ The sign changes: $p(0)>0$ and $p(1)<0$. By the Intermediate Value Theorem, a continuous function taking values $+1$ and $-2$ must hit every value between — including $0$ — somewhere in $(0,1)$. Therefore $p$ has at least one root there. $■$ The property relied on: continuity on a closed interval — the pen crosses the $x$-axis without lifting, so it cannot skip zero.

Step 1: $x=1$ satisfies $x\le 1$, so $f(1)=1^2=1$ (defined). Step 2:

• $L^{-}={\lim }_{x\to 1^{-}}{x}^2=1$.
• $L^{+}$: for $x>1$, $\frac{x^2-1}{x-1}=\frac{(x-1)(x+1)}{x-1}=x+1\to 2$. So $L^{-}=1$, $L^{+}=2$ — both finite, unequal. Verdict: jump, size $|2-1|=1$. Subtlety: the right piece looks removable on its own (a $\frac{0}{0}$ hole), but glued to the left piece the two sides disagree → the combined function has a jump at $x=1$, and $f(1)=1$ cannot fix a whole gap. Recall Differentiability implies continuity: since $f$ isn't even continuous here, it certainly can't be differentiable at $x=1$.

How do you read the type straight from a formula? ::: Zero denominator with a cancelling numerator → removable; non-cancelling → infinite; different-height pieces → jump. What is the first question to ask at any $\frac{0}{0}$? ::: Do the two one-sided limits agree and stay finite? Only then removable. What does continuity for an unknown constant require? ::: $L^{-}=L^{+}=f(a)$ at every seam — one equation per boundary. What does the IVT need to guarantee a root? ::: Continuity on a closed interval, plus a sign change at the endpoints.