Introduction to Limits

(a) For $x<2$ the rule is $x+1$: $\lim_{x\to2^-}f(x)=2+1=3$.

(b) For $x>2$ the rule is $5-x$: $\lim_{x\to2^+}f(x)=5-2=3$.

(c) Both one-sided limits exist and equal $3$, so $\lim_{x\to2}f(x)=3$.

(d) The middle line applies only at $x=2$: $f(2)=5$.

The limit is computed from $x\neq2$ and never sees $f(2)$, so $3\neq5$ is perfectly legal. Redefining $f(2)=3$ would make $f$ continuous there — that is why the discontinuity is called removable.

Left piece: through $(-3,-1)$ and $(1,3)$, slope $1$, so $h(x)=x+2$ for $-3\le x<1$. Right piece: through $(1,1)$ and $(4,4)$, so $h(x)=x$ for $1<x\le4$.

$\lim_{x\to1^-}h(x)=1+2=3$ — the circle at $(1,3)$ being open is irrelevant. $\lim_{x\to1^+}h(x)=1$. The one-sided limits differ, so $\lim_{x\to1}h(x)$ does not exist. $h(1)=0$ from the solid dot: left limit, right limit and value are three independent numbers.

$\lim_{x\to4^-}h(x)=4$. A two-sided limit needs $h$ defined on both sides of $4$; there is no graph for $x>4$, so only the left-hand limit is posed.

(a) $\lim_{x\to1^-}f(x)=c+3$ and $\lim_{x\to1^+}f(x)=1-c$. The two-sided limit exists exactly when $c+3=1-c\quad\Longrightarrow\quad c=-1,$ and then both sides equal $2$.

(b) The second rule covers $x=1$: $f(1)=1-(-1)(1)=2$. Limit equals value, so $f$ is continuous at $1$ when $c=-1$.

(c) With $c=0$: $f(x)=3$ for $x<1$ and $f(x)=x^2$ for $x\ge1$, so the one-sided limits are $3$ and $1$. Both exist but disagree — a jump, not a blow-up or an oscillation.

(a) As $x\to2^-$, $x-2$ is negative and tiny, so $F(x)\to-\infty$; as $x\to2^+$ it is positive and tiny, so $F(x)\to+\infty$. Neither one-sided limit is a real number, and they run to opposite infinities, so $\lim_{x\to2}F(x)$ does not exist; $x=2$ is a vertical asymptote.

(b) Let $M>0$ and put $\delta=\dfrac{1}{\sqrt{M}}$. If $0<|x-2|<\delta$ then $0<(x-2)^2<\delta^2=\frac{1}{M}\quad\Longrightarrow\quad G(x)=\frac{1}{(x-2)^2}>M.$ Since $M>0$ was arbitrary, $\lim_{x\to2}G(x)=\infty$.

(c) No. Existence requires a real $L$; the symbol $\infty$ records that $G$ exceeds every bound near $2$.

$|f(x)-5|=|3x-3|=3|x-1|$.

(a) $3|x-1|<0.06\iff|x-1|<0.02$, so the largest $\delta$ is $0.02$ (any smaller positive $\delta$ also works). (b) $|x-1|<0.001$: $\delta=0.001$.

(c) Let $\varepsilon>0$ be given. Put $\delta=\varepsilon/3>0$. Suppose $0<|x-1|<\delta$. Then $|(3x+2)-5|=3|x-1|<3\delta=\varepsilon.$ Since $\varepsilon>0$ was arbitrary, $\lim_{x\to1}(3x+2)=5$. Check: $\varepsilon=0.06$ gives $\delta=0.02$, matching (a).

Scratch: $|x^2-4|=|x-2|\,|x+2|$. If $|x-2|<1$ then $1<x<3$, so $|x+2|<5$ and $|x^2-4|<5|x-2|$.

Proof. Let $\varepsilon>0$. Put $\delta=\min\{1,\varepsilon/5\}>0$ and suppose $0<|x-2|<\delta$. Since $\delta\le1$, $|x+2|=|(x-2)+4|\le|x-2|+4<5$. Since $\delta\le\varepsilon/5$, $|x-2|<\varepsilon/5$. Hence $|x^2-4|=|x-2|\,|x+2|<\frac{\varepsilon}{5}\cdot5=\varepsilon.$ Since $\varepsilon>0$ was arbitrary, $\lim_{x\to2}x^2=4$.

The restriction is used only for the bound $|x+2|<5$. Without it $x$ could be far from $2$, making $|x+2|$ huge and the product large even when $|x-2|$ is small. Any restriction works: $\delta\le\tfrac12$ gives $|x+2|<4.5$ and $\delta=\min\{\tfrac12,\tfrac{2\varepsilon}{9}\}$.