Introduction to Limits

(a) Left-hand limit. For $x<2$ the active rule is $f(x)=x+1$, so $\lim_{x\to 2^-}f(x)=\lim_{x\to 2^-}(x+1)=2+1=3.$

(b) Right-hand limit. For $x>2$ the active rule is $f(x)=5-x$, so $\lim_{x\to 2^+}f(x)=\lim_{x\to 2^+}(5-x)=5-2=3.$

(c) Two-sided limit. Both one-sided limits exist and both equal $3$. By the theorem "$\lim_{x\to a}f(x)=L$ if and only if both one-sided limits exist and equal $L$", $\lim_{x\to 2}f(x)=3.$

(d) Value. The middle line of the definition applies exactly at $x=2$, so $f(2)=5$.

Comment. The limit is $3$ and the value is $5$; they are different, and that is perfectly legal. The limit is computed from a punctured interval about $2$ and never looks at $x=2$ itself. Because the limit exists but differs from the value, $f$ has a removable discontinuity at $2$: redefining $f(2)=3$ would make $f$ continuous there.

The domain of $g$ is every real $x$ with $x\ne 3$, so a punctured interval about $3$ is available and the question makes sense.

(a) Split on the sign of $x-3$.

If $x>3$ then $x-3>0$, so $|x-3|=x-3$ and $g(x)=\frac{x-3}{x-3}=1 \quad\text{for all } x>3, \qquad\text{hence}\qquad \lim_{x\to 3^+}g(x)=1.$

If $x<3$ then $x-3<0$, so $|x-3|=-(x-3)$ and $g(x)=\frac{-(x-3)}{x-3}=-1 \quad\text{for all } x<3, \qquad\text{hence}\qquad \lim_{x\to 3^-}g(x)=-1.$

(b) Both one-sided limits exist, but $-1\ne 1$. Therefore $\lim_{x\to 3}g(x)$ does not exist. This is the jump failure mode.

(c) Proof from the definition. Suppose, for contradiction, that some real number $L$ satisfies $\lim_{x\to 3}g(x)=L$. Apply the definition with $\varepsilon=1$: there is a $\delta>0$ such that $0<|x-3|<\delta \implies |g(x)-L|<1.$ The punctured interval contains points on both sides of $3$. Take $x_1=3+\frac{\delta}{2}>3 \qquad\text{and}\qquad x_2=3-\frac{\delta}{2}<3,$ both of which satisfy $0<|x-3|=\frac{\delta}{2}<\delta$. By part (a), $g(x_1)=1$ and $g(x_2)=-1$, so $|1-L|<1 \qquad\text{and}\qquad |-1-L|<1.$ Now apply the triangle inequality: $2=|1-(-1)|=|(1-L)+(L+1)|\le |1-L|+|L+1|<1+1=2,$ which says $2<2$: a contradiction. Hence no real $L$ can be the limit, and $\lim_{x\to 3}g(x)$ does not exist.

Recover the formulas.

Left piece: through $(-3,-1)$ and $(1,3)$, slope $=\dfrac{3-(-1)}{1-(-3)}=\dfrac{4}{4}=1$, so the line is $y=x+2$. Check: at $x=-3$, $y=-1$; at $x=1$, $y=3$. Hence $h(x)=x+2$ for $-3\le x<1$.

Right piece: through $(1,1)$ and $(4,4)$, slope $=\dfrac{4-1}{4-1}=1$, so the line is $y=x$. Check: at $x=1$, $y=1$; at $x=4$, $y=4$. Hence $h(x)=x$ for $1<x\le 4$.

(a) $\lim_{x\to 1^-}h(x)=\lim_{x\to 1^-}(x+2)=1+2=3$. The circle at $(1,3)$ being open is irrelevant: the limit records the height the branch heads toward, not whether the point is attained.

(b) $\lim_{x\to 1^+}h(x)=\lim_{x\to 1^+}x=1$.

(c) The one-sided limits are $3$ and $1$. They both exist but $3\ne 1$, so $\lim_{x\to 1}h(x)$ does not exist; the graph jumps by $2$ at $x=1$.

(d) $h(1)=0$, read off the solid dot. Notice that the three numbers $3$, $1$ and $0$ are all different: left limit, right limit and value are three independent pieces of information.

(e) The point $-1$ lies strictly inside the interval $[-3,1)$, so on both sides of $-1$ the rule is $h(x)=x+2$. Therefore $\lim_{x\to -1}h(x)=-1+2=1, \qquad h(-1)=-1+2=1.$ Limit equals value, so $h$ is continuous at $-1$.

(f) $\lim_{x\to 4^-}h(x)=\lim_{x\to 4^-}x=4$, which also equals $h(4)=4$. The two-sided limit at $4$ is not posed because $h$ is undefined for $x>4$: there is no punctured interval about $4$ contained in the domain, so there is no right-hand branch to approach along. At a domain endpoint only the one-sided limit is meaningful.

(a) Reading in from both ends, the values approach $0.16666\ldots$ from above on the left and from below on the right. The natural guess is $\lim_{x\to 0}f(x)=\frac{1}{6}\approx 0.166667.$

(b) Exact confirmation. Take any $x$ with $x>-9$ and $x\ne 0$. Since $\sqrt{x+9}+3\ge 3>0$, multiplying by the conjugate is legal: $\begin{aligned} \frac{\sqrt{x+9}-3}{x} &= \frac{\sqrt{x+9}-3}{x}\cdot\frac{\sqrt{x+9}+3}{\sqrt{x+9}+3} \\ &= \frac{(\sqrt{x+9})^2-3^2}{x(\sqrt{x+9}+3)} \\ &= \frac{(x+9)-9}{x(\sqrt{x+9}+3)} \\ &= \frac{x}{x(\sqrt{x+9}+3)} \\ &= \frac{1}{\sqrt{x+9}+3}. \end{aligned}$ Cancelling $x$ is valid because $x\ne 0$ everywhere on the punctured interval that the limit uses. As $x\to 0$ we have $x+9\to 9$ and $\sqrt{x+9}\to 3$, so $\lim_{x\to 0}\frac{\sqrt{x+9}-3}{x}=\frac{1}{3+3}=\frac{1}{6}.$ This matches the table: for instance at $x=0.1$, $\dfrac{1}{\sqrt{9.1}+3}=\dfrac{1}{6.016621}=0.166206$, exactly the tabulated entry.

(c) Why tables are not proofs. Evaluate $g(x)=\sin\dfrac{\pi}{x}$ at the suggested points: $g(1)=\sin\pi=0,\quad g(0.1)=\sin(10\pi)=0,\quad g(0.01)=\sin(100\pi)=0,\quad g(0.001)=\sin(1000\pi)=0.$ The table reads $0,0,0,0$ and appears to shout "the limit is $0$". But take $x_n=\dfrac{2}{4n+1}$ for a positive integer $n$. Then $\dfrac{\pi}{x_n}=\dfrac{(4n+1)\pi}{2}=2\pi n+\dfrac{\pi}{2}$, so $g(x_n)=1$, and $x_n\to 0$ as $n\to\infty$. So arbitrarily close to $0$ there are inputs with output $1$ as well as inputs with output $0$, and $\lim_{x\to 0}g(x)$ does not exist. A table samples finitely many points; a limit is a claim about all points near $a$. Tables suggest, definitions prove.

(a) For $x<1$ the rule is the polynomial $cx+3$, so $\lim_{x\to 1^-}f(x)=c\cdot 1+3=c+3.$ For $x>1$ the rule is the polynomial $x^2-cx$, so $\lim_{x\to 1^+}f(x)=1^2-c\cdot 1=1-c.$ The two-sided limit exists exactly when these two numbers are equal: $\begin{aligned} c+3 &= 1-c \\ c+c &= 1-3 \\ 2c &= -2 \\ c &= -1. \end{aligned}$ So $c=-1$ is the only value. Verify both sides with $c=-1$: the left-hand limit is $-1+3=2$ and the right-hand limit is $1-(-1)=2$. They agree, so $\lim_{x\to 1}f(x)=2.$

(b) The second line of the definition covers $x\ge 1$, so it is the one that applies at $x=1$: $f(1)=1^2-(-1)(1)=1+1=2.$ The limit equals the value, so $f$ is in fact continuous at $x=1$ for $c=-1$.

(c) With $c=0$ the pieces become $f(x)=3$ for $x<1$ and $f(x)=x^2$ for $x\ge 1$. Then $\lim_{x\to 1^-}f(x)=0+3=3, \qquad \lim_{x\to 1^+}f(x)=1^2-0=1.$ Both one-sided limits exist, but $3\ne 1$, so $\lim_{x\to 1}f(x)$ does not exist. The failure is a jump of size $|3-1|=2$; note that the function is bounded and perfectly tame on each side, so nothing here is about blowing up or oscillating.

(a) As $x\to 2^-$ we have $x-2<0$ with $|x-2|$ shrinking to $0$, so $\dfrac{1}{x-2}$ is negative with unbounded magnitude: $\lim_{x\to 2^-}\frac{1}{x-2}=-\infty.$ As $x\to 2^+$ we have $x-2>0$ and small, so $\lim_{x\to 2^+}\frac{1}{x-2}=+\infty.$ Neither one-sided limit is a real number, so $\lim_{x\to 2}F(x)$ does not exist. Moreover, because the two sides run to opposite infinities, we cannot even record a single symbol $+\infty$ or $-\infty$ for the two-sided limit. The line $x=2$ is a vertical asymptote of $F$.

(b) Proof. Let $M>0$ be given. Work backwards from the target. For $x\ne 2$ we have $(x-2)^2>0$, so $\frac{1}{(x-2)^2}>M \iff 1>M(x-2)^2 \iff (x-2)^2<\frac{1}{M} \iff |x-2|<\frac{1}{\sqrt{M}}.$ So set $\delta=\frac{1}{\sqrt{M}}>0.$ Now suppose $0<|x-2|<\delta$. Squaring the positive quantities preserves the inequality: $0<(x-2)^2<\delta^2=\frac{1}{M}.$ Taking reciprocals of positive numbers reverses the inequality: $G(x)=\frac{1}{(x-2)^2}>M.$ Since $M>0$ was arbitrary, $\lim_{x\to 2}G(x)=\infty$.

Numerical check: for $M=10000$ the recipe gives $\delta=\frac{1}{100}=0.01$. At $x=2.005$ we have $0<|x-2|=0.005<0.01$ and $G(x)=\dfrac{1}{0.000025}=40000>10000$, as promised.

(c) No. The statement "$=\infty$" is not an existence claim: $\infty$ is not a real number, and the definition of "the limit exists" requires a real $L$. The equation is a precise description of the failure - $G$ eventually exceeds every bound near $2$. If asked "does the limit exist?", answer no; if asked "describe the behaviour", answer $\infty$.

First simplify the quantity that must be made small: $|f(x)-5|=|(3x+2)-5|=|3x-3|=|3(x-1)|=3|x-1|.$

(a) We need $3|x-1|<0.06$, i.e. $|x-1|<\dfrac{0.06}{3}=0.02$. So the implication is true precisely when $\delta\le 0.02$, and the largest such $\delta$ is $\delta=0.02.$ Any smaller positive $\delta$ also works; the definition never requires the largest one.

(b) We need $3|x-1|<0.003$, i.e. $|x-1|<\dfrac{0.003}{3}=0.001$, so the largest is $\delta=0.001.$

(c) General proof. Let $\varepsilon>0$ be given. Put $\delta=\frac{\varepsilon}{3},$ which is positive since $\varepsilon>0$. Suppose $0<|x-1|<\delta$. Then $\begin{aligned} |(3x+2)-5| &= |3x-3| \\ &= 3|x-1| \\ &< 3\delta \\ &= 3\cdot\frac{\varepsilon}{3} \\ &= \varepsilon. \end{aligned}$ Since $\varepsilon>0$ was arbitrary, the definition is satisfied and $\lim_{x\to 1}(3x+2)=5$.

Consistency check against parts (a) and (b): the formula gives $\delta=\dfrac{0.06}{3}=0.02$ and $\delta=\dfrac{0.003}{3}=0.001$, matching both answers exactly.

Plausibility check. At $x=-2$: $4-7(-2)=4+14=18$, so $L=18$ is the right target.

Scratch work. Simplify what must be made small: $|(4-7x)-18|=|-7x-14|=|-7(x+2)|=|-7|\cdot|x+2|=7|x+2|.$ This is below $\varepsilon$ exactly when $|x+2|<\dfrac{\varepsilon}{7}$.

(a) Proof. Let $\varepsilon>0$ be given. Put $\delta=\frac{\varepsilon}{7}>0.$ Suppose $0<|x-(-2)|=|x+2|<\delta$. Then $\begin{aligned} |(4-7x)-18| &= |-7x-14| \\ &= 7|x+2| \\ &< 7\delta \\ &= 7\cdot\frac{\varepsilon}{7} \\ &= \varepsilon. \end{aligned}$ Since $\varepsilon>0$ was arbitrary, $\lim_{x\to -2}(4-7x)=18$.