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Since the natural logarithm is increasing, comparison with an integral givesThe lower bound is $n\log n-n+1$. Dividing all three expressions by $n\log n$ and applying the squeeze theorem gives

Let ${\beta }_{n+1}>0$ be the complete quotient after $a_n$. Part (a) writes the irrational number $\theta$ asso $\theta$ lies between the adjacent continued fraction convergents $p_n/q_n$ and $p_{n-1}/q_{n-1}$. Their separation isBecause all $a_j$ are positive integers, the recurrence $q_n=a_n{q}_{n-1}+q_{n-2}$ implies $q_n\to \infty$. Hence this separation tends to zero, and the squeeze theorem gives

LetContinuity at zero gives $m_n,M_n\to f(0)$. Part (a), with the positive weight $n{e}^{-nx}$, givesThe squeeze theorem therefore yields

Since $f$ is bounded on $[0,1]$, say $|f|\le K$, the omitted tail satisfiesAdding the tail proves

Put $u=c^{-1}-1>0$, so $c=(1+u{)}^{-1}$. By Bernoulli's inequality, $(1+u{)}^n\ge 1+nu$, and henceMore explicitly, for $\epsilon >0$ choose an integer $N>1/(u\epsilon )$. For $n\ge N$, the right-hand side is less than $\epsilon$. Thus the geometric sequence has limit of a sequence zero, either directly by the definition or by the squeeze theorem.

Given $\epsilon >0$, the two assumed limits of sequences supply thresholds $N_1,N_2$ such that $|a_n-\ell |<\epsilon$ for $n\ge N_1$ and $|c_n-\ell |<\epsilon$ for $n\ge N_2$. For $n\ge \max \{N_1,N_2\}$,Thus $|b_n-\ell |<\epsilon$ eventually, proving the squeeze theorem:

On $(0,1)$, the product rule and chain rule give $f^{\prime }(x)=2x\sin (1/x)-\cos (1/x)$. On $(-1,0)$, the function is constant and has derivative zero. At zero,Its absolute value is at most $|h|$, so the squeeze theorem gives $f^{\prime }(0)=0$. ThereforeFor $x_n=1/(2\pi n)$, we have $x_n\to 0$ but $f^{\prime }(x_n)=-1$ for every $n$. This differs from $f^{\prime }(0)=0$, so $f^{\prime }$ is not continuous at zero. Indeed, $y_n=1/((2n+1)\pi )$ gives $f^{\prime }(y_n)=1$, so the right-hand limit of a function does not exist. Thus $f$ is differentiable everywhere on $(-1,1)$, while $f^{\prime }$ is discontinuous at zero.

Use the polar radius $r=at\in [0,1]$ as the parameter of the Archimedean spiral. Its angle in polar coordinates is $r/a$. The radial and angular contributions to its arc length are orthogonal, soThere is no need to evaluate this integral. The simple bounds $r\le \sqrt{a^2+r^2}\le r+a$ implyThe squeeze theorem therefore gives the scaled length of an Archimedean spiral:Geometrically, the radial contribution becomes negligible after multiplying the arc length by $a$; the remaining contribution is the area under the function graph $r\mapsto r$ on $[0,1]$.

For the first sequence, $|a_n|=n^{-3}\to 0$. The squeeze theorem gives $a_n\to 0$, and every subsequence of a convergent sequence has the same limit of a sequence. Thus its set of subsequential limits is $\boxed{\{0\}}$.

For the second sequence, the four possible sine values giveThe even-indexed subsequence is constantly one, and $b_{4k+3}\to 0$, so both zero and one occur as subsequential limits. A convergent subsequence is bounded, so it contains only finitely many terms with $n\equiv 1(\mathrm{mod}4)$. Of the remaining three residue classes, at least one occurs infinitely often. The corresponding further subsequence has limit of a sequence zero or one; the subsequence rule forces the original convergent subsequence to have that same limit of a sequence. Therefore the exact answer is $\boxed{\{0,1\}}$. Divergence to $+\infty$ along $n\equiv 1(\mathrm{mod}4)$ is not a finite subsequential limit.

Put $u_n=11^n\log n/(2^{2n+1}n!)$ for $n\ge 2$. The factorial givesThe logarithm identityfollows from continuity of the logarithm at one and $\log n\to +\infty$. The product rule for limits of sequences therefore makes the displayed ratio tend to zero. In particular, some $N$ satisfies $0<u_{n+1}\le u_n/2$ for all $n\ge N$. By mathematical induction, $0<u_{N+j}\le u_N{2}^{-j}$. A geometric sequence with ratio in $(0,1)$ tends to zero, so the squeeze theorem proves

In level $n$, each group isand there are $2^n$ such groups. The partial sums at completed groups are therefore zero. Inside a group of level $n$, the only other partial sums are $2^{-(n+1)}$ and $2^{-n}$. Every level contains finitely many terms, so the level containing the $N$th term tends to infinity with $N$. Thus all partial sums lie between zero and a bound tending to zero. By the squeeze theorem, the series has sum $\boxed{0}$.

For the next two parts, label the positive occurrences of size $2^{-(n+1)}$ by $P_n$ and the negative occurrences of size $-2^{-n}$ by $N_n$. There are exactly $2^{n+1}$ occurrences in $P_n$ and $2^n$ in $N_n$. Occurrences with equal value retain separate labels, so checking their counts verifies a bijection of terms.

Although $\cos (1/x)$ oscillates without a limit, its magnitude is at most one. Therefore $|x\cos (1/x)|\le |x|\to 0$. The squeeze theorem givesThe shrinking factor $x$ suppresses the oscillation from either side of zero.

For fixed $p>0$, the partial sums are increasing and bounded above by $1+1/p$. The monotone convergence theorem for sequences therefore proves convergence to a finite sum $S(p)$. Passing to the limit $N\to \infty$ in the two bounds givesThe squeeze theorem now yieldsSince $S(p)=\zeta (1+p)$, this is the real-axis singular limit of the Riemann zeta function at one. The limiting sum is taken at fixed $p$ before sending $p$ to zero; setting $p=0$ inside the convergent-series argument would instead give the divergent harmonic series.

For $n=1,2,3$, the difference quotient at zero is $x^{n-1}\sin (1/x)$. It oscillates without a limit for $n=1$, while the squeeze theorem gives limit zero for $n=2,3$. The other two cases fail continuity and therefore cannot be differentiable. Hence

For every $n\ge 1$, $|x^n\sin (1/x)|\le |x{|}^n\to 0$, so the squeeze theorem gives limit zero. For $n=0$, sequences $x_k=(\pi /2+2k\pi {)}^{-1}$ and ${\tilde{x}}_k=(3\pi /2+2k\pi {)}^{-1}$ give function values $1$ and $-1$, so no limit exists. HenceThe value assigned at zero when $n=4$ does not affect this punctured limit.

For $b>0$, split the interval into $n$ equal widths $\Delta x=b/n$. Since $x^2$ increases on this interval, the left and right rectangles give lower and upper Riemann sums:Using the finite sum of squares,The rectangle bounds differ by $U_n-L_n=b^3/n\to 0$. The squeeze theorem therefore gives, directly from elementary areas,For $b=0$ the result is zero. If negative $b$ is allowed, reverse the integration limits and use the reflection $x=-u$; the same signed formula holds. No antiderivative is needed for this first-principles calculation.

The binomial series gives $(1+ϵ{)}^{-\alpha }=1-\alpha ϵ+O({ϵ}^2)$. ThereforeTo justify the limiting approximation over the infinite interval, also use the right-endpoint sum. It is $L_{ϵ}=(1+ϵ{)}^{-1-\alpha }{U}_{ϵ}$, and monotonicity gives $L_{ϵ}\le I\le U_{ϵ}$. Both bounds have the same limit. The geometric partition bounds for a power-law improper integral and the squeeze theorem thus proveThis bounding argument matters because the geometric partition has unbounded widths far along the half-line; a finite-interval maximum-mesh argument cannot simply be applied to the entire infinite partition at fixed $ϵ$.

For $m=0$, continuity fails. Take $x_k=(\pi /2+2\pi k{)}^{-1}$ and ${\tilde{x}}_k=(3\pi /2+2\pi k{)}^{-1}$. Both tend to zero, but the corresponding function values are $1$ and $-1$. Thus there is no limit, and the function is not differentiable either, since differentiability would imply continuity.

For $m=1$, the bound $|x\sin (1/x)|\le |x|$ and the squeeze theorem give continuity at zero. Its difference quotient is $\sin (1/h)$, however, which has no limit by the same two sequences. It is therefore not differentiable there.

For $m=2$, the bound $|x^2\sin (1/x)|\le x^2$ proves continuity. Its difference quotient is $h\sin (1/h)$, whose absolute value is bounded by $|h|$, so the derivative exists and equals zero. The complete classification is

Use the root limit of a finite sum of powers. Put $M=\max \{1,a,a^2/2\}\ge 1$. For every $x>0$, at least one of the three terms is $M^x$ and none is larger, soTaking the positive $x$th root givesSince $3^{1/x}\to 1$, the squeeze theorem givesIndeed, for $a\le 1$ both remaining bases are at most $1$; for $1\le a\le 2$, $a\ge a^2/2$; and for $a\ge 2$, $a^2/2\ge a$. The endpoint cases $a=0,1,2$ are included in the same bound.

The original PDF has the absolute-value hypothesis $|g(\xi )|<C{e}^{-s\xi }$, which is missing from the local OCR. From the Green's function formula and $|\sin |\le 1$,Because $s>a>0$, the upper bound tends to zero. The squeeze theorem proves $\boxed{y(x)\to 0\text{ as }x\to \infty }$.

The answers, in the order completeness, sequential compactness, connectedness, and path connectedness, areDefine $F(0)=0$ and $F(x)=\sqrt{x}\sin (1/x)$ for $x>0$. The estimate $|F(x)|\le \sqrt{x}$ proves continuity at zero by the squeeze theorem; elsewhere it follows from the usual continuous functions. Thus $X$ is the graph of a function $F$ on the closed set $[0,\infty )$.

To check closedness directly, if $(x_n,F(x_n))\to (x,y)$ in the Euclidean plane, then $x\ge 0$ and continuity gives $y=F(x)$. Therefore $X$ is a closed subset of the complete metric space ${\mathbb{R}}^2$, hence is a complete metric space. The sequence $(n,F(n))$ has no convergent subsequence, since its first Cartesian coordinates tends to infinity, so sequential compactness fails.

For any two points $(u,F(u)),(v,F(v))$, the mapis a continuous path in $X$ joining them. Thus $X$ is path-connected, and consequently connected. The factor $\sqrt{x}$ is crucial: the oscillations shrink to the single point $(0,0)$.