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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.

A sequence $(a_n)$ of real numbers is a convergent sequence with limit of a sequence $\ell$ whenDivergence to positive infinity meansIn both definitions the same $N$ must control every later term. The second condition is not convergence to a real number.

False. Take the positive geometric sequence $\boxed{a_n=2^{-n}}$. It tends to zero, whileA positive limit of a sequence would make the claim true since $a_{n+1}/a_n$ would then tend to $\ell /\ell =1$, but positivity of each individual term does not exclude a zero limit of a sequence.

False. Take the positive sequence $\boxed{a_n=n}$. Thenbut $a_n\to +\infty$, so it has no finite limit of a sequence. Successive ratios close to one do not prevent cumulative growth.

False. Choose the positive sequenceIt has limit of a sequence one, but Bernoulli's inequality givesThus $(a_n^n)$ is not a bounded sequence.

Let $T_N={\sum }_{n=1}^N|a_n|$ and $S_N={\sum }_{n=1}^N{a}_n$. If $(T_N)$ converges, it is a Cauchy sequence. Thus, for every $\epsilon >0$, some $N$ makeswhere the triangle inequality gives the bound. Therefore $(S_N)$ is a Cauchy sequence in the complex numbers. These form a complete metric space, so $(S_N)$ has a finite limit of a sequence. Equivalently, apply the Cauchy criterion for series. Hence absolute convergence implies convergence.

For a series of complex numbers, form the partial sums $S_N={\sum }_{n=1}^N{a}_n$. The series converges when these partial sums have a finite limit of a sequence $S\in \mathbb{C}$:The series has absolute convergence when ${\sum }_{n=1}^{\infty }|a_n|$ converges as a series of nonnegative real numbers. A series beginning at $n=0$ uses the same definitions with that starting index.

Choose $h_n=\min \{1/n,(b-c)/2\}>0$. Then $h_n\to 0$ and $[c,c+h_n]\subset (a,b)$. The mean value theorem states that a continuous function on a closed interval, differentiable in its interior, has an interior derivative equal to its endpoint difference quotient. It gives $x_n\in (c,c+h_n)$ withHere $x_n\ne c$, $x_n\to c$, and the right-hand side tends to $f^{\prime }(c)$ by differentiability at $c$. If $L={\lim }_{x\to c}{f}^{\prime }(x)$ exists, the same values $f^{\prime }(x_n)$ must tend to $L$. Uniqueness of the limit of a sequence givesThus a derivative cannot have a removable discontinuity at a point where the original function is differentiable. The proof also excludes an infinite limit of a function there, since the displayed difference quotients have a finite limit of a sequence.

A sequence $(a_n)$ has limit of a sequence $L$ ifFor the reciprocal rule, the required condition is $\boxed{L\ne 0}$. Indeed, eventually $|a_n-L|<|L|/2$, so $|a_n|\ge |L|/2>0$. The reciprocals are then defined on this tail, andA finite number of zero initial terms does not affect this tail limit of a sequence; if a reciprocal sequence is required at every index, those terms must be excluded or assigned arbitrary values separately.

The tail $(a_n{)}_{n\ge 1}$ is a monotone sequence bounded below by zero. The bounded monotone sequence theorem therefore gives $a_n\to L\ge 0$. Taking the limit of a sequence in $a_n^2\ge c$ yields $L^2\ge c>0$, so $L>0$. The reciprocal rule from part (a) permits passage to the limit of a sequence in the recurrence relation:Multiplication by $2L$ givesThis proves existence of the positive square root without assuming that existence when constructing the sequence.

For fixed $z\ne 0$, the ratio of successive absolute values isThe ratio test says that a series converges absolutely when this ratio has a limit of a sequence strictly below one. Thus this power series converges absolutely for every complex $z$; at zero the conclusion is immediate. Its radius of convergence is $\boxed{R=\infty }$.

Yes. By uniform convergence, choose $N$ so that $|p_n(x)-f(x)|<1$ for every $n\ge N$ and every $x\in \mathbb{R}$. ThenThe polynomial $p_n-p_N$ is bounded on the whole real line. The supplied growth property of nonconstant polynomials implies that this difference must be constant; write $p_n=p_N+c_n$. Evaluating at zero gives $c_n=p_n(0)-p_N(0)\to c=f(0)-p_N(0)$. For each $x$, take the limit of a sequence in $p_n(x)=p_N(x)+c_n$ to obtainThis is a polynomial. The uniform limits of polynomials on the real line therefore have a tail in which all nonconstant coefficients have already stabilized; only the constant coefficients may continue to change.

No. On $(-1,1)$ take $c=0$ andAt zero, its difference quotient has absolute value $|x\sin (1/x)|\le |x|\to 0$, so $f^{\prime }(0)=0$. Away from zero, the product rule and chain rule apply. DefineThese are distinct positive inputs tending to zero. Since $\sin A_n=0$ and $\sin (A_n+B_n)=\sin B_n$,The even-indexed subsequence tends to $-2/\pi$, whereas the odd-indexed subsequence tends to $2/(3\pi )$. Thus the limit of a sequence does not exist. The two points lie on the same side of $c$, so the nonnegative-weight estimate in part (b) is unavailable. This is a concrete instance of differentiability does not imply continuity of the derivative.

Start with $[a_0,b_0]=[a,b]$. At stage $n$, form the midpoint $c_n=(a_n+b_n)/2$. If $f(c_n)=0$, stop; for a sequence defined for all later indices, repeat $c_n$ forever. Otherwise retain the half-interval with opposite endpoint signs: set $[a_{n+1},b_{n+1}]=[a_n,c_n]$ if $f(a_n)f(c_n)<0$, and $[a_{n+1},b_{n+1}]=[c_n,b_n]$ otherwise. This is the bisection method.

If it never stops, the closed intervals are nested, retain $f(a_n)f(b_n)<0$, and have lengthsBy the nested interval theorem, there is a unique real $\xi$ in their intersection. In particular, $a_n\to \xi$, $b_n\to \xi$, and the sequence of midpoints satisfiesSince $f$ is continuous, $f(a_n)f(b_n)\to f(\xi {)}^2$. All these products are negative, so their limit of a sequence is nonpositive. Therefore $f(\xi {)}^2\le 0$, forcing $f(\xi )=0$. We have proved that the sequence is a convergent sequence whose limit of a sequence is a zero of a function:The error bound applies at every stage before exact termination; after termination the error is zero. The intermediate value theorem also guarantees a zero of a function in each retained interval, but the nested-interval argument identifies the particular one to which the sequence of midpoints converges.

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.

If $L<1$, set $\epsilon =(1-L)/2>0$. The definition of limit of a sequence gives $a_n<L+\epsilon =(1+L)/2<1$ for all sufficiently large $n$, contradicting $a_n\ge 1$. Hence $\boxed{L\ge 1}$.

Factor the difference of cubes. Since both cube roots are at least one,Given $\epsilon >0$, the definition of limit of a sequence supplies $N$ with $|a_n-L|<3\epsilon$ for $n\ge N$. The displayed inequality then gives $|a_n^{1/3}-L^{1/3}|<\epsilon$, proving directly that $\boxed{a_n^{1/3}\to L^{1/3}}$.

The sequence does not converge. For $n=2^k$, the largest prime factor is $p(n)=2$, while the sum of prime factors with multiplicity is $s(n)=2k$. This subsequence therefore has ratio $1/k\to 0$.

There are infinitely many prime numbers: if there were only $q_1,\dots ,q_r$, a prime factor of $q_1\cdots q_r+1$ would be absent from the list. Listing the prime numbers in increasing order gives a subsequence of indices $n=q_j$ for which $p(n)=s(n)=q_j$, so the ratio is constantly one. Every subsequence of a convergent sequence has the same limit of a sequence. The two distinct subsequential limits $\boxed{0\text{ and }1}$ rule out convergence.

A series ${\sum }_{n=1}^{\infty }{a}_n$ converges if its partial sums $s_N={\sum }_{n=1}^N{a}_n$ have a finite limit of a sequence $S$ as $N\to \infty$; its sum is then $S$. It has absolute convergence if the series ${\sum }_{n=1}^{\infty }|a_n|$ converges, where $|a_n|$ is the absolute value of the term.

A series ${\sum }_{n=1}^{\infty }{a}_n$ converges when its partial sums $s_N={\sum }_{n=1}^N{a}_n$ have a finite limit of a sequence.

The comparison test says that if $0\le a_n\le b_n$ eventually and $\sum b_n$ converges, then $\sum a_n$ converges. Indeed, the partial sums of the nonnegative terms $a_n$ are increasing, and are bounded above by a fixed initial contribution plus the sum of the comparison series. The bounded monotone sequence theorem proves their convergence. Finite initial terms do not affect convergence. Applying this result to $|a_n|$ also proves absolute convergence whenever $|a_n|\le b_n$ eventually with $\sum b_n<\infty$.

All $a_n$ are positive, and $a_1=e^{-1}$. The logarithm turns consecutive ratios into differences, producing the finite telescoping sumThus, already for every finite $n$,If the partial sums converge to $S$, the continuity of the exponential function givesThe shift from $a_{n+1}$ to $a_n$ preserves the limit of a sequence. Convergence of the partial sums is established in part (f), so no convergence assumption is left unsupported.