Solving the linear ordinary differential equation and imposing the initial condition givesAs $\lambda >0$, the requested limit is

Near zero, $x^3-x^5=x^3(1-x^2)$ has the sign of $x$, so $f(x)<1$ just to the left and $f(x)>1$ just to the right. Thus the critical point is neither a local minimum nor a local maximum. Moreover the Taylor series gives $f^{\prime \prime }(x)=6x+O(x^3)$, which changes sign at zero. ThereforeThis is the point called a “saddle point” in the question; stationary inflection point specifies the one-variable meaning.

Use the Taylor series $e^u=1+u+u^2/2+\cdots$ with $u=x^3-x^5$. Since $u$ begins at cubic order and $u^2$ begins at sixth order, through cubic terms the Taylor series isThe linear and quadratic coefficients are zero. The stronger remainder $O(x^5)$, in big O notation, follows because the fourth-order term is also zero; the next term is $-x^5$.

The chain rule givesThe exponential function never vanishes, so all critical points areIn particular $f^{\prime }(0)=0$.

The second coordinate after two iterations satisfieswhenever those two divisions are needed, by the same remainder bound, now applied to $b_i>b_{i+1}>0$. Let $m=\lfloor {\log }_{2}b\rfloor$. If the Euclidean algorithm has not already stopped after $2m$ iterations, then its positive second coordinate is at most $1$: for $m\ge 1$,For $m=0$ the original $b$ is already $1$. A second coordinate of $1$ requires just one more division. ConsequentlyThis removes dependence on the potentially much larger input $a$. It improves the scale ${\log }_2(ab)$ to one involving only the smaller input; the rounded numerical bound need not be smaller for every pair. One may always take the minimum of the bounds in (ii) and (iii).

Count the final division producing zero as an iteration, and let the total be $n$. The terminal product $a_n{b}_n=0$ alone gives no useful bound. Instead apply (i) to each nonterminal product, which is a positive integer and hence at least $1$.

If $n\ge 2$, the penultimate product satisfiesAs $n$ is an integer, this impliesFor $n=1$, the same bound holds since $a>b\ge 1$ implies $ab\ge 2$. In particular the Euclidean algorithm uses $O(\log a+\log b)$ divisions, in big O notation. The convention that includes the final zero-producing division explains the one-step offset in the argument.

Consider one iteration with $a_i>b_i>0$ and $r_i=a_i\mathrm{mod}{b}_i$. If $b_i\le a_i/2$, then $r_i<b_i\le a_i/2$. If $b_i>a_i/2$, the quotient is $1$, so $r_i=a_i-b_i<a_i/2$. In either case $r_i<a_i/2$, whenceRepeated application, or mathematical induction, provesThis also holds for the final iteration, whose left side is zero. For zero iterations the correct relation is equality $a_0{b}_0=ab$, rather than a strict inequality.

Set $(a_0,b_0)=(a,b)$. At each iteration with $b_i>0$, compute the remainder $r_i=a_i\mathrm{mod}{b}_i$, then setStop at the first index $n$ with $b_n=0$ and return $a_n$. The positive second coordinates strictly decrease, so algorithm termination follows from the well-ordering principle. The equality of greatest common divisors established in (ii) is a loop invariant, and at termination it gives

Write $a=qb+r$ using the Euclidean division. If $d\mid a$ and $d\mid b$, then $d\mid (a-qb)=r$. Conversely, if $d\mid b$ and $d\mid r$, then $d\mid (qb+r)=a$. Therefore $(a,b)$ and $(b,r)$ have exactly the same positive common divisors, provingFor $r=0$, this agrees with $\gcd (b,0)=b$.

The greatest common divisor is the largest positive integer dividing both inputs:This maximum exists because $1$ belongs to the set and all its elements are at most $b$.

The Euclidean division gives $a=qb+r$ with $q\in {\mathbb{N}}_0$ and $0\le r<b$. Writing $a=nu,b=nv$ givesSince $u-qv$ is an integer, this proves the divisibilityincluding $r=0$.

By divisibility, write $a=nu$ and $b=nv$ for positive integers $u,v$. ThenThe factor $u^2-v^2$ is an integer, so

A direct mathematical induction proves $0\le F_n\le 2^n$. The base cases $F_0=0\le 1$ and $F_1=1\le 2$ hold. If the bound holds at $n-1$ and $n$, then the Fibonacci recurrence givesThus the definition of big O notation holds with $C=1$, proving

Suppose the definition of big O notation held with constants $C>0,N$. Choose an odd integer $n>\max (C,N)$. At this input, $f(n)=n$ and $g(n)=1$, so the required inequality would be $n\le C$, a contradiction. ThusThe infinitely many even inputs, where $f(n)/g(n)=1/n$, cannot repair the failure on odd inputs because the bound must hold for every sufficiently large input.

For $n\ge 1$, $\log 3<2$ implies $3^{\log n}=n^{\log 3}\le n^2$, and also $n\le n^2$. ConsequentlyThe definition of big O notation holds with $C=4,N=1$:

For functions $f,g:\mathbb{N}\to \mathbb{R}$, big O notation meansUsing the natural logarithm and exponential function, $3^{\log n}=n^{\log 3}$. Since $\log 3>1$, for every $n\ge 1$ we have $n\le n^{\log 3}$. The witnesses $C=1,N=1$ therefore prove

Use mathematical induction with two base cases. At $n=0$, the formula gives $(1-1)/\sqrt{5}=0=F_0$. At $n=1$, it gives $(\varphi -\psi )/\sqrt{5}=1=F_1$.

If it holds at $n-1$ and $n$, then the Fibonacci recurrence givesThe identities ${\varphi }^2=\varphi +1$ and ${\psi }^2=\psi +1$ turn the numerator into ${\varphi }^{n+1}-{\psi }^{n+1}$. Thus

The linearity of the Fibonacci recurrence allows a linear combinationThe initial conditions give $A+B=0$ and $A\varphi +B\psi =1$. Since $\varphi -\psi =\sqrt{5}$, we obtain $A=1/\sqrt{5}$ and $B=-1/\sqrt{5}$. Thus Binet's formula is

For a nonzero geometric sequence, substitute $H_n=c{q}^n$ into the linear recurrence and cancel $c{q}^{n-1}$:The characteristic equation of a linear recurrence therefore has rootsThe zero sequence $c=0$ is a degenerate solution for any $q$ for which the displayed expression is defined. A nonzero sequence with $q=0$ cannot satisfy the recurrence at $n=1$, since it would require $0=0+H_0$.

In the intended range $0\le k\le n$, division by the positive binomial coefficient $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ givesConsequently the intended remainder isThere is a necessary qualification to the printed unrestricted domain. If binomial coefficients outside $0\le k\le n$ are extended by zero, then every $k>n$ satisfies $g(k)=\left(\genfrac{}{}{0ex}{}{n}{k}\right)=0$, regardless of $n\mathrm{mod}3$. For example, $n=3,k=4$ satisfies the literal equality but has $n\mathrm{mod}3=0$. Thus the literal extended-domain premise alone does not determine a remainder; the boxed conclusion requires $0\le k\le n$, or equivalently here $\left(\genfrac{}{}{0ex}{}{n}{k}\right)\ne 0$.