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If $|h(x)|\le C$ near $c$ and $a(x)\to 0$ as $x\to c$, then $a(x)h(x)\to 0$ by the squeeze theorem, even if $h$ has no limit. Defining the product to be zero at $c$ gives continuity there. For example, $\sqrt{|x|}\sin (1/\sin x)$ extends continuously at zero; the same oscillation near a nonzero multiple of $\pi$ has nonvanishing amplitude and does not have a continuous zero extension.

Away from integer multiples of $\pi$, the function is a composition of continuous functions. At zero, the bound $|g(x)|\le \sqrt{|x|}$ gives $g(x)\to 0=g(0)$ by the squeeze theorem. This is oscillation multiplied by a vanishing amplitude.

Fix $n\ne 0$ and set $t_k=\pi /2+2\pi k$ and $x_k=n\pi +(-1{)}^n\arcsin (1/t_k)$. Then $x_k\to n\pi$, $\sin x_k=1/t_k$, andSo $g$ is discontinuous at every nonzero integer multiple of $\pi$. Its set of continuity points is

On $[0,\pi ]$, $g(x)\le \sqrt{x}<\sqrt{\pi }$ for $x<\pi$, while $g(\pi )=0$. With $x_k=\pi -\arcsin (1/t_k)$ as above, $g(x_k)=\sqrt{x_k}\to \sqrt{\pi }$. Consequently the supremum is $\sqrt{\pi }$ and is not attained. This illustrates that supremum can fail to be attained at an oscillatory endpoint.

On $[\pi ,3\pi /2]$, set $x_0=\pi +\arcsin (2/(3\pi ))$. It lies strictly inside the interval, and $1/\sin x_0=-3\pi /2$, so $g(x_0)=\sqrt{x_0}>\sqrt{\pi }$. Choose $0<\delta <x_0-\pi$. On $[\pi ,\pi +\delta ]$,On $[\pi +\delta ,3\pi /2]$, $g$ is a continuous function, so the extreme value theorem gives a maximum there, at least $g(x_0)$. It exceeds every value on the discarded interval. Therefore $g$ does attain its supremum on $[\pi ,3\pi /2]$. This is the interior-value criterion for an attained maximum.

For $0<x<\pi /2$, the usual unit-circle area comparison, with angles measured in radians, gives $\sin x\le x\le \tan x$. HenceThe cosine is a continuous function at zero, so the squeeze theorem yields the right-hand limit of a function $1$. The quotient $\sin x/x$ is even, so the left-hand limit of a function is the same. Thus the limit is $1$.

For $x>0$, the exponent $x/(1+x)$ lies in $(0,1)$, and $0\le {\cos }^{4}x\le 1$. ConsequentlyThe last inequality uses the positive-term power series $e^x={\sum }_{k\ge 0}{x}^k/k!\ge x^2/2$ for $x>0$. The final bound tends to zero, so the squeeze theorem gives limit $0$.

We first prove by induction thatIt holds for $n=1$, and if it holds at $n$, thenConsequentlyPart (c) and the squeeze theorem now give

Because $(a_n)$ is a monotone sequence, every one of its first $n$ terms is at most $a_n$, soOn the other hand, every term from $a_n$ through $a_{2n}$ is at least $a_n$, and henceBoth the lower bound and upper bound tend to $x$ because the Cesaro means converge to $x$. The squeeze theorem gives

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