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If $f$ is continuous at $a$, then for every $\epsilon >0$ there is $r_0>0$ such that $d(x,a)<r_0$ implies $|f(x)-f(a)|<\epsilon$. In particular, for $0<r\le r_0$,The squeeze theorem giveswhich implies condition (1). Thus continuity implies upper semicontinuity.

For $p>0$, integral comparison for $x^{-1-p}$ bounds its convergent positive series by $1/p\le {\sum }_{n\ge 1}{n}^{-1-p}\le 1+1/p$. Multiplication by $p$ gives $1\le p\zeta (1+p)\le 1+p$, so the squeeze theorem proves the displayed real-axis limit. This establishes the leading $1/p$ divergence by real integration alone. The convergence of the series at each fixed positive $p$ follows from bounded increasing partial sums via the monotone convergence theorem for sequences.

For finitely many fixed nonnegative real numbers $b_1,\dots ,b_m$, with $m\ge 1$,Let $M$ be that maximum. If $M=0$, every term is zero. Otherwise $M^p\le \sum b_j^p\le m{M}^p$ for $p>0$. Taking roots gives $M\le (\sum b_j^p{)}^{1/p}\le m^{1/p}M$, and $m^{1/p}\to 1$. The squeeze theorem proves the result, including ties between the largest bases. For example, $(1+a^p+(a^2/2{)}^p{)}^{1/p}$ tends to $\max (1,a,a^2/2)$ for $a\ge 0$. The finiteness and fixed number of terms supply the upper factor $m$; an infinite sum needs further hypotheses.

Let an Archimedean spiral have polar radius $r=at$ and angle $t$ in polar coordinates, with $0\le t\le R/a$, where $R>0$ is fixed. Parameterizing by $r$ gives the arc lengthThe bounds $r\le \sqrt{a^2+r^2}\le r+a$ yieldThe squeeze theorem proves the displayed limit. Thus the arc length grows like $R^2/(2a)$ as the turns become closely spaced, without requiring an explicit antiderivative.