Let $f:[0,1] \rightarrow R$ with $1 \leq f \leq 2$ set $$N(p)=\left( \int_0^1 f^p dx \right)^{\frac{1}{p}} \qquad p \neq 0$$ To find the three limits $\lim_{p\rightarrow \pm \infty} N(p)$ and $\lim_{p\rightarrow 0} N(p)$.
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One thing is clear: $\liminf_{p\rightarrow \infty} N(p) \geq 1$.
For $p>0$ $$ 1\leqslant N(p)=\left( \int_0^1 f^p dx \right)^{\frac{1}{p}} \leqslant (2^p)^{1/p}=2 $$ For $p<0$, let $q=-p$. $$ 1\leqslant N(p)=N(-q)=\frac1{\left( \int_0^1 \frac1{f^q} dx \right)^{\frac{1}{q}}} \leqslant 2 $$ Since $$ \liminf_{p\to0^+}N(p)\geqslant 1\quad\text{and }\quad\limsup_{p\to0^-}N(p)\leqslant 1 $$ We have $$ \liminf_{p\to0}N(p)=\limsup_{p\to0}N(p)=1\quad\text{or }\quad\lim_{p\to0}N(p)=1 $$ Moreover we have $$ \liminf_{p\to{\pm\infty}}N(p)=1 \quad\text{and }\quad\limsup_{p\to{\pm\infty}}N(p)=2 $$ It is easy to find particular $f$ to attain above limits.
If $f$ is continuous, then $$ \lim_{p\to{\pm\infty}}N(p)=\sup{\{f(x):x\in[0,1]\}}=2 $$