Suppose $m(E)<\infty$ and $f\in\mathcal{L}^{\infty}(E)$ and $1\leq p_1<p_2\leq \infty$. The goal of the two problems below is to show \begin{align*} \lim_{p_1\to p_2}\|f\|_{p_1}=\|f\|_{p_2}. \end{align*}
(a) Prove that \begin{align*} \limsup_{p_1\to p_2}\|f\|_{p_1}\leq\|f\|_{p_2}. \end{align*}
(b) Prove that \begin{align*} \liminf_{p_1\to p_2}\|f\|_{p_1}\geq\|f\|_{p_2}. \end{align*} Hint: Apply $\log$ to the definition of $\|f\|_{p_1}$ and compute $\liminf$ using Fatou's Lemma. Note: both $\log$ and the exponential function are continuous and they can move in and out of a limit.
Here's my attempt at part (b):
Proof: By definition, we have that $\displaystyle{\|f\|_{p_1}=\left(\int_E|f|^{p_1}dm\right)^{1/p_1}}$. So, \begin{align*} \ln{(\|f\|_{p_1})}=\ln{\left(\left(\int_E|f|^{p_1}dm\right)^{1/p_1}\right)}=\frac{1}{p_1}\ln{\left(\int_E|f|^{p_1}dm\right)}. \end{align*} Since the natural log is a continuous function, then we get that \begin{align*} \liminf_{p_1\to p_2}{\ln{(\|f\|_{p_1})}}=\ln{\left(\liminf_{p_1\to p_2}\|f\|_{p_1}\right)}. \end{align*} Thus \begin{align*} \liminf_{p_1\to p_2}{\ln{(\|f\|_{p_1})}}&=\liminf_{p_1\to p_2}\frac{1}{p_1}\ln\left(\int_E|f|^{p_1}dm\right)\\ &=\frac{1}{p_2}\ln\left(\liminf_{p_1\to p_2}\int_E|f|^{p_1}dm\right). \end{align*} We have that $|f|_1^p$ is a nonnegative function for all $p_1$. By Fatou's Lemma, \begin{align*} \liminf_{p_1\to p_2}\int_E|f|^{p_1}dm\geq\int_E\liminf_{p_1\to p_2}|f|^{p_1}dm. \end{align*} Hence \begin{align*} \frac{1}{p_2}\ln\left(\liminf_{p_1\to p_2}\int_E|f|^{p_1}dm\right)&\geq \frac{1}{p_2}\ln\left(\int_E\liminf_{p_1\to p_2}|f|^{p_1}dm\right)\\&=\frac{1}{p_2}\ln\left(\int_E|f|^{p_2}dm\right)\\&=\ln\left(\left(\int_E|f|^{p_2}dm\right)^{1/p_2}\right). \end{align*} Therefore, \begin{align*} \liminf_{p_1\to p_2}\|f\|_{p_1}\geq\left(\int_E|f|^{p_2}dm\right)^{1/p_2}=\|f\|_{p_2}. \end{align*}
To prove part (a), if $p_2<\infty$ apply Holder's inequality with exponent $q=p_2/p_1>1$ to get \begin{align}\left(\int_E|f|^{p_1}dm\right)^{1/p_1}&\le \left(\int_E(|f|^{p_1})^{q}dm\right)^{1/(qp_1)}\left(\int_E1^{q'}dm\right)^{1/(q'p_1)}\\ &=\left(\int_E|f|^{p_2}dm\right)^{1/p_2}\left(m(E)\right)^{\frac{1}{p_1}-\frac{1}{p_2}},\end{align} since $\frac1{p_1q'}=\frac{1}{p_1}-\frac1{qp_1}=\frac{1}{p_1}-\frac{1}{p_2}$. If you let $p_1\to p_2$ you get part (a). The case $p_2=\infty$ is even simpler.