Find $ \lim_{n \to \infty}n \log\left(1+ \left(\frac{f(x)}{n}\right)^p\right)$ Find $$\lim_{n \to \infty}n \log\left (1+ \left(\frac{f(x)}{n}\right)^p\right)$$  where $0<p<1$
my attempt :By using L-Hospital Rule i got
$$\lim_{n\to \infty} \dfrac  {\dfrac{1}{1+\left( \dfrac{f(x)}{n}\right)^p} 
\cdot\dfrac {f(x)^p}{-n^{2p}} }{\dfrac{-1}{n^2}}$$
After that I’m   not able to   proceed further
 A: Let $y = f(x)$. Consider the essential case $y \neq 0$.
\begin{align*}
n\log\left(1 + \left(\frac{y}{n}\right)^p\right) &
= y^pn^{1-p}\log \left(\left(  1 + \left( \frac{1}{\left(\frac{n}{y} \right)^p}\right)\right)^{\left(\frac{n}{y}\right)^p}\right).
\end{align*}
Since $\left\{\left(\frac{n}{y}\right)^p \right\}_{n \in \mathbb{N}}$ is monotonic,
\begin{align*}
\log \left(\left(  1 + \left( \frac{1}{\left(\frac{n}{y} \right)^p}\right)\right)^{\left(\frac{n}{y}\right)^p}\right) \to 1.
\end{align*}
But
\begin{align*}
y^pn^{1-p} \to +\infty \ or \ -\infty.
\end{align*}
Thus the limit is $+\infty$ or $-\infty$.
A: There are problems of defintion in $f(x)^p$ if $f(x)<0,$ so we should assume $f(x)\ge 0.$
If $f(x)=0,$ then the expression equals $0$ for all $n,$ so the limit is $0.$
If $f(x)>0,$ we can use
$$\lim_{u\to 0}\ln(1+u)/u= 1,$$
which is nothing more than the definition of $\ln'(1),$ which equals $1.$
Our expression equals
$$n(f(x)/n)^{-p}\cdot\frac{\ln(1+(f(x)/n))^p}{(f(x)/n)^p}.$$
The first factor equals $n^{1-p}f(x)^{-p},$ which has limit $\infty.$ The second factor $\to 1$ by $(1).$ Thus the limit equals $\infty$ in the case $f(x)>0.$
