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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

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  • $\begingroup$ Taylor series ? $\endgroup$ – Thomas Jan 18 at 7:06
  • $\begingroup$ $\lim_{n \to \infty}n{{(\frac{f(x)}n)}^p} \dfrac{\log\left (1+ \left(\frac{f(x)}{n}\right)^p\right)}{{(\frac{f(x)}n)}^p}$ and the last factor goes to 1. $\endgroup$ – PNDas Jan 18 at 7:33
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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 $-\infty$.

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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 $f(x)>0.$

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