# 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

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

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

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 $$-\infty$$.

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 $$f(x)>0.$$