Limit of a sequence of functions in the Lebesgue Dominated Convergence Theorem I am currently studying the Lebesgue Dominated Convergence Theorem and, to apply it, I must calculate the following.
$$
\lim_{n\to\infty} \frac{1+xn}{(1+x)^n}$$
where $x\in (0,1]$.
I know the limit exists using the ratio test however I thought the limit was 0, which my lecturer has queried. Would someone be able to indicate how to calculate the limit please.
 A: It should be valid only for $x\in(0,1]$.
Use that $(1+x)^{n}\geq 1+nx+\dfrac{n(n-1)}{2}x^{2}$, then
\begin{align*}
\dfrac{1+nx}{(1+x)^{n}}\leq\dfrac{1+nx}{1+nx+\dfrac{n(n-1)}{2}x^{2}}=\dfrac{\dfrac{1}{n}+x}{\dfrac{1}{n}+x+\dfrac{n-1}{2}x^{2}}\rightarrow 0.
\end{align*}
A: We will show an alternative approach.
As others have pointed out the limit $l$ equals $0$ when $x\in(0,1]$.
In that case it is true that $1+x\gt1$ and thus $$\lim_{n\rightarrow\infty}(1+x)^n=\infty$$ It now remains to show that $(1+x)^n$ is an increasing sequence for every $n$ so that we can use the Cesàro-Stolz lemma to calculate the limit: $$\lim_{n\rightarrow\infty}\frac{1+nx}{(1+x)^n}=\lim_{n\rightarrow\infty}\frac{(1+(n+1)x)-(1+nx)}{(1+x)^{n+1}-(1+x)^n}=\lim_{n\rightarrow\infty}\frac{x}{(1+x)^n((1+x)-1)}=\\x\lim_{n\rightarrow\infty}\frac{1}{(1+x)^n}=0$$
When $x=0$ it holds $(1+x)^n=1 \ \forall n\in\mathbb{N}$ and $1+nx=1$ so the limit $l=1$ when $x=0$
A: If dominated convergence is what you want, you may want to use the following well known result: $\lim_n\frac{n^p}{(1+q)^n}=0$ for all $p\in\mathbb{R}$ and $q>0$.
To apply dominated convergence then, it is enough to find an appropriate  dominating function, for
$$f_n(x)=\frac{1+nx}{(1+x)^n}=\frac{1-n}{(x+1)^n}+\frac{n}{(1+x)^{n-1}}\xrightarrow{n\rightarrow\infty}0\qquad \text{if} \quad x\neq0,$$
that is, $f_n(x)\xrightarrow{n\rightarrow\infty}0$ a.s. in $(0,1]$.
The binomial theorem is handy here: for $n>2$,
$$(1+x)^n\geq 1+\binom{n}{1}x=1+nx$$
Hence $$ |f_n(x)|\leq1$$ and $\int^1_0\,dx=1<\infty$. Consequently $\lim_n\int^1_0f_n(x)\,dx=0$.

Notice also that  limit of $\int^1_0f_n(x)\,dx$ of the sequence of functions in your problem can be  estimated directly:
\begin{align}
\int^1_0\frac{1+nx}{(1+x)^n}\,dx&=\int^1_0\frac{dx}{(1+x)^n}+ n\int^1_0\frac{x+1-1}{(x+1)^n}\,dx\\
&=(1-n)\frac{(1+x)^{-n+1}}{-n+1}\Big|^1_0 +n\frac{(x+1)^{-n+2}}{-n+2}\Big|^1_0\\
&=-(2^{-n+1}-1)-\frac{n}{n-2}\big( 2^{-n+2}-1)\\
&=1-\frac{n}{n-2}-2^{-n+1}-\frac{n}{n-1}2^{-n+2}\xrightarrow{n\rightarrow\infty}0
\end{align}
This is one instance in which dominated convergence gives a cleaner answer (once the dominating function and the convergence of the integrands $f_n$ have been established.)
