$\lim_ {n\to \infty}\ \frac{n}{n+1}=1$
I'd write it like this:
$\lim_ {n\to \infty}\ \frac{n}{n}+\frac{n}{1}=\lim_ {n\to \infty}\ 1+n=\infty$
What am I missing?
$\lim_ {n\to \infty}\ \frac{n}{n+1}=1$
I'd write it like this:
$\lim_ {n\to \infty}\ \frac{n}{n}+\frac{n}{1}=\lim_ {n\to \infty}\ 1+n=\infty$
What am I missing?
Since no one has said so explicitly, to handle a limit of this form (taking a limit at infinity of a rational expression), start of by dividing every additive term by the highest power of $n$ that you see. More precisely, multiply both top and bottom by the reciprocal of the highest power of $n$ that you see; which, in this case is just $n$:
$ \lim_{n\rightarrow\infty}{n\over n+1}= \lim_{n\rightarrow\infty}\Bigl[\underbrace{{1/ n}\over{1/ n}}_{=1}\cdot{n\over n+1} \Bigr] = \lim_{n\rightarrow\infty}{ {1\over n} \cdot n \over {{1\over n} (n+1)}} = \lim_{n\rightarrow\infty}{ 1 \over {1+{1\over n} }} ={1\over 1+0}=1. $
What you are missing is that, in general, $\frac{a}{b+c}\neq\frac{a}{b}+\frac{a}{c}.$ That is, your two functions are not equal.
Note that $\frac{n}{n}+\frac{n}{1} = 1+n$, whereas $\frac{n}{n+1}\lt 1$ for all $n\gt 0$.
What you can do is to instead consider the reciprocal: $\lim_{n\to\infty}\frac{n}{n+1} = \lim_{n\to\infty}\frac{1}{\quad\frac{n+1}{n}\quad} = \lim_{n\to\infty}\frac{1}{\frac{n}{n}+\frac{1}{n}} = \lim_{n\to\infty}\frac{1}{1+\frac{1}{n}}.$ Now notice that $\frac{1}{n}\to 0$ as $n\to\infty$.
You can better transform your limit, doing:
n+1 = x, so x-1=n, then your new limit pass from:
$\lim_ {n\to \infty}\ \frac{n}{n+1}=1$
to:
$\lim_ {x\to \infty}\ \frac{x-1}{x}=\lim_ {x\to \infty}\ \frac{x}{x}-\frac{1}{x}=1+0=1$
Yout problem is in the way you separate the terms.
Yet another way...
$\frac{n}{n+1}$
can be written as:
$\frac{n+(1-1)}{n+1}$
which is:
$\frac{n+1}{n+1} -\frac{1}{n+1} = 1 - \frac{1}{n+1}$
so
$\lim_ {n\to \infty}\ \frac{n}{n+1} =\lim_ {n\to \infty}\ (1 - \frac{1}{n+1})$
since limit of 1/(n+1) as n approaches infinity is 0, and since the limit of 1 as n approaches infinity is 1, then the desired limit is = $1 - 0 =1$