Does $\displaystyle\lim_{n\to\infty}\frac{n}{n+1} = 1$? $\displaystyle\lim_{n\to\infty}\frac{n}{n+1} = 1$? Is that true? If you plug in infinity for $n$, it would be infinity/infinity and that equals $1$? Please help by letting me know why in a very simple way. Thank you.
 A: Hint: Note that
$$
\frac{n}{n+1}=1-\frac1{n+1}
$$
A: The $\displaystyle\lim_{n\to\infty}\frac{n}{n+1} = 1$. When you see $\infty \over \infty$ do not treat it like $n\over n$ where $n\in \mathbb{R}$. So ${\infty \over \infty} \neq1$, it is an indeterminate form, we do not know what it is. However, we can still approach this problem in a few different ways to determine that the limit does equal $1$. Consider $$\lim_{n\rightarrow\infty}{n\over n+1}.$$ We can multiply the numerator and the denominator by $1\over n$ and obtain $$\lim_{n\rightarrow\infty}{1\over 1+{1\over n}}.$$ Now we can see that as $n\rightarrow\infty$, ${1\over n}\rightarrow 0$ and so  $\displaystyle\lim_{n\to\infty}\frac{1}{1+{1\over n}} = 1$. Thus $\displaystyle\lim_{n\to\infty}\frac{n}{n+1} = 1$. Another approach is to notice that $\displaystyle\lim_{n\to\infty}\frac{n}{n+1}$ has the indeterminate form $\infty\over \infty$. So we can apply L'Hospital's Rule to simplify this limit. Taking the derivative of the numerator and the denominator we see that $\displaystyle\lim_{n\to\infty}{1\over 1} = 1$.
