Find the limit of this sequence $\lim_{n\to \infty}\frac{n}{1 + \frac{1}{n}} - n$ Find the limit of this sequence $$\lim_{n\to \infty}\frac{n}{1 + \frac{1}{n}} - n$$
First I tried dividing everything by $n$ but that would leave me with $$\lim_{n\to \infty}\frac{1}{\frac{1}{n} + \frac{1}{n^2}} - 1$$
and as $n\to \infty$ i'd be left with $\frac{1}{0} - 1$. Would I be correct in saying that the limit is -1 or does the $\frac{1}{0}$ mess that up?
 A: $$\frac{n}{1 + \frac{1}{n}} - n=\frac{n^2}{n+1}-n=\frac{n^2-n(n+1)}{n+1}=\frac{-n}{n+1}\to-1$$
A: The limit is $-1$ but the $\frac{1}{0} $ does mess it up.
Just add the fractions:
you get $$\frac{n^2}{n+1} - n = \frac{n^2 - n^2 - n}{n} = \frac{-n}{n+1}$$
Finding the limit should now be easy
A: Note that the individual terms are negative - something a bit smaller than $n$ minus $n$ - always worth doing a reality check on the kind of answer you are expecting.
Simplifying the fraction - as others have noted - gives you $$-\frac n{n+1} = -1+ \frac 1{n+1}$$ I have sometimes found the second form here useful, as it makes the limit explicit - but sometimes it is more calculation and prone to error as a result.
A: First of all, you can't divide everything by $n$.  You can divide the numerator and denominator by $n$ as that doesn't change the value of a fraction.  In this case, though, all you need to do is put both terms over the same denominator.
$$\cfrac n{1+\frac1n}-n=\cfrac{n-(n+1)}{1+\frac1n}=-\cfrac1{1+\frac1n}$$
Now it should be easy to take the limit as $n$ goes to infinity.
