Compute $\lim_{n\to \infty} \frac{(n^2+1)}{(n+1)^2} $ I am having some trouble understanding a concept with limits.
The question is $$\lim_{n\to \infty} \frac{(n^2+1)}{(n+1)^2} $$
I understand the method of dividing by $n^2$, and coming to the answer of $1$. However why is it not valid to write that this limit is equal to $$\lim_{n\to \infty} \frac{(n^2+1)}{n^2+2n+1} = 
\lim_{n\to \infty} \frac{(n^2+1)}{(n^2+1)+2n} $$ and then to to cancel leaving with $$\lim_{n\to \infty} \frac{1}{2n} $$ which would be equal to zero not 1? I feel like I may be missing some basic arithmetic or some silly error?
Thanks all
 A: You cant cancel the way you did in the last step, for example:$$\frac{x}{x +2} \neq \frac{1}{2}$$
Just put $x=1$ to see that the $2$ terms are a different thing.
A: I believe I would categorize what you have done as a "silly error."
The numerator of the fraction is $n^2+1$, and the denominator of the fraction is $n^2+2n+1$.  You then inserted parentheses, but there is an addition after the closing parenthesis.  You can only "cancel" a factor from the numerator and denominator if it is indeed a FACTOR.  In this case numerator and denominator do not share any common factors because the numerator is unfactorable over the real numbers and the denominator does not have the numerator as a factor.  Therefore the fraction cannot be simplified.
A: I think that beyond an algebraic understanding, one should address the more fundamental issue of what happens to the expression as $n \to \infty$.  For instance, if $n = 1$, then $$\frac{n^2 + 1}{(n^2+1) + 2n} = \frac{2}{2+2},$$ and it seems like the comparison to $\frac{1}{2n}$ is okay.  But when $n = 10$, then $$\frac{n^2 + 1}{(n^2+1) + 2n} = \frac{101}{101 + 20}$$ whereas $\frac{1}{2(10)} = \frac{1}{20}.$  And when $n = 1000$, we can now see what is going wrong much more clearly:  $$\frac{n^2 + 1}{(n^2+1) + 2n} = \frac{1000001}{1000001 + 2000}.$$  As $n$ gets larger, the growth of $n^2 + 1$ far outpaces that of $2n$, to the extent that if $n$ is enormous, that the addition of $2n$ in the denominator is trivial; thus the numerator and denominator are approximately equal for such large values.  To properly cancel out the term $n^2+1$, we must write $$\frac{n^2 + 1}{(n^2+1) + 2n} = \frac{1}{1 + \frac{2n}{n^2+1}},$$ and again, because $n^2+1$ grows much faster than $2n$, the fraction $\frac{2n}{n^2+1}$ will tend toward zero.
A formal proof of these claims is straightforward and have largely been addressed in other answers, but I felt it important to revisit the idea of a limit at infinity as being like a process of choosing ever larger values of $n$ and seeing what happens as a result.
A: The expanded form is :
$$
\frac{n^2+1}{(n+1)^2}=\frac{n^2}{(n+1)^2}+\frac{1}{(n+1)^2} \neq \frac{1}{2n} 
$$
Shadock :)
