Limit of the sequence of intersection point OR intersection point of the limit of the graph My classmate ask me a question as follow.
We don't know that where is the question come from.
For all $n\in \Bbb{N}$.
Let $P_n$ be the parabola $y=x^2-nx+1$
and $L_n$ be the straight line $y=nx$.
Find the intersection points when $n$ tends to infinity.
Solution: 
Substitute $y=nx$ to $y=x^2-nx+1$, 
we get $nx=x^2-nx+1$ and $x^2-2nx+1=0$.
Solve $x$, 
we have $$x=\frac{2n\pm \sqrt{4n^2-4}}{2}=n\pm \sqrt{n^2-1}.$$
We pick the $n-\sqrt{n^2-1}$ 
because the other one stands for the $x$-coordinate 
of the intersection point at infinity.
Then $$\lim_{n\to \infty}x=\lim_{n\to \infty}(n-\sqrt{n^2-1})=\lim_{n\to \infty}\frac{(n-\sqrt{n^2-1})(n+\sqrt{n^2-1})}{n+\sqrt{n^2-1}}\\=\lim_{n\to \infty}\frac{1}{n+\sqrt{n^2-1}}=0.$$
The difference between our solutions is here.
My classmate substitute $x=0$ to $y=x^2-nx+1$
and get $y=1$.
He explain that the order  pair $(x,y)=(0,1)$ 
is the intersection point of the two graph $\Gamma_1$ and $\Gamma_2$,
where $\Gamma_1:x=0$, which is the "limit" of $y=nx$ when  $n$ tends to infinity
and $\Gamma_2$ is a graph which always pass through $(0,1)$,
$\Gamma_2$ is the "limit" of $y=x^2-nx+1$ when  $n$ tends to infinity.
My solution is:
Substitute $x=n-\sqrt{n^2-1}$ to $y=nx$,
we get $y=n(n-\sqrt{n^2-1})$.
Then $$\lim_{n\to \infty}y=\lim_{n\to \infty}n(n-\sqrt{n^2-1})=\lim_{n\to \infty}\frac{n}{n+\sqrt{n^2-1}}\\=\lim_{n\to \infty}\frac{n/n}{n/n+\sqrt{\frac{n^2-1}{n^2}}}=\lim_{n\to \infty}\frac{1}{1+\sqrt{1-\frac{1}{n^2}}}=\frac{1}{2}.$$
I think my explanation of the problem is: 
"$(0,\frac{1}{2})$ is the limit of the sequence of intersection points of $y=x^2-nx+1$ and $y=nx$ when $n$ tends to infinity.
Who is right?
 A: You are right. Denote by $(x_n, y_n)$ the intersection you write about. You have computed $$\lim_{n\to \infty} x_n = 0 \quad \text{and}\quad \lim_{n\to \infty} y_n = \frac{1}{2} $$
without any suspicious steps. He, on the other hand, computes $\lim_{n\to \infty} y_n = 1$ as
$$\lim_{n\to \infty} y_n =\lim_{n\to \infty}\left[ 1 +(\lim_{m\to\infty} x_m)^2 - n\lim_{m\to\infty} x_m\right] = \lim_{n\to \infty}\left[1 + 0^2 - n\cdot 0\right] = 1\text{,}  $$
where he makes a mistake, since he should have computed the limit as
$$\lim_{n\to \infty} y_n =\lim_{n\to \infty} 1 + x_n^2 - n x_n \quad\left(\text{or}\quad \lim_{n\to\infty} nx_n\right)$$
like you did. Doing his way, he (indirectly) claims that $n x_n \to 0$, but it is known, that limits of the type $\infty\cdot 0$ cannot be computed in the way he did it. For example, if $a_n = n$, $b_n = 1/n^2$, $c_n = 1/n$ and $d_n = 1/ \sqrt{n}$, then $\lim_{n\to \infty} a_n b_n = 0$, $\lim_{n\to \infty} a_nc_n=1$ and $\lim_{n\to \infty} a_n d_n = \infty$, eventhough $\lim_{n\to \infty} b_n=\lim_{n\to \infty} c_n=\lim_{n\to \infty} d_n=0$.
A: Not a detailed solution, just an intuitive one. The polynomial $y=x^2-nx+1$ irrevelant of $n$, passes through the point $A(0,1)$. The line $y=\lambda x$ as $\lambda \to \infty$ tends to become $x=0$. Combining these, the intersection point is $A(0,1)$.
Judging by the other answer, I leave this here to contradict the popular belief that an intuitive answer not mathematically proven is correct.
