Orthogonality of limits in arbitrary inner product space

Question

This is a question I'm having quite a bit of trouble with.

Let $(X,\langle\dot{},\dot{}\rangle)$ be an inner product space and $\{{x_n}\}$ be a sequence in $X$ that converges to some $x\in X$ in the norm $||\dot{} ||$ induced by $ \langle \dot{} , \dot{} \rangle $; i.e. $\lim_{n\to \infty} ||x- {x_n } ||=0 $. If $x_n$ is orthogonal to a fixed nonzero element $y \in X $ for all $n$ then how do I prove that the limit $x$ is also orthogonal to $y$.

I can't make the jump from showing that all elements of $\{ x_n \}$ are orthogonal to $y$ to proving that its limit is as well. The limit gives us information about the norm of the limit, but I can't see the connection that allows us to say that if two things have the same norm and one of those things is orthogonal to another, then the other of those two is as well.

Answer 1

$$ |\langle x,y \rangle |= |\langle x - x_n,y \rangle| \le \|x_n - x\| \|y\| \to 0.$$

Answer 2

$\langle x_{n}{},y\rangle = 0 $ for all $x_{n}$ hence $\lim\langle x_{n}{},y\rangle = 0 $ which by continuity of the scalar product means that the limit also is orthagonal to $y$.

Also if the norm is zero then the element must be zero.