$\displaystyle y_{n}(x)=\sin(x), 0\leq |x| \leq 2\pi n$ and
$\displaystyle y_{n}(x)=0, 2\pi n< |x| <\infty$
$\displaystyle y_n(x)$ converges pointwise to $\displaystyle \sin(x)$ ,but for all $\displaystyle n$ there exists $\displaystyle x_n$ such that $\displaystyle |y_n(x_n)-\sin(x_n)|=1$.