Q. Let $p_n(x)=a_n x^2+b_n x$ be a sequence of quadratic polynomials where $a_n,b_n \in \Bbb R$ $\forall n \ge 1$. Let $\lambda_0$,$\lambda_1$ be distinct nonzero real numbers such that $\lim_{n \to \infty}p_n(\lambda_0)$ and $\lim_{n\to \infty}p_n(\lambda_1)$ exist.
Then,
$1. \lim_{n\to \infty} p_n(x)$ exists for all $x \in \Bbb R$.
$2. \lim_{n\to \infty} p_n'(x)$ exists for all $x\in \Bbb R$.
$3. \lim_{n\to \infty}p_n(\frac {\lambda_0+\lambda_1}2)$ does not exist.
$4. \lim_{n\to \infty}p_n'(\frac {\lambda_0+\lambda_1}2)$ does not exist.
$p_n(\frac {\lambda_0+\lambda_1}2)=\frac {a_n}{4}(\lambda_0^2+\lambda_1^2)+\frac {b_n}{2} (\lambda_0+\lambda_1)+\frac {\lambda_0\lambda_1}2 \ldots\ldots (1)$
Also,
$p_n(\lambda_0)=a_n \lambda_0^2 +b_n \lambda_0 \ldots\ldots(2)$
and
$p_n(\lambda_1)=a_n \lambda_1^2 +b_n \lambda_1 \dots\ldots(3)$ which exist as $n \to \infty$ .
Hence $(1)$ exists as $n\to \infty$ (using $(2)$ and $(3)$) $\Rightarrow$ option 3 is false.
Option 4 seems to be true because $p_n'(\frac {\lambda_0+\lambda_1}2)=a_n \lambda_0+a_n \lambda_1+b_n$ and it is possible that $(b_n)$ may be divergent.
I have no idea about about options 1 and 2 :( Can you help?
Let $u_n=p_n(\lambda_0)$ and $v_n=p_n(\lambda_1)$. Then $(u_n)$ converges to some limit $u$ and $(v_n)$ converges to some limit $v$.
Then
$$ a_n\lambda_0^2+b_n\lambda_0=u_n, \ a_n\lambda_1^2+b_n\lambda_1=v_n \tag{1} $$
Solving this $2\times 2$ system for $a_n,b_n$, we obtain
$$ a_n=\frac{-\lambda_1 u_n +\lambda_0 v_n}{(\lambda_1-\lambda_0)\lambda_0\lambda_1}, b_n=\frac{\lambda_1^2 u_n -\lambda_0^2 v_n}{(\lambda_1-\lambda_0)\lambda_0\lambda_1} \tag{2} $$
whence
$$ p_n(x)=\frac{(-\lambda_1 u_n +\lambda_0 v_n)x^2+(\lambda_1^2 u_n -\lambda_0^2 v_n)x}{(\lambda_1-\lambda_0)\lambda_0\lambda_1} \tag{3} $$
So that $(p_n(x))$ converges to $\frac{(-\lambda_1 u +\lambda_0 v)x^2+(\lambda_1^2 u-\lambda_0^2 v)x}{(\lambda_1-\lambda_0)\lambda_0\lambda_1}$. This shows that (1) is true. Similarly (2) is true