Consider the basic linear model:
$$y_i =\beta_0 +\beta_1 x_i +\varepsilon_i$$
I am aware that if $E[\varepsilon_i|x_i ]=0$ then $E[\hat{\beta_1}]=\beta_1$ (unbiasedness) and also that if $Cov(x_i,\varepsilon_i)=0$ then $plim \hat{\beta_1} = \beta_1$ (consistency).
I am hoping to prove that if $Cov(x_i,\varepsilon_i) \ne 0$ then $E[\hat{\beta_1}]\ne\beta_1$. Can anyone help? My attempt begins below.
$$\hat{\beta_1}=\frac{\sum_{i=1}^n (x_i-\bar{x})(y_i-\bar{y})}{\sum_{i=1}^n (x_i-\bar{x})^2}$$
$$\hat{\beta_1}=\beta_1 + \frac{\sum_{i=1}^n (x_i-\bar{x})(\varepsilon_i-\bar{\varepsilon})}{\sum_{i=1}^n (x_i-\bar{x})^2}$$
$$E[\hat{\beta_1}]=\beta_1 + E\left[\frac{\sum_{i=1}^n (x_i-\bar{x})(\varepsilon_i-\bar{\varepsilon})}{\sum_{i=1}^n (x_i-\bar{x})^2}\right]$$
My goal is to show the expectation on the right must be non-zero as a result of $Cov(x,\varepsilon)\ne 0$. The difficulty is that the expectation operator can be separately applied to the numerator and denominator. The numerator is in expectation non-zero, but if the sample covariance of $x$ and $\varepsilon$ is not independent of the sample variance of $x$, I don't know how to proceed.
Thanks in advance!