I want to find the tangent on point $(x_0, y_0)$ to an ellipse with equation $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$.
We assume $y_0$ is positive. We can derive $y=\frac b a \sqrt{a^2-x^2}$.
In order to find the slope I use the limit $$ \lim_{x \to x_0}\frac b a \frac{\sqrt{a^2-x^2}-\sqrt{a^2-x_0^2}}{x-x_0}=\lim_{x \to x_0}\frac b a \frac{a^2-x^2-a^2+x_0^2}{(x-x_0)(\sqrt{a^2-x^2}+\sqrt{a^2-x_0^2})} $$ The amount of this limit is obviously zero. Where's the problem?
In the RHS of that equality, the numerator should be $(a^2-x^2)-(a^2-x_0^{\,2})$, which is equal to $-x^2+x_0^{\,2}$. This, in turn, is equal to $-(x-x_0)(x+x_0)$. So, your limit becomes$$-\lim_{x\to x_0}\frac ba\frac{x+x_0}{\sqrt{a^2-x^2}+\sqrt{a^2-x_0^{\,2}}}.$$Can you take it from here?