Vtyjkyuk

Because the partial derivatives exist in $G$, the mean-value theorem says that there is a number $x^*$ between $x_0$ and $x_0 + Delta x$ such that

$u(x_0 + Delta x, y_0 + Delta y)-u(x_0,y_0 + Delta y) = Delta x fracpartial upartial x(x*, y_0 + Delta y)$

Furthermore, since the partial derivatives are continuous at $(x_0,y_0)$, we can write

$fracpartial upartial x(x^*, y_0 + Delta y)=fracpartial upartial x(x_0, y_0) + epsilon$

I'm having trouble understanding the following statement

$fracpartial upartial x(x^*, y_0 + Delta y)=fracpartial upartial x(x_0, y_0) + epsilon$

How do we derive the previous equality using continuity?

By definition of continuity
$$lim_(x,y)to(x_0,y_0) fracpartial upartial x(x,y) =fracpartial upartial x(x_0,y_0)$$ since $fracpartial upartial x$ is continuous at $(x_0,y_0)$.

Regardless of continuity we can define the function $$epsilon = fracpartial upartial x(x^*,y_0+Delta y)-fracpartial upartial x(x_0,y_0) .$$
By the above mentioned continuity you get $epsilonto 0$ if $x^*to 0$ and $Delta yto 0$.

This is simply an alternate expression for continuity of $partial_x u$ at $x_0, y_0$. Rewrite the definition of continuity.