The problem:
Find the linear approximation of the function $$ f(x,y) = \ln(e+x+y) $$ at point $(0, 0)$. Use it to approximate the value of the function at $(0.1, 0.2)$
What I have so far:
I found the gradient of $f$ to be $$ \nabla f = \left\langle \frac{1}{x}, \frac{1}{y} \right\rangle. $$
However, I end up dividing by $0$ when plugging the point $(0, 0)$ into the gradient.
What property am I missing? Or have I taken a wrong direction altogether?
First of all, $$ \nabla f = \left\langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right\rangle = \left\langle \frac{1}{e + x + y}, \frac{1}{e + x + y} \right\rangle. $$
Evaluating both the function and its gradient at $(x, y) = (0, 0)$ gives $$ f(0, 0) = 0, \qquad \nabla f(0, 0) = \bigl\langle \tfrac{1}{e}, \tfrac{1}{e} \bigr\rangle, $$ so for nearby $(x, y)$, the linear approximation $L$ is given by $$ \begin{align} L(x, y) &= f(0, 0) + \nabla f(0, 0) \cdot \langle x - 0, y - 0 \rangle \\ &= 0 + \bigl\langle \tfrac{1}{e}, \tfrac{1}{e} \bigr\rangle \cdot \langle x, y \rangle \\ &= \frac{x}{e} + \frac{y}{e} \\ &= \frac{x + y}{e}. \end{align} $$
So, $$ L(0.1, 0.2) = \frac{0.3}{e} \approx 0.1104. $$