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\textbf{Partial Differentiation}

\textit{\textbf{Limits}}
\end{center}

\textbf{Question}

Evaluate the given limit. If the limit does not
exist, explain why.

$$ \lim_{(x,y)\to(1,2)} \frac{2x^2-xy}{4x^2-y^2}$$


\textbf{Answer}

The fraction is not defined on points of the line $y=2x$.

Therefore it cannot have a limit at $(1,2)$. But cancelling $2x-y$ gives 

$\ds \lim_{(x,y)\to(1,2)}
\frac{2x^2-xy}{4x^2-y^2}=\lim_{(x,y)\to(1,2)} \frac{x}{2x+ y}=\frac{1}{4}$


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