I want to prove $\lim_{(x,y) \to (0,0)} \frac{ e^{2x}+1-2e^{x}\cos{y} -2(xe^{x}\cos{y} + ye^{x}\sin{y} - x)}{x^2+y^2} = -1$
I divide it to two terms and wolfram alpha says that, $\lim_{(x,y) \to (0,0)} \frac{ e^{2x}+1-2e^{x}\cos{y}}{x^2+y^2} = 1$, $\lim_{(x,y) \to (0,0)}\frac{xe^{x}\cos{y} + ye^{x}\sin{y} - x}{x^2+y^2} = 1$
It looks too complicate to use epsilon-delta argument. Also polar transformation seems not reasonable since we have $x$ and $y$ separately into exponential and $\cos$, $\sin$ function. How can I prove these limits?
You can use the first few terms of the Taylor series to simplify the expressions.
$ \begin{align} &\lim_{(x,y) \to (0,0)} \frac{ e^{2x}+1-2e^{x}\cos{y}}{x^2+y^2}\\ \\ &=\lim_{(x,y) \to (0,0)} \frac{1+2x+\frac{(2x)^2}{2}+1-2\left(1+x+\frac{x^2}{2}\right)\left(1-\frac{y^2}{2}\right)}{x^2+y^2}\\ \\ &=\lim_{(x,y) \to (0,0)} \frac{x^2+y^2+H.O.T.}{x^2+y^2}\\ \\ &=1 \end{align} $ $$\\$$ $ \begin{align} &\lim_{(x,y) \to (0,0)}\frac{xe^{x}\cos{y} + ye^{x}\sin{y} - x}{x^2+y^2}\\ \\ &=\lim_{(x,y) \to (0,0)}\frac{x(1+x)(1)+y(1)(y)-x}{x^2+y^2}\\ \\ &=\lim_{(x,y) \to (0,0)}\frac{x^2+y^2}{x^2+y^2}\\ \\ &=1 \end{align} $
We take only as many terms of the Taylor series as are required. The higher order terms vanish as the limit tends to the origin.