Show that a limit with natural logarithm does not exist (two variables)

Question

I can't prove this limit below doesn't exist. $$\lim_{x \to 1 \; y \to 0} \frac {\ln(x + y)} {y}$$ I already know that $$\lim_{x \to 0} \frac {\ln(1 + x)} {x} = 1$$ It seems absurd but the limit of two variables is different from ones. Have you guys got any ideas. Tks.

Answer 1

Using the standard limit you quote, we can write $$\frac {\ln(x + y)} {y}=\underbrace{\frac {\ln(1+(x-1 + y))}{x-1+y}}_{\substack{\downarrow\\ \strut1}}\frac {x -1+ y} {y},$$ so the limit, if any, is the same as that of $$\frac{x-1+y}{y}=1+\frac{x-1}{y}.$$ However, using polar coordinates: $x=1+r\cos\theta,\; y=r\sin\theta$, it is easy to see that the limit of $$\frac{x-1}{y}=\frac{\cos\theta}{\sin\theta}$$ does not exist since its value would depend on $\theta$.

Answer 2

Put $x=1+\frac 1 n$ and $y =\frac 1 {n^{2}}$. Note that $\frac {\ln (1+\frac 1 n+\frac 1 {n^{2}})} {1/n^{2}}=\frac {\ln (1+\frac 1 n+\frac 1 {n^{2}})} {\frac 1 n+\frac 1 {n^{2}}} \frac {\frac 1 n+\frac 1 {n^{2}}} {1/n^{2}}\to (1)(\infty)=\infty$.