# Limit by polar coordinates $\lim_{(x,y)\to(1,0)} \frac{y^2\log(x)}{(x-1)^2+y^2}=0$

I need to demonstrate the following limit

$$\lim_{(x,y)\to(1,0)} \frac{y^2\log(x)}{(x-1)^2+y^2}=0$$

By polar coordinates.

I apply the substitution $$x=1+\rho\cos(\theta)$$, $$y=\rho\sin(\theta)$$.

$$\lim_{\rho\to0} \frac{\rho^2\sin^2(\theta)\log(1+\rho\cos(\theta))}{\rho^2}=\lim_{\rho\to0} \sin^2(\theta)\log(1+\rho\cos(\theta))$$ Here I have some issues with the following steps.

$$|\sin^2(\theta)\log(1+\rho\cos(\theta))|\leq\rho|\sin^2(\theta)\cos(\theta)|=\rho\to0$$

I'm not sure about if that step is correct $$|\log(1+\rho\cos(\theta))|\leq\rho\cos(\theta)$$

My thought process was that I observed that

$$0\leq|\log(x)|\leq x-1$$ for $$x\ge1$$. Thus, by setting $$t=x-1 \to x=t+1$$, I conclude that

$$0\leq|\log(t+1)|\leq t$$

For $$t\ge0$$. Hence since $$\cos(\theta)$$ is a bounded value I conclude that $$|\log(1+\rho\cos(\theta))|\leq\rho\cos(\theta)$$ For $$\rho\ge0$$.

And with this inequality I conclude that the limit is $$0$$ as I've show above.

Is the demonstration correct?

Since $$t=\rho\cos(\theta)$$ could be negative as $$\rho\to 0^+$$, it is not sufficient to consider the inequality $$|\log(1+t)|\leq t$$ which holds for $$t\geq 0$$. However your proof works as soon as you replace the above identity with $$|\log(1+t)|\leq 2|t|$$ which holds in a full neighbourhood of $$t=0$$, i.e. $$|t|\leq \frac{1}{2}$$.

P.S. Proof of the inequality: for $$|t|\leq\frac{1}{2}$$ we find $$|\log(1+t)|=\left|\sum^{\infty}_{n=1}\frac{(-1)^nt^n}{n}\right|\leq|t|\sum^{\infty}_{n=1}\frac{1}{1\cdot2^{n-1}}=2|t|.$$

As you said, we have the inequality

$$\ln(x)1}$$

By substituting $$x\mapsto1/x$$, we then get

$$\ln(1/x)<\frac1x-1=\frac{1-x}x\tag{0

Using log properties and multiplying both sides by $$-1$$, we get

$$\ln(x)>\frac{x-1}x\tag{0

However, as $$x\to1$$, the denominator becomes negligible, hence why it can be omitted when taking the limit.