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

Question

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.

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Is the demonstration correct?

Answer 1

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 $$ |\log(1+t)|=\left|\sum^{\infty}_{n=1}\frac{(-1)^nt^n}{n}\right|\leq|t|\sum^{\infty}_{n=1}\frac{1}{1\cdot2^{n-1}}=2|t|. $$

Answer 2

As you said, we have the inequality

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

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

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

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

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

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