Finding $\lim_{(x,y)\to (1,0)} \frac{(x-1)^2\log(x)}{(x-1)^2 + y^2}$ if it exists

Question

Show that $\displaystyle\lim_{(x,y)\to (1,0)} \frac{(x-1)^2\log(x)}{(x-1)^2 + y^2}$ exists. Also find the limit.

Method 1

$\frac{(x-1)^2}{(x-1)^2 + y^2}$ is less than $1$. Therefore, $\frac{(x-1)^2\log(x)}{(x-1)^2 + y^2}<\log(x)$. Also, $\frac{(x-1)^2\log(x)}{(x-1)^2 + y^2} \geq 0$ for all $(x,y)$ in the domain.

Therefore, we have,

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

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

Method 2

Let $\varepsilon>0$. Choose $\delta<e^{\varepsilon}-1 \implies\log(\delta+1)<\varepsilon$.

If,

$(x-1)^2+y^2<\delta^2 \implies |x-1|<\delta \implies |x|-1<\delta$

$\implies x<\delta +1$ $ (\because$ $\log(x)$ is not defined for $x<0)\implies \log(x)<\log(\delta+1)$,

then, $$\left|\frac{(x-1)^2\log(x)}{(x-1)^2 + y^2}-0\right| \leq |\log(x)|< |\log(\delta+1)|<|\varepsilon| = \varepsilon$$

Hence, by $\varepsilon$-$\delta$ definition of limits, $\displaystyle \lim_{(x,y)\to (1,0)} \frac{(x-1)^2\log(x)}{(x-1)^2 + y^2}$ exists and is equal to $0$.

Are both my methods correct?

Answer 1

You have the main idea correct. But there are details that you need to be careful as some of the comments already point out. For instance $\log(x)\ge 0$ is not always true.

Also, your "method 2" would be less cumbersome if you apply continuity of the logarithm instead of trying to solve an inequality in a sharp way to get $\delta$.

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

Here is a short proof.

Proof. Observe that $$ 0\le \left|\frac{(x-1)^2\log(x)}{(x-1)^2+y^2}\right|= \left|\frac{(x-1)^2}{(x-1)^2+y^2}\right||\log(x)| \le |\log(x)|\;. $$ Also $\displaystyle \lim_{(x,y)\to(1,0)}|\log(x)|=0$. Thus by the "squeeze-type" theorem, $$ \lim_{(x,y)\to(1,0)}\frac{(x-1)^2\log(x)}{(x-1)^2+y^2} = 0\;. $$

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I don't see an explicit "squeeze theorem" of the multivariable function version. Though the idea is of course the same. One may write out the detail of $$\lim_{(x,y)\to(1,0)}|\log(x)|=0\tag{1}$$ as follows.

Let $\epsilon>0$. By continuity of the logarithm (and the absolute value function) at $x=1$, there exists $\delta>0$ so that $$ |\log(x)|<\epsilon\quad\textrm{whenever}\quad |x-1|<\delta\;. $$ Now, if $|(x,y)-(1,0)|<\delta$, then $$ |x-1|<\sqrt{|x-1|^2+|y|^2}<\delta $$ and hence $|\log(x)|<\epsilon$.

The estimates above could also give you an $\epsilon$-$\delta$ proof of the limit.