Find limit $\lim_{x \to 0}\frac{\arctan 3x}{\ln((1-x)^2)}$

Question

I have to find the following limit:

$$\lim_{x \to 0}\frac{\arctan3x}{\ln((1-x)^2)}$$

L’Hôpital’s rule only makes things worse. Transforming the numerator and the denominator in Taylor series doesn’t help either. I’ve seen in another question that $\arctan \left(\frac{x}{x}\right) \to 1$, but then I don’t see what to do with $x/(\ldots)$.

Thanks for your help!

Edit: I think I made an error while differentiating, L’Hôpital’s rule works. Thanks for your help.

Answer 1

By L’Hospital ’s rule, one has $$ \lim_{x \to 0}\frac{\arctan(3x)}{\ln((1-x)^2)}=\lim_{x \to 0}\frac{\dfrac 3{1+9x^2}}{\dfrac {-2}{1-x}}. $$Can you finish it?

Answer 2

L’Hospital doesn’t make things worse.

$$\lim_{x\to0}\frac{\dfrac3{1+9x^2}}{-\dfrac{2(1-x)}{(1-x)^2}}=-\frac32.$$

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Note that it is wiser to draw the exponent $2$ out of the logarithm first.

Answer 3

"Transforming the numerator and the denominator in taylor series doesn’t help too"

I do not agree $$\frac{\tan ^{-1}(3 x)}{\log \left((1-x)^2\right)}=\frac{\tan ^{-1}(3 x)}{2\log \left(1-x\right)}=\frac 12\frac{3 x-9 x^3+O\left(x^4\right)}{-x-\frac{x^2}{2}-\frac{x^3}{3}+O\left(x^4\right) }$$ Simplify and use long division to find the limit and how it is approached.

Answer 4

$$\lim_{x\to0}\frac{\arctan(3x)}{\log^2((1-x))}$$

Take the limit from the left: $$\lim_{x\to0^-}\frac{\arctan(3x)}{\log^2((1-x))}$$ With l’Hopital’s rule $$\lim_{x\to0^-}\frac{\arctan(3x)}{\log^2((1-x))}=\lim_{x \to 0^-}\frac{3(x-1)}{2(9x^2+1)\log(1-x)}.$$ Now \begin{align} \lim_{x\to0^-}\frac{3(x-1)}{2(9x^2+1)\log(1-x)}&=\frac{3\lim\limits_{x \to 0^-}\dfrac{x-1}{9x^2+1}\lim\limits_{x \to 0^-}\dfrac{1}{\log(1-x)}}{2}\\&=\dfrac{3(-1)\lim\limits_{x \to 0^-}\dfrac{1}{\log(1-x)}}{2}\\&=\frac{-3(\infty)}{2}\\&=-\infty. \end{align}

Following the same steps for the limit from the right you will get \begin{align} \lim_{x\to0^+}\frac{3(x-1)}{2(9x^2+1)\log(1-x)}&=\frac{3\lim\limits_{x\to0^+}\dfrac{x-1}{9x^2+1}\lim\limits_{x\to0^+}\dfrac{1}{\log(1-x)}}{2}\\&=\frac{3(-1)\lim\limits_{x\to0^+}\dfrac{1}{\log(1-x)}}{2}\\&=\frac{-3(-\infty)}{2}\\&=\infty. \end{align} Since $\lim_{x\to0^+}\frac{1}{\log(1-x)}=-\infty$ and $\lim_{x\to0^-}\frac{1}{\log(1-x)}=\infty$.

Hence, the two-sided limit does not exist.