Limit, L’Hopital, inverse trig function

Question

The question is: evaluate $$\lim_{x\to 0}\left(\frac{1}{x}-\frac{1}{\arctan(x)}\right).$$ Please check my work: $$\lim_{x\to 0}\left(\frac{1}{x}-\frac{1}{\arctan(x)}\right) =\lim_{x\to 0}\frac{\arctan(x)-x}{x\arctan(x)}=\lim_{x\to 0}\frac{\arctan(x)-x}{\arctan(x)}\cdot \lim_{x\to 0}\frac{1}{x} \\=\lim_{x\to 0}\frac{\frac{1}{1+x^2}-1}{\frac{1}{1+x^2}}\cdot \lim_{x\to 0}\frac{1}{x}=0\cdot \infty.$$

Answer 1

After reading your work, I think that it is easier if you proceed in this way $$\lim_{x\to 0}\left(\frac{1}{x}-\frac{1}{\arctan(x)}\right)= \lim_{x\to 0}\frac{\arctan(x)-x}{x\arctan(x)}= \lim_{x\to 0}\frac{\arctan(x)-x}{x^2}\cdot \lim_{x\to 0}\frac{x}{\arctan(x)}.$$ Now apply L'Hopital's rule.

Answer 2

$$\bigg[x\tan^{-1}x\bigg]' = x\cdot\frac{1}{1+x^2}+\tan^{-1}x=\frac{x+(x^2+1)\tan^{-1}x}{1+x^2}$$ $$\bigg[x+(x^2+1)\tan^{-1}x\bigg]'=1+1+2x\tan^{-1}x$$ On applying L' Hopital twice, $$l=\lim_{x\to0}\frac{-x^2}{x+(x^2+1)\tan^{-1}x}=\lim_{x\to0}\frac{-2x}{2x\tan^{-1}x+2}=\cdots$$

Answer 3

$$\lim _{x\to 0}\left(\frac{1}{x}-\frac{1}{\arctan\left(x\right)}\right)$$ Step 1:Simplify $$\lim _{x\to \:0}\left(\frac{\arctan \left(x\right)-x}{x\arctan \left(x\right)}\right)$$ Step 2: L'Hopitale's rule $$\lim _{x\to \:0}\left(\frac{\frac{1}{x^2+1}-1}{\arctan \left(x\right)+\frac{x}{x^2+1}}\right)$$ Step 3:Simplify $$\lim _{x\to \:0}\left(-\frac{x^2}{\arctan \left(x\right)\left(x^2+1\right)+x}\right)$$ Step 4:L'Hopitale's Rule $$\lim _{x\to \:0}\left(\frac{-2x}{2x\arctan \left(x\right)+2}\right)$$ Step 5: Plug in value of $x$. You will get answer as $0$

Answer 4

While using L'Hospital 's rule you missed part of the denominator. Check your work and continue with taking common denominator and one more L'Hospital to get the answer. My answer is $0$ see if I have the right answer.