# Find $\lim_{x\to0}{\frac{\arctan (2x)}{3x}}$ without using '0/0=1'

Find limit as $$\lim_{x \to 0} \frac{\arctan (2x)}{3x}$$ without using $\frac{0}{0} = 1$.

I wanted to use $$\frac{2}{3} \cdot \frac{0}{0} = \frac{2}{3} \cdot 1,$$ but our teacher considers $\frac{0}{0}$ a "dangerous case" and we are not allowed to use this method.

We have not studied integrals yet, so I cannot use integral formulas. We have only studied a bit of derivative taking...

I tried many substitutions, but failed :( Any suggestion?

• @user283294 What do you mean "without 0/0 = 1" ? What are the rules of calculating this limit? Oct 23, 2015 at 21:35
• Did you do L'Hospital rules yet?
– 5xum
Oct 23, 2015 at 21:35
• The argument that "$0/0 = 1$" is not only a dangerous method, it is utterly and completely FALSE. It is never allowed and your teacher should say so clearly. Oct 23, 2015 at 21:42
• @user283294 I agree with 5xum - you should use L'Hospital's Rule as it is a $0/0$ case. Oct 23, 2015 at 21:43

Let $u=\arctan(2x)$ Then as $x \rightarrow 0$ we have $u \rightarrow 0$ So $\lim_{x \rightarrow 0} \frac{\arctan(2x)}{3x}=\lim_{u \rightarrow 0} \frac{u}{3 \frac{tan(u)}{2}}=\frac{2}{3}\lim_{u \rightarrow 0 } \cos(x) \cdot \frac{u}{\sin(u)}=\frac{2}{3} \lim_{u \rightarrow 0} \frac{u}{\sin(u)} \cdot \lim_{u \rightarrow 0} \cos(u)$

• This is the best approach, in my view. Exactly how I did it. +1 for you.
– MPW
Oct 23, 2015 at 22:00

This is really just the derivative of the function $f(x) = \frac{1}{3} \arctan 2x$ at $x = 0$, written down using the definition. Since you know that $\frac{d}{dx} \frac{1}{3} \arctan 2x = \frac{1}{3} \frac{2}{1 + (2x) ^2}$ (check this!), you obtain the result $\frac{2}{3}$ by substituting $x = 0$.

You may use L'Hospital's rule: $$\lim_{x\to0}{\frac{\arctan (2x)}{3x}}=\lim_{x\to0}{\frac{\frac{2}{1+4x^2}}{3}}=\frac23.$$

With equivalents:

We know $\arctan u\sim_0u$, hence $$\frac{\arctan 2x}{3x}\sim_0\frac{2x}{3x}=\frac23.$$