solve the following limit without L'Hospital rule or series:

$\underset{x\to 0}{lim}\frac{\mathrm{arctan}2x}{\mathrm{sin}3x}$

$\underset{x\to 0}{lim}\frac{\mathrm{arctan}2x}{\mathrm{sin}3x}$

Sovardipk
2022-07-13
Answered

solve the following limit without L'Hospital rule or series:

$\underset{x\to 0}{lim}\frac{\mathrm{arctan}2x}{\mathrm{sin}3x}$

$\underset{x\to 0}{lim}\frac{\mathrm{arctan}2x}{\mathrm{sin}3x}$

You can still ask an expert for help

Johnathan Morse

Answered 2022-07-14
Author has **18** answers

First you want to show that

$\underset{y\to 0}{lim}\frac{\mathrm{arctan}y}{y}=1$

To do this, I would let $\theta =\mathrm{arctan}y$. Then $\theta \to 0$ as $y\to 0$, so

$\underset{y\to 0}{lim}\frac{\mathrm{arctan}y}{y}=\underset{\theta \to 0}{lim}\frac{\theta}{\mathrm{tan}\theta}=\underset{\theta \to 0}{lim}\frac{\theta}{\mathrm{sin}\theta}\cdot \mathrm{cos}\theta =1\cdot 1=1$

(the limit $\underset{\theta \to 0}{lim}\frac{\mathrm{sin}\theta}{\theta}=1$ is one of the basic limits you can usually invoke without proof).

Now

$\frac{\mathrm{arctan}2x}{\mathrm{sin}3x}=\frac{{\displaystyle \frac{\mathrm{arctan}2x}{2x}}}{{\displaystyle \frac{\mathrm{sin}3x}{3x}}}\cdot \frac{2}{3}$

$\underset{y\to 0}{lim}\frac{\mathrm{arctan}y}{y}=1$

To do this, I would let $\theta =\mathrm{arctan}y$. Then $\theta \to 0$ as $y\to 0$, so

$\underset{y\to 0}{lim}\frac{\mathrm{arctan}y}{y}=\underset{\theta \to 0}{lim}\frac{\theta}{\mathrm{tan}\theta}=\underset{\theta \to 0}{lim}\frac{\theta}{\mathrm{sin}\theta}\cdot \mathrm{cos}\theta =1\cdot 1=1$

(the limit $\underset{\theta \to 0}{lim}\frac{\mathrm{sin}\theta}{\theta}=1$ is one of the basic limits you can usually invoke without proof).

Now

$\frac{\mathrm{arctan}2x}{\mathrm{sin}3x}=\frac{{\displaystyle \frac{\mathrm{arctan}2x}{2x}}}{{\displaystyle \frac{\mathrm{sin}3x}{3x}}}\cdot \frac{2}{3}$

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