Limit without l'Hospital

Question

need help with this limit without l'Hospital

$$\lim\limits_{x \to \infty }\frac{\arctan\left(-\frac{4}{x}\right)}{\left|\arctan\frac{3}{x}\right|}$$

Answer 1

Hint: $$f(x)\to 0\implies\frac{\arctan(f(x))}{f(x)}\to 1.$$ Also important: the sign of $\arctan\frac{3}{x}$.

$$ \lim_{x \to \infty }\frac{\arctan(-\frac{4}{x})}{\left|\arctan\frac{3}{x}\right|}= \lim_{x \to \infty }\frac{-\arctan(\frac{4}{x})\frac{4}{x}}{\left|\arctan\frac{3}{x}\right|\frac{4}{x}} \lim_{x \to \infty }\frac{-\arctan(\frac{4}{x})}{\frac{4}{x}}\frac{\frac{4}{x}}{\left|\arctan\frac{3}{x}\right|}=\cdots $$

Answer 2

Using "$\arctan x \sim x$ as $x\to 0$"

So $$\lim_{x\to \infty}\frac{\arctan \frac{-4}{x}}{|\arctan \frac3x|} = \lim_{x\to \infty}\frac{\frac{-4}{x}}{| \frac3x|} = \lim_{x\to \infty}\frac{-4}{3}\cdot\frac{|x|}{x}$$

As $x\to +\infty$, it is 1; and As $x\to -\infty$, it is -1. So its limit does not exist!

Answer 3

Almost completely without calculus: The $|\cdot|$ in the denominator is irrelevant. We are interested in $\frac\alpha\beta$ where $\tan\alpha=-\frac4x$ and $\tan\beta=\frac3x$ with $x\gg0$ (i.e. $\frac3x$ and $-\frac4x$ are small). For small angles we know from the addition theorems for $\tan$ that $\tan(\pm kx)\approx\pm k\tan x$, hence $\alpha\approx -4\gamma$ and $\beta\approx 3\gamma$ for $\gamma$ with $\tan \gamma=\frac1x$. Conclusion: $\frac{\alpha}{\beta}\approx -\frac43$.