# Calculating $\lim\limits_{x \to -\infty}(1+\tan(-\frac{4}{x}))^\frac{1}{\arctan \frac{3}{x}}$

Can you help me with solving this limit?

$\lim\limits_{x \to -\infty}(1+\tan(-\frac{4}{x}))^\frac{1}{\arctan \frac{3}{x}}$

Thank you. Edit: Is there any solution without L'Hospital?

In cases where you need to evaluate the limit of an expression of type $\{f(x)\}^{g(x)}$ (i.e. when both base and exponent are variables rather than constants), it is much simpler to take logs and then evaluate the limit. Let $L$ be the desired limit and then we can proceed as follows \begin{align} \log L &= \log\left(\lim_{x \to -\infty}\left(1 + \tan\left(-\frac{4}{x}\right)\right)^{1/\arctan(3/x)}\right)\notag\\ &= \lim_{x \to -\infty}\log\left(1 + \tan\left(-\frac{4}{x}\right)\right)^{1/\arctan(3/x)}\text{ (via continuity of log)}\notag\\ &= \lim_{x \to -\infty}\dfrac{1}{\arctan\left(\dfrac{3}{x}\right)}\cdot\log\left(1 + \tan\left(-\frac{4}{x}\right)\right)\notag\\ &= \lim_{t \to 0^{+}}\frac{\log(1 + \tan 4t)}{\arctan(-3t)}\text{ (putting }x = -1/t)\notag\\ &= -\lim_{t \to 0^{+}}\frac{\log(1 + \tan 4t)}{\tan 4t}\cdot\frac{\tan 4t}{4t}\cdot\frac{4t}{3t}\cdot\frac{3t}{\arctan 3t}\notag\\ &= (-1)\cdot 1\cdot 1\cdot\frac{4}{3}\cdot 1 = -\frac{4}{3}\notag \end{align} It follows that $L = e^{-4/3}$.

HINT: $$\lim_{x \to -\infty}(1+\tan(-\frac{4}{x}))^\frac{1}{\arctan \frac{3}{x}}$$ $$\left(\lim_{\tan(-\frac{4}{x}) \to 0}(1+\tan(-\frac{4}{x}))^{\frac{1}{\tan(-\frac{4}{x})}}\right)^{\frac{\tan(-\frac{4}{x})}{\arctan \frac{3}{x}}}$$ $$=e^{\lim_{x \to -\infty}\frac{\tan(-\frac{4}{x})}{\arctan \frac{3}{x}}}$$

Now apply L'Hospital's rule..

• you started off well but at the end suggested needlessly the L'Hospital's Rule. It is much simpler to put $t = 1/x$ and use the fact that $(1/t)\tan t \to 1$ and $(1/t)\arctan t \to 1$ as $t \to 0$. – Paramanand Singh Nov 3 '15 at 4:17

\begin{aligned} \lim _{t\to 0}\left(\frac{1}{\arctan\left(3t\right)}\cdot \ln\left(\left(1+\tan\left(-4t\right)\right)\right)\right) & = \lim _{t\to 0}\left(\frac{1}{\arctan\left(3t\right)}\cdot \ln\left(\left(1-4t+o\left(t\right)\right)\right)\right) \\& = \lim _{t\to 0}\left(\frac{-4t+o\left(t\right)}{3t+o\left(t\right)}\right) \\& = -\frac{4}{3} \end{aligned}

So

$$\lim _{t\to 0}\left(1+\tan\left(-4t\right)\right)^{\frac{1}{\arctan\left(3t\right)}} = \color{red}{\frac{1}{e^{\frac{4}{3}}}}$$ Solved with Taylor expansion