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?
 A: 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..
A: 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 $L = e^{-4/3}$.
A: $$
\begin{aligned}
\lim _{t\to 0}\left(\frac{1}{\arctan\left(3t\right)}\cdot \ln\left(\left(1+\tan\left(-4t\right)\right)\right)\right)
& = \lim _{t\to 0}\left(\frac{1}{\arctan\left(3t\right)}\cdot \ln\left(\left(1-4t+o\left(t\right)\right)\right)\right)
\\& = \lim _{t\to 0}\left(\frac{-4t+o\left(t\right)}{3t+o\left(t\right)}\right)
\\& = -\frac{4}{3}
\end{aligned}
$$
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
