Can someone advise me or show me a certain method for solving these kind of limits without using L'Hospital rule, I've used all the methods that I know and I ended up with a dead end.
Here is what I've reached after simplifying, limit of
$$e^{x\ln \left( 1+\tan \frac{1}{x} \right)}$$
as $x$ approaches infinity, I got stuck here can someone give me some tips on what do when I reach a dead end like this?
Herein, we will rely only on the squeeze theorem and a set of inequalities that we can show using pre-calculus tools only. To that end, we begin with the following primer.
PRIMER:
I showed in THIS ANSWER using only the limit definition of the exponential function that the logarithm function satisfies the inequalities
$$\bbox[5px,border:2px solid #C0A000]{\frac{x-1}{x}\le \log(x)\le x-1 }\tag 1$$
for $x>0$.
In addition, recall from elementary geometry the inequalities SEE THIS ANSWER
$$\theta\cos(\theta)\le \sin(\theta)\le \theta \tag 2$$
for $0<\theta<\pi/2$.
Dividing $(2)$ by $\cos(\theta)$ for $0\le \theta <\pi/2$, we see that
$$\theta \le \tan(\theta)\le \frac{\theta}{\cos(\theta)} \tag 3$$
Squaring the right-hand side inequality in $(2)$, setting $\sin^2(\theta)=1-\cos^2(\theta)$ and isolating $\cos(\theta)$ reveals $\cos(\theta)\ge \sqrt{1-\theta^2}$ for $0\le \theta<1$. Using this inequality on the right-hand side of $(3)$, we obtain
$$\theta\le \tan(\theta)\le \frac{\theta}{\sqrt{1-\theta^2}} \tag 4$$
for $0\le \theta <1$.
Finally, letting $\theta=1/x$ in $(4)$ yields
$$\bbox[5px,border:2px solid #C0A000]{\frac{1}{x} \le \tan(1/x)\le \frac{1}{\sqrt{x^2-1}}} \tag 5$$
for $1<x$.
Applying $(1)$ and $(5)$, we find that $x\log(1+\tan(1/x))$ is bounded as
$$ \frac{\sqrt{x^2-1}}{1+\sqrt{x^2-1}}\le \frac{x\tan(1/x)}{1+\tan(1/x)}\le x\log(1+\tan(1/x))\le x\tan(1/x) \le \frac{x}{\sqrt{x^2-1}}$$
for $x>1$, whereupon application of the squeeze theorem reveals
$$\lim_{x\to \infty}x\log(1+\tan(1/x))=1$$
Therefore,
$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to \infty}\left(1+\tan(1/x)\right)^x=\lim_{x\to \infty}e^{x\log(1+\tan(1/x))}=e}$$
$$\lim_{x\to\infty}(1+\tan\left({1\over x}\right))^x=e^{\lim_{x\to\infty}\tan\left({1\over x}\right)x}=e$$
$$\left[\lim_{x\to\infty}\left(1+\tan1/x\right)^{\frac1{\tan 1/x}}\right]^{\lim_{x\to\infty} x\tan 1/x}$$
The inner limit converges to $e$
The exponent $$\lim_{x\to\infty} \dfrac x{\tan 1/x}=\lim_{h\to0}\dfrac{\sin h}h\cdot\lim_{h\to0}\dfrac1{\cos h}=?$$
