Finding $\lim_{x\to 0} \frac{(1+\tan x)^{\frac{1}{x}}-e}{x}$ How would I go about solving this following limit?
$$\lim_{x\to 0} \frac{(1+\tan x)^{\frac{1}{x}}-e}{x}$$
My attempts:
Direct substitution yields the limit to be undefined, also ruling out the possibility of using L'Hospital's Rule.
I don't see any clever substitutions that can be made with this limit.
Would squeeze theorem help here? Maybe using the trig. identities:
$$-1 \le \cos x \le 1$$
and
$$-1 \le \cos x \le 1$$
EDIT
I attempted to break the limit down term by term.
So, for the first one:
$$y = \lim_{x\to 0} (1 + \tan x)^{1/x}$$
Taking the natural log of both sides:
$$\ln y = \lim_{x\to 0} \frac{\ln(1+\tan x)}{x}$$
Direct sub. yields $0/0$. Using L'Hospital's rule:
$$\ln y = \lim_{x\to 0} \frac{\frac{\sec^2{x}}{1+\tan x}}{1} = \frac{\sec^2{x}}{1+\tan x} = 1$$
Thus, $\ln y = 1$, so $y= e$
EDIT #2
Thanks to a random comment, it actually does help me:
$$\lim_{x\to 0} \frac{(1+\tan x)^{\frac{1}{x}}-e}{x}$$
$$\lim_{x\to 0} \frac{e-e}{0} = \frac{0}{0}$$
Thus, we can use L'Hospitals here:
$$\lim_{x\to 0} \frac{(\tan x+1)^{1/x} \left(\frac{\sec^2 x}{x(\tan(x)+1)}-\frac{\ln(\tan(x)+1))}{x^2}\right)}{1} = (\tan x+1)^{1/x} \left(\frac{\sec^2 x}{x(\tan(x)+1)}-\frac{\ln(\tan(x)+1))}{x^2}\right)$$
I haven't made any further progress, sadly.
Any help would be appreciated.

Let's proceed in the following manner $$\begin{aligned}L &= \lim_{x \to 0}\frac{(1 + \tan x)^{1/x} - e}{x}\\
&= \lim_{x \to 0}\dfrac{\exp\left(\dfrac{\log(1 + \tan x)}{x}\right) - e}{x}\\
&= e\cdot\lim_{x \to 0}\dfrac{\exp\left(\dfrac{\log(1 + \tan x)}{x} - 1\right) - 1}{x}\\
&= e\cdot\lim_{x \to 0}\dfrac{\exp z - 1}{x}\text{ (putting }z = \dfrac{\log(1 + \tan x)}{x} - 1)\\
&= e\cdot\lim_{z \to 0}\dfrac{\exp z - 1}{z}\cdot\lim_{x \to 0}\frac{z}{x}\\
&= e\cdot 1\cdot\lim_{x \to 0}\frac{z}{x}\\
&= e\cdot\lim_{x \to 0}\frac{\log(1 + \tan x) - x}{x^{2}}\\
&= e\cdot\lim_{x \to 0}\left\{\frac{\log(1 + \tan x) - \tan x}{x^{2}} + \frac{\tan x - x}{x^{2}}\right\}\\
&= e\cdot\lim_{x \to 0}\frac{\log(1 + \tan x) - \tan x}{x^{2}} + e\cdot\lim_{x \to 0}\frac{\tan x - x}{x^{2}}\\
&= e\cdot\lim_{x \to 0}\frac{\log(1 + \tan x) - \tan x}{x^{2}} + e\cdot 0\\
&= e\cdot\lim_{x \to 0}\frac{\log(1 + \tan x) - \tan x}{\tan^{2}x}\cdot\frac{\tan^{2}x}{x^{2}}\\
&= e\cdot\lim_{x \to 0}\frac{\log(1 + \tan x) - \tan x}{\tan^{2}x}\cdot 1\\
&= e\cdot\lim_{t \to 0}\frac{\log(1 + t) - t}{t^{2}}\text{ (putting }t = \tan x)\\
&= e\cdot\lim_{t \to 0}\dfrac{\left(t - \dfrac{t^{2}}{2} + \cdots\right) - t}{t^{2}} = -\frac{e}{2}\end{aligned}$$ I have used the following limits $$\lim_{x \to 0}\frac{\log(1 + \tan x)}{x} = \lim_{x \to 0}\frac{\log(1 + \tan x)}{\tan x}\cdot \frac{\tan x}{x} = 1$$ so that $z \to 0$ $$\lim_{x \to 0}\frac{\tan x - x}{x^{2}} = 0$$ which can be proved using inequalities $\sin x < x < \tan x$ for $x \in (0, \pi/2)$ (although the proof is slightly tricky but easily available in MSE). Finally in the last step Taylor series for $\log (1 + t)$ is used.
Update: Proof for $$\lim_{x \to 0}\frac{\tan x - x}{x^{2}} = 0$$ Clearly we can consider only the case $x \to 0^{+}$ because the function under consideration is odd and the letting $x \to 0^{-}$ will only change the sign of the answer (which wont matter as the answer would come out as $0$). Now we can see that $$\begin{aligned}A &= \lim_{x \to 0^{+}}\frac{\tan x - x}{x^{2}}\\
&= \lim_{x \to 0^{+}}\frac{\sin x - x\cos x}{x^{2}\cos x}\\
&= \lim_{x \to 0^{+}}\frac{\sin x - x\cos x}{x^{2}\cdot 1}\\
&= \lim_{x \to 0^{+}}\frac{\sin x - x}{x^{2}} + \lim_{x \to 0^{+}}\frac{x - x\cos x}{x^{2}}\\
&= \lim_{x \to 0^{+}}\frac{\sin x - x}{x^{2}} + \lim_{x \to 0^{+}}x\cdot\frac{1 - \cos x}{x^{2}}\\
&= \lim_{x \to 0^{+}}\frac{\sin x - x}{x^{2}} + 0 \cdot\frac{1}{2}\\
&= \lim_{x \to 0^{+}}\frac{\sin x - x}{x^{2}}\\\end{aligned}$$ Next we have the inequality $$\sin x < x < \tan x = \frac{\sin x}{\cos x}$$ for $0 < x < \pi/2$ and hence $$\cos x < \frac{\sin x}{x} < 1$$ or $$\frac{\cos x - 1}{x} < \frac{\sin x - x}{x^{2}} < 0$$ Now taking limits as $x \to 0^{+}$ and noting that $$\frac{\cos x - 1}{x} = -2\cdot\frac{\sin^{2}(x/2)}{(x/2)^{2}}\cdot\frac{(x/2)^{2}}{x} \to 0$$ we get $$A = \lim_{x \to 0^{+}}\frac{\sin x - x}{x^{2}} = 0$$
