1. Simplify the summand

Let

$$\theta_r=\frac{x}{2^{r+1}}.$$

Then the summand is

$$\frac{\tan\theta_r+\tan^3\theta_r}{1-\tan^2\theta_r} =\frac{\tan\theta_r(1+\tan^2\theta_r)}{1-\tan^2\theta_r}.$$

Using

$$$$

and the identity

$$\tan 2\theta=\frac{2\tan\theta}{1-\tan^2\theta},$$

we get

$$\frac{\tan\theta(1+\tan^2\theta)}{1-\tan^2\theta} =\frac{\tan\theta\sec^2\theta}{1-\tan^2\theta}.$$

But it is easier to rewrite directly in sine-cosine form:

$$\frac{\tan\theta+\tan^3\theta}{1-\tan^2\theta} =\frac{\tan\theta(1+\tan^2\theta)}{1-\tan^2\theta} =\frac{\tan\theta\sec^2\theta}{1-\tan^2\theta}.$$

Now,

$$\tan\theta=\frac{\sin\theta}{\cos\theta},\quad \sec^2\theta=\frac1{\cos^2\theta},\quad 1-\tan^2\theta=\frac{\cos2\theta}{\cos^2\theta}.$$

So

$$\frac{\tan\theta\sec^2\theta}{1-\tan^2\theta} =\frac{\sin\theta}{\cos\theta\cos2\theta}.$$

Using

$$\sin2\theta=2\sin\theta\cos\theta,$$

this becomes

$$\frac{\sin2\theta}{2\cos^2\theta\cos2\theta}$$

but the cleanest observation is:

$$\frac{\tan\theta+\tan^3\theta}{1-\tan^2\theta} =\frac{\tan\theta}{\cos2\theta} =\frac{\sin\theta}{\cos\theta\cos2\theta}.$$

Now compare with

$$\tan2\theta=\frac{2\tan\theta}{1-\tan^2\theta}.$$

Hence

$$\frac{\tan\theta+\tan^3\theta}{1-\tan^2\theta} =\frac{1}{2}\tan2\theta\,(1+\tan^2\theta)$$

which is still not telescoping directly. So instead, let us simplify more carefully:

Since

$$\tan\theta+\tan^3\theta=\tan\theta(1+\tan^2\theta)=\tan\theta\sec^2\theta,$$

and

$$1-\tan^2\theta=\frac{\cos2\theta}{\cos^2\theta},$$

we get

$$\frac{\tan\theta\sec^2\theta}{1-\tan^2\theta} =\tan\theta\cdot \frac{1}{\cos^2\theta}\cdot \frac{\cos^2\theta}{\cos2\theta} =\frac{\tan\theta}{\cos2\theta}.$$

Now,

$$\tan2\theta=\frac{2\tan\theta}{1-\tan^2\theta}$$

and

$$1+\tan^2\theta=\sec^2\theta.$$

A better route is to test whether this equals $\frac12\tan2\theta$:

$$\frac12\tan2\theta=\frac{\tan\theta}{1-\tan^2\theta},$$

which is not our summand.

So let us proceed by series expansion near $x=0$, since the required limit is only as $x\to0$.

---

2. Asymptotic expansion of the summand for small argument

Let $u=\frac{x}{2^{r+1}}.$ For small $u$,

$$\tan u=u+\frac{u^3}{3}+O(u^5),$$

so

$$\tan^2 u=u^2+O(u^4),\qquad \tan^3 u=u^3+O(u^5).$$

Thus

$$\tan u+\tan^3 u=u+\frac{4}{3}u^3+O(u^5).$$

Also,

$$\frac{1}{1-\tan^2u}=1+u^2+O(u^4).$$

Therefore,

$$\frac{\tan u+\tan^3u}{1-\tan^2u} =\left(u+\frac{4}{3}u^3+O(u^5)\right)(1+u^2+O(u^4)) =u+\frac{7}{3}u^3+O(u^5).$$

Hence

$$f(x)=\sum_{r=0}^{\infty}\left(\frac{x}{2^{r+1}}+O\left(\frac{x^3}{2^{3r}}\right)\right).$$

So

$$f(x)=x\sum_{r=0}^{\infty}\frac1{2^{r+1}}+O(x^3) =x+O(x^3),$$

because

$$\sum_{r=0}^{\infty}\frac1{2^{r+1}}=1.$$

Thus, near $x=0$,

$$f(x)\to 0 \quad \text{and} \quad f(x)=x+O(x^3).$$

---

3. Evaluate the required limit

We need

$$L=\lim_{x\to0}\frac{e^x-e^{f(x)}}{x-f(x)}.$$

Apply the Mean Value Theorem to $e^t$ between $t=x$ and $t=f(x)$:

$$e^x-e^{f(x)}=e^{\xi_x}(x-f(x))$$

for some $\xi_x$ lying between $x$ and $f(x)$. Since both $x\to0$ and $f(x)\to0$, we have $\xi_x\to0$. Therefore,

$$L=\lim_{x\to0} e^{\xi_x}=e^0=1.$$

So the required integer is

$$\boxed{1}.$$

---

4. Comparison with stored answer

Stored correct answer = $1$.

Our derived answer = $1$.

Hence they agree.