Evaluate the limit

lim_{x → 0} (tan(tan x) − tan(sin x)) / (tan x − sin x)

To evaluate the limit:
$\underset{x\to 0}{lim}\frac{tan(tanx)-tan(sinx)}{tanx-sinx}$

Let us simplify the expression. We can use the trigonometric identity for the difference of tangents:
$tanA-tanB=\frac{sin(A-B)}{cosAcosB}$

Applying this to the numerator with $A=tanx$ and $B=sinx$, we get:
$tan(tanx)-tan(sinx)=\frac{sin(tanx-sinx)}{cos(tanx)cos(sinx)}$

Now, substitute this back into the limit expression:
$\underset{x\to 0}{lim}\frac{sin(tanx-sinx)}{cos(tanx)cos(sinx)(tanx-sinx)}$

We can separate this limit into the product of two simpler limits:
$\underset{x\to 0}{lim}\left(\frac{1}{cos(tanx)cos(sinx)}\right)·\underset{x\to 0}{lim}\left(\frac{sin(tanx-sinx)}{tanx-sinx}\right)$

First, evaluate the limit of the cosine term. As $x\to 0$, both $tanx\to 0$ and $sinx\to 0$. Since $cos\left(0\right)=1$, we have:
$\underset{x\to 0}{lim}\frac{1}{cos(tanx)cos(sinx)}=\frac{1}{1·1}=1$

Second, evaluate the remaining limit. Let $u=tanx-sinx$. As $x\to 0$, we have $u\to 0$. Thus, the limit becomes:
$\underset{u\to 0}{lim}\frac{sinu}{u}=1$

Multiplying these two results together:
$1·1=1$

Therefore, the correct answer is 1.