(tanx-x)/(x-sinx)=(sinx-xcosx)/(xcosx-sinxcosx). As x approaches 0 the expression approaches 2. Why?
To find out, let's use some calculus. Let's suppose that sinx=a[0]+a[1]x+a[2]x^2+a[3]x^3+...+a[n]x^n, a polynomial series in x, with real coefficients a[0], etc. We are going to use a value for x<1 so the series would converge. When x=0, sinx=0 so a[0]=0.
The derivative of sinx is cosx, so cosx=a[1]+2a[2]x+3a[3]x^2+...+na[n]x^(n-1).
When x=0 cosx=1, so a[1]=1 and sinx=x+a[2]x^2+...
The second derivative of sinx is -sinx=2a[2]+6a[3]x+... When x=0, sinx=0, so 2a[2]=0 and a[2]=0 and sinx becomes x+a[3]x^3. Moving on to the next derivative which is -cosx we have 6a[3]+... But -cosx when x=0 is -1, so a[3]=-1/6, making sinx=x-x^3/6+...
We could keep on going through more derivatives, but it's not necessary because we can establish right now what x-sinx tends to as x approaches zero; x-sinx=x^3/6.
Now, let tanx=b[0]+b[1]x+b[2]x^2+... just like we did with sinx. We can see that b[0]=0 because tanx=0 when x=0.
The first derivative of tanx=sec^2x=b[1]+2b[2]x+3b[3]x^2+... secx is the reciprocal of cosx which is 1 when x=0, so b[1], like a[1], is 1 and tanx=x+b[2]x^2. The derivative of sec^2x is the derivative of cos^-2x=-2cos-3x*(-sinx)=2tanxsec^2x=0, when x=0, making b[2]=0. We need to go to the next derivative to find b[3]. But 2tanxsec^2x=2tanx(1+tan^2x)=2tanx+2tan^3x, which differentiates to 2sec^2x+6tan^2xsec^2x, and this derivative is 2 when x=0. Therefore 2=6b[3] and b[3]=1/3 and tanx-x=x^3/3. So the limit as x approaches zero of (tanx-x)/(x-sinx)=(x^3/3)/(x^3/6)=2.