(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+ax+ax^2+ax^3+...+a[n]x^n, a polynomial series in x, with real coefficients a, etc. We are going to use a value for x<1 so the series would converge. When x=0, sinx=0 so a=0.
The derivative of sinx is cosx, so cosx=a+2ax+3ax^2+...+na[n]x^(n-1).
When x=0 cosx=1, so a=1 and sinx=x+ax^2+...
The second derivative of sinx is -sinx=2a+6ax+... When x=0, sinx=0, so 2a=0 and a=0 and sinx becomes x+ax^3. Moving on to the next derivative which is -cosx we have 6a+... But -cosx when x=0 is -1, so a=-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+bx+bx^2+... just like we did with sinx. We can see that b=0 because tanx=0 when x=0.
The first derivative of tanx=sec^2x=b+2bx+3bx^2+... secx is the reciprocal of cosx which is 1 when x=0, so b, like a, is 1 and tanx=x+bx^2. The derivative of sec^2x is the derivative of cos^-2x=-2cos-3x*(-sinx)=2tanxsec^2x=0, when x=0, making b=0. We need to go to the next derivative to find b. 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 and b=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.
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