Question 1020705
I'm assuming the original limit is
{{{lim(x->0,   (cos(2x) - cos(x))/(x))}}}



If so, then here are the steps

<table border=1 cellpadding = 5>
<tr><th>Step Number</th><th>Expression</th></tr>
<tr><td>1. </td><td>{{{lim(x->0,   (cos(2x) - cos(x))/(x))}}}</td></tr>
<tr><td>2. </td><td>{{{lim(x->0,   (2cos^2(x) - 1 - cos(x))/(x))}}}</td></tr>
<tr><td>3. </td><td>{{{lim(x->0,   (2cos^2(x) - cos(x) - 1)/(x))}}}</td></tr>
<tr><td>4. </td><td>{{{lim(x->0,   ((2cos(x)+1)(cos(x)-1))/(x))}}}</td></tr>
<tr><td>5. </td><td>{{{lim(x->0,   (2cos(x)+1)*(((cos(x)-1))/(x)))}}}</td></tr>
<tr><td>6. </td><td>{{{lim(x->0,   (2cos(x)+1))*lim(x->0,   (((cos(x)-1))/(x)))}}}</td></tr>
<tr><td>7. </td><td>{{{lim(x->0,   (2cos(x)+1))*0}}}</td></tr>
<tr><td>8. </td><td>{{{(2*cos(0)+1)*0}}}</td></tr>
<tr><td>9. </td><td>{{{(2*1+1)*0}}}</td></tr>
<tr><td>10. </td><td>{{{(2+1)*0}}}</td></tr>
<tr><td>11. </td><td>{{{3*0}}}</td></tr>
<tr><td>12. </td><td>{{{0}}}</td></tr>
</table>


------------------------------------------------
Notes: 


In step 2, I used the <a href = "http://tutorial.math.lamar.edu/pdf/Trig_Cheat_Sheet.pdf">trig identity</a> {{{cos(2x) = 2cos^2(x)-1}}} (see page 2 under "Double Angle Formulas")


In step 6, I broke up the limit using the limit law {{{lim(x->a,   (f(x)*g(x)))=lim(x->a,   (f(x)))*lim(x->a,   (g(x)))}}}


In step 7, I used the <a href = "http://tutorial.math.lamar.edu/Classes/CalcI/ProofTrigDeriv.aspx">special trig limit</a> {{{lim(x->0,   ((cos(x)-1)/x)) = 0}}} (scroll down to bottom half of the page)


From step 8 and beyond, I used the substitution rule and then evaluated/simplified to get the final answer of 0

------------------------------------------------


So in the end, 
{{{lim(x->0,   (cos(2x) - cos(x))/(x))=0}}}


Final Answer: <font color=red size = 5>0</font>