+118723 im x → 1, ((sin(x+1))-(sin2x))/(x-1)
\(\displaystyle \lim_{ x → 1} \; \frac{sin(x+1)-sin2x}{x-1}\\ \mbox{This is the indeterminate form of 0/0 so use l'hopital's rule}\\ =\displaystyle \lim_{ x → 1} \; \frac{cos(x+1)-2cos2x}{1}\\ =\displaystyle \lim_{ x → 1} \; cos(x+1)-2cos2x\\ = cos(2)-2cos2\\ =-cos(2)\)
