Limitless Limits

$\lim _{ x\rightarrow -1 }{ \frac { \cos { 2 } -\cos { 2x } }{ { x }^{ 2 }-|x| } }$ As when $x<0$ , $|x|=-x$ , $\lim _{ x\rightarrow -1 }{ \frac { \cos { 2 } -\cos { 2x } }{ { x }^{ 2 }+x } }$ Applying L'Hópital Rule, $\lim _{ x\rightarrow -1 }{ \frac { 2\sin { 2x } }{ 2x +1 } }$ $2\sin {2}$

$\lim _{ x\rightarrow -1 }{ \frac { \cos { 2 } -\cos { 2x } }{ { x }^{ 2 }-|x| } }$

Using $\cos { 2x }$ identity,

$\lim _{ x\rightarrow -1 }{ \frac { \cos { 2 }+\sin ^{ 2 }{ x } -\cos ^{ 2 }{ x } }{ { x }^{ 2 }-|x| } }$

Applying L'Hôpital's rule and using $\left( { u }^{ 2 } \right) '=2uu'$ ,

$\lim _{ x\rightarrow -1 }{ \frac { 0+\left( 2\sin { x } \cos { x } \right) -\left( 2\left( -\sin { x } \right) \cos { x } \right) }{ 2x-\frac { x }{ |x| } } }$

Using $2\sin { x } \cos { x }$ identity,

$\lim _{ x\rightarrow -1 }{ \frac { 2\sin { 2x } }{ 2x-\frac { x }{ |x| } } }$

Substituting x by -1.

$\lim _{ x\rightarrow -1 }{ \left( -2\sin { \left( -2 \right) } \right) }$

$\approx \boxed { 1.818 }$