Find
$$\lim_{x \to \frac{\pi}{2}} \frac{\tan 11x}{\tan 13x}$$
Using L'Hospital by brute force we get $13/11$. However we can only use the rule when both functions $\tan 11x$ and $\tan 13x$ are defined on a common interval $I$ with a further condition that for each $x \in I$, $\tan 13x \neq 0$. How should I choose this interval or redefine the functions in order to guarantee that the conditions for using L'Hospital's rule are satisfied? Is there a better approach to this problem?
Best Answer
let $x-\frac { \pi }{ 2 } =t$
$$\lim _{ x\to \frac { \pi }{ 2 } } \frac { \tan 11x }{ \tan 13x } =\lim _{ t\rightarrow 0 } \frac { \tan 11\left( \frac { \pi }{ 2 } +t \right) }{ \tan 13\left( \frac { \pi }{ 2 } +t \right) } =\\ =\lim _{ t\rightarrow 0 } \frac { \tan \left( \frac { 11\pi }{ 2 } +11t \right) }{ \tan \left( \frac { 13\pi }{ 2 } +13t \right) } =\lim _{ t\rightarrow 0 } \frac { \tan \left( \frac { 3\pi }{ 2 } +11t \right) }{ \tan \left( \frac { \pi }{ 2 } +13t \right) } =\\ =\lim _{ t\rightarrow 0 } \frac { -\cot { \left( 11t \right) } }{ -\cot { \left( 13t \right) } } =\lim _{ t\rightarrow 0 } \frac { \tan { \left( 13t \right) } }{ 13t } \frac { 11t }{ \tan { \left( 11t \right) } } \frac { 13 }{ 11 } =\frac { 13 }{ 11 } \\ $$