Explanation
$$\mathop {\lim }\limits_{x \to 0} {\log _{{{\tan }^2}x}}\left( {{{\tan }^2}2x} \right)$$
Let,
$$\begin{array}{l} y={ \log _{ { { \tan }^{ 2 } }x } }{ \tan ^{ 2 } }2x \\ =\log { \tan ^{ 2 } } 2x={ \left( { { { \tan }^{ 2 } }x } \right) ^{ y } } \\ =2\log \tan 2x=2y\log \tan x \\ \Rightarrow y=\dfrac { { \log \tan 2x } }{ { \log \tan x } } \\ \Rightarrow { \lim }_{ x\to 0 }y={ \lim }_{ x\to 0 }\dfrac { { \log \tan 2x } }{ { \log \tan x } } \end{array}$$
Using L-Hospital Rule
$$\begin{array}{l} \Rightarrow { \lim }_{ x\to 0 }y={ \lim }_{ x\to 0 }\dfrac { { \dfrac { 1 }{ { \tan 2x } } \times { { \sec }^{ 2 } }2x\times 2 } }{ { \dfrac { 1 }{ { \tan x } } { { \sec }^{ 2 } }x } } \\ \Rightarrow { \lim }_{ x\to 0 }y={ \lim }_{ x\to 0 }\dfrac { { 2{ { \sec }^{ 2 } }2x } }{ { \tan 2x } } \times \dfrac { { \tan x } }{ { { { \sec }^{ 2 } }x } } \\ \Rightarrow { \lim }_{ x\to 0 }y={ \lim }_{ x\to 0 }\dfrac { { 2\tan x } }{ { \tan 2x } } \, \, \, \, \, \, \left( { { \lim }_{ x\to 0 }\sec x=4 } \right) \end{array}$$
$$\begin{array}{l} \Rightarrow { \lim }_{ x\to 0 }y={ \lim }_{ x\to 0 }\dfrac { { 2\dfrac { { \tan x } }{ x } \times x } }{ { \dfrac { { \tan 2x } }{ { 2x } } \times 2x } } \\ \Rightarrow { \lim }_{ x\to 0 }y={ \lim }_{ x\to 0 }\, \, \, 1\, \, \left( { { \lim }_{ x\to 0 }\dfrac { { \tan x } }{ x } =1 } \right) \\ \Rightarrow { \lim }_{ x\to 0 }{ \log _{ { { \tan }^{ 2 } }x } }\left( { { { \tan }^{ 2 } }2x } \right) =1 \end{array}$$
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