How to solve this limit: $\lim\limits_{x \to 0}\left(\frac{(1+2x)^\frac1x}{e^2 +x}\right)^\frac1x$ $$\lim\limits_{x \to 0}\left(\frac{(1+2x)^\frac{1}{x}}{e^2 +x}\right)^\frac{1}{x}=~?$$
Can not solve this limit, already tried with logarithm but this is where i run out of ideas. Thanks.

Using L'Hospital rule twice we get $$\lim _{ x\to 0 } \left( \frac { (1+2x)^{ \frac { 1 }{ x }  } }{ e^{ 2 }+x }  \right) ^{ \frac { 1 }{ x }  }=~ { e }^{ \lim _{ x\rightarrow 0 }{ \frac { 1 }{ x } \ln { \left( \frac { (1+2x)^{ \frac { 1 }{ x }  } }{ e^{ 2 }+x }  \right)  }  }  }={ e }^{ \lim _{ x\rightarrow 0 }{ \frac { 1 }{ x } \left[ \frac { 1 }{ x } \ln { \left( 1+2x \right) -\ln { \left( { e }^{ 2 }+x \right)  }  }  \right]  }  }=\\ ={ e }^{ \lim _{ x\rightarrow 0 }{ \frac { \ln { \left( 1+2x \right) -x\ln { \left( { e }^{ 2 }+x \right)  }  }  }{ { x }^{ 2 } }  }  }\overset { L'Hospital }{ = } { e }^{ \lim _{ x\rightarrow 0 }{ \frac { \frac { 2 }{ 1+2x } -\ln { \left( { { e }^{ 2 }+x } \right) -\frac { x }{ { e }^{ 2 }+x }  }  }{ 2{ x } }  }  }\overset { L'hospital }{ = } { e }^{ \lim _{ x\rightarrow 0 }{ \frac { -\frac { 4 }{ { \left( 1+2x \right)  }^{ 2 } } -\frac { 1 }{ { e }^{ 2 }+x } -\frac { { e }^{ 2 } }{ { \left( { e }^{ 2 }+x \right)  }^{ 2 } }  }{ 2 }  }  }={ e }^{ -\frac { 4{ e }^{ 2 }+2 }{ 2{ e }^{ 2 } }  }$$
