How to evaluate $\lim\limits_{x \to +\infty} \frac{(x+2)!+4^x}{((2x+1)^2+\ln x)x!}$? I have a problem with this limit, I don't know what method to use. I have no idea how to compute it.
Can you explain the method and the steps used?
$$\lim\limits_{x \to +\infty} \frac{(x+2)!+4^x}{((2x+1)^2+\ln x)x!}$$

$$\lim_{ x\to +\infty  } \frac { (x+2)!+4^{ x } }{ ((2x+1)^{ 2 }+\ln { x } )x! } =\lim_{ x\to +\infty  } \frac { x!\left( x+1 \right) (x+2)+4^{ x } }{ ((2x+1)^{ 2 }+\ln { x } )x! } =\lim_{ x\to +\infty  } \frac { x!\left( \left( x+1 \right) (x+2)+\frac { { 4 }^{ x } }{ x! }  \right)  }{ ((2x+1)^{ 2 }+\ln { x } )x! } =\\ =\lim_{ x\rightarrow \infty  }{ \frac { { x }^{ 2 }+3x+2+\frac { { 4 }^{ x } }{ x! }  }{ 4{ x }^{ 2 }+4x+1+\ln { x }  } = } \lim_{ x\rightarrow \infty  }{ \frac { { x }^{ 2 }\left( 1+\frac { 3 }{ x } +\frac { 2 }{ { x }^{ 2 } } +\frac { { 4 }^{ x } }{ { x }^{ 2 }x! }  \right)  }{ { x }^{ 2 }\left( 4+\frac { 4 }{ x } +\frac { 1 }{ { x }^{ 2 } } +\frac { \ln x }{ x }  \right)  } = } \frac { 1 }{ 4 } $$
