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?
$$\lim\limits_{x \to +\infty} \frac{(x+2)!+4^x}{((2x+1)^2+\ln x)x!}$$
 A: The answer is 1/4 and you can calculate it by without a pen/paper.
At infinity $(x + 2)!$ grows way faster than $4^x$ also $(2x + 1)^2$ grows way faster than $\ln x$. So you can rewrite your limit as:
$$\lim\limits_{x \to +\infty} \frac{(x+2)!}{(2x+1)^2 \cdot x!} = \lim\limits_{x \to +\infty} \frac{(x+2)(x+1)}{(2x+1)^2 } = 1/4$$
A: With limits going to infinity, usually factoring out the greatest term is the way to go, since then you'll have a lot of terms going to $0$. Here the term that grows faster is $x!$. So:
\begin{align}
\require{cancel}
\lim_{x \to +\infty} \frac{(x+2)(x+1)x!\left(1+\frac{4^x}{(x + 2)!}\right)}{((2x+1)^2+\ln x)x!} &= \lim_{x \to +\infty} \frac{(x^2+3x+2)\left(1+\frac{4^x}{(x + 2)!}\right)}{4x^2\left(1 + \frac1x + \frac1{4x^2}+\frac{\ln x}{4x^2}\right)} =\\
&= \lim_{x \to +\infty} \frac{\cancel{x^2}\left(1+\frac3x+\frac2{x^2}\right)\left(1+\frac{4^x}{(x + 2)!}\right)}{4\cancel{x^2}\left(1 + \frac1x + \frac1{4x^2}+\frac{\ln x}{4x^2}\right)} = \frac14.
\end{align}
A: $$\lim\limits_{x \to +\infty} \frac{(x+2)!+4^x}{(2x+1)^2(1+\frac{\ln x}{(2x+1)^2})(x)!} = $$
$$\lim\limits_{x \to +\infty} \frac{(x+2)!+4^x}{(2x+1)^2(x)!} = $$
$$\lim\limits_{x \to +\infty} \frac{(x+2)!}{(2x+1)^2(x)!} + \lim\limits_{x \to +\infty} \frac{4^x}{(2x+1)^2(x)!} = $$
$$\lim\limits_{x \to +\infty} \frac{(x+2)(x+1)}{(2x+1)^2} + \lim\limits_{x \to +\infty} \frac{4^x}{(2x+1)^2(x)!} = $$
$$\lim\limits_{x \to +\infty} \frac{x^2+3x+1}{4x^2+4x+1} + 0 =$$ 
$$\lim\limits_{x \to +\infty} \frac{1+\frac{3}{x}+\frac{1}{x^2}}{4+\frac{4}{x}+\frac{1}{x^2}} =\frac{1}{4}$$
The only missing part is 
$$\lim\limits_{x \to +\infty} \frac{4^x}{(2x+1)^2(x)!} = 0$$
For this you need to notice that
$$\frac{4^{x+1}}{(x+1)!} = \frac{4 \cdot 4^{x}}{(x+1)x!}=\frac{4}{x+1}\frac{4^{x}}{x!} < \frac{4}{x}\frac{4^{x}}{x!}$$
Since $\lim\limits_{n \to +\infty} \frac{4^n}{x^n}=0$ we have $\lim\limits_{x \to +\infty} \frac{4^{x+1}}{(x+1)!}=0$ as well
A: Divide numerator and denominator by $(x+2)!$, which seems to be the dominant factor:
$$
\lim_{x\to\infty}
\frac{
  1+\dfrac{4^x}{(x+2)!}
 }{
  \dfrac{(2x+1)^2}{(x+2)(x+1)}+\dfrac{\ln x}{(x+2)(x+1)}
 }
$$
Can you compute the limits of the three fractions?
A: Use the Stirling approximation for the factorial function (when the argument goes to infinity):
$$x! \approx \sqrt{2\pi x}\left(\frac{x}{e}\right)^x$$
as $x\to +\infty$
Also you can use this in your denominator for $x!$ and on the numerator for $(x+2)!$ in this way:
$$(x+2)! \approx \sqrt{2\pi(x+2)}\left(\frac{x+2}{e}\right)^{x+2} = \sqrt{2\pi(x+2)}\left(\frac{x+2}{e}\right)^x\cdot\left(\frac{(x+2)^2}{e^2}\right)$$
Then try to substitute those in your limit and manage a bit!
A: Rewrite the expression as:
$$\frac{(x+2) (x+1)}{(2 x+1)^2+\log (x)}+\frac{4^x}{x! \left((2 x+1)^2+\log (x)\right)}$$
The second term  is positive, yet less than $4^x/x!$, which is easy to show becomes 0.
Now consider the inverse of the first term and expand it:
$$\frac{4 x^2}{(x+1) (x+2)}+\frac{4 x}{(x+1)
   (x+2)}+\frac{1}{(x+1) (x+2)}+\frac{\log (x)}{(x+1) (x+2)}$$
All of these should be easy :)
A: $$\lim_{ x\to +\infty  } \frac { (x+2)!+4^{ x } }{ ((2x+1)^{ 2 }+\ln { x } )x! } =\lim_{ x\to +\infty  } \frac { x!\left( x+1 \right) (x+2)+4^{ x } }{ ((2x+1)^{ 2 }+\ln { x } )x! } =\lim_{ x\to +\infty  } \frac { x!\left( \left( x+1 \right) (x+2)+\frac { { 4 }^{ x } }{ x! }  \right)  }{ ((2x+1)^{ 2 }+\ln { x } )x! } =\\ =\lim_{ x\rightarrow \infty  }{ \frac { { x }^{ 2 }+3x+2+\frac { { 4 }^{ x } }{ x! }  }{ 4{ x }^{ 2 }+4x+1+\ln { x }  } = } \lim_{ x\rightarrow \infty  }{ \frac { { x }^{ 2 }\left( 1+\frac { 3 }{ x } +\frac { 2 }{ { x }^{ 2 } } +\frac { { 4 }^{ x } }{ { x }^{ 2 }x! }  \right)  }{ { x }^{ 2 }\left( 4+\frac { 4 }{ x } +\frac { 1 }{ { x }^{ 2 } } +\frac { \ln x }{ x }  \right)  } = } \frac { 1 }{ 4 } $$
