# 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?

$$\lim\limits_{x \to +\infty} \frac{(x+2)!+4^x}{((2x+1)^2+\ln x)x!}$$

• Use derivatives,l hospitals rule also $x!$ is derivative of $x^n$ – Archis Welankar Jan 2 '16 at 13:35
• @ArchisWelankar $x!$ is not the derivative of $x^n$. Its derivative is $n\cdot x^{n-1}$ – Von Neumann Jan 2 '16 at 13:38
• Amarildo: being rather simple-minded, I'd encourage you to use $n$ for integers, and keep $x$ for real-valued variables. For your limit, as a first question: do you know which goes to infinity "the fastest", between factorial and exponential? I.e., $\lim_{n\to\infty} \frac{c^n}{n!}$, for any constant $c > 0$? – Clement C. Jan 2 '16 at 13:41
• No like see $x^2$ so its nth derivative will be 2 which us $2!$ i would rather add word nth derivative of $x^n$ – Archis Welankar Jan 2 '16 at 13:47

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:

$$\lim\limits_{x \to +\infty} \frac{(x+2)!}{(2x+1)^2 \cdot x!} = \lim\limits_{x \to +\infty} \frac{(x+2)(x+1)}{(2x+1)^2 } = 1/4$$

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}

$$\lim\limits_{x \to +\infty} \frac{(x+2)!+4^x}{(2x+1)^2(1+\frac{\ln x}{(2x+1)^2})(x)!} =$$ $$\lim\limits_{x \to +\infty} \frac{(x+2)!+4^x}{(2x+1)^2(x)!} =$$ $$\lim\limits_{x \to +\infty} \frac{(x+2)!}{(2x+1)^2(x)!} + \lim\limits_{x \to +\infty} \frac{4^x}{(2x+1)^2(x)!} =$$ $$\lim\limits_{x \to +\infty} \frac{(x+2)(x+1)}{(2x+1)^2} + \lim\limits_{x \to +\infty} \frac{4^x}{(2x+1)^2(x)!} =$$ $$\lim\limits_{x \to +\infty} \frac{x^2+3x+1}{4x^2+4x+1} + 0 =$$ $$\lim\limits_{x \to +\infty} \frac{1+\frac{3}{x}+\frac{1}{x^2}}{4+\frac{4}{x}+\frac{1}{x^2}} =\frac{1}{4}$$

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

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?

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!

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 :)

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