$\lim_\limits{x\to 3}\left(\frac{\left(x!-2x\right)}{x-3}\right)$

I tried substituting $x(x-1)$ in place for $x!$ as the limit approaches $3$, giving the limit $3$ by factorization.
But while plotting graph on Desmos calculator, it showed a weird gamma function graph, giving the answer around 5.5 like this
But when I plotted graph substituting $x(x-1)$ in place for $x!$, the graph was a straight line and gave limit $3$. like this
Please help me to get the correct answer.

  • 1
    $\begingroup$ How would you define $x!$ without the gamma function? Keeping in mind that the function needs to be defined for all $x$ in an interval like $(2.99,3.01)$? Anyway, looks like the derivative of gamma should come out. Either by l'Hospital (if allowed) or possibly by using the functional equation of gamma. $\endgroup$ – Jyrki Lahtonen May 5 '20 at 4:19
  • $\begingroup$ How did you get your original answer by factorisation? You still get an indeterminate form. Better check your working. You need an extension of the factorial valid for non-integers for this, and the most commonly used one is the gamma function. $\endgroup$ – Deepak May 5 '20 at 4:47
  • $\begingroup$ I tried substituting x(x−1) in place for x!, giving x^2-x $\endgroup$ – Aatmaj May 5 '20 at 5:00

I believe the first answer is correct. Mathematica gives the answer as $9 - 6\gamma$, where $\gamma$ is the Euler-Mascheroni constant. The reason for this has to do with the fact that the (generalized) factorial can be expressed in terms the Gamma function \begin{align*} \Gamma(z) = \int_0^\infty x^{z-1}e^{-x}dx, \end{align*} when $\Re z > 0$. This function has the property that $\Gamma(n) = (n-1)!$ for all positive integers $n$ and so it is the natural extension of the factorial. Using this, we can replace $x! = \Gamma(x+1) = \int_0^\infty y^{x}e^{-y}dy$. The Wikipedia page for the gamma function presents a nice form for its derivative when $x$ is an integer:

\begin{align*} \Gamma'(x+1) =x!\left(-\gamma+\sum_{k=1}^x\frac{1}{k}\right) \end{align*}

Now applying L'Hopital's rule and use the above formula for the derivative we get: \begin{align*} \lim_{x \rightarrow 3} \frac{x!-2x}{x-3} &= \lim_{x \rightarrow 3} \frac{\Gamma(x+1)-2x}{x-3} = \lim_{x \rightarrow 3} \Gamma'(x+1)-2 = \Gamma'(4) - 2\\ &= 3!\left(-\gamma + 1 + \frac{1}{2} + \frac{1}{3}\right) - 2 = 9-6\gamma \approx 5.5. \end{align*}


Not sure why you're using $x(x-1)$ in place of $x!$. It is much easier to consider the gamma function $\Gamma$ and, its "logarithmic derivative", the digamma function $\psi$. In particular, the property

$$\psi(x) = \frac{\Gamma'(x)}{\Gamma(x)},$$

holds. This gives us the quite handy identity that

$$\Gamma'(x) = \Gamma(x) \psi(x).$$

Let the limit be denoted as $L$. Then,

$$L = \lim_{x \to 3} \frac{\Gamma(x+1) - 2x}{x - 3}.$$

This has a $0/0$ indeterminate form, so you can use L'Hopital's rule to obtain

$$L = \lim_{x \to 3} (\Gamma'(x+1) - 2),$$

with which we have

$$L = \lim_{x \to 3} (\psi(x+1) \Gamma(x+1) - 2) = \psi(4)\Gamma(4) - 2.$$

Here, $\Gamma(4) = 3! = 6$. The digamma function is given by

$$\psi(x) = -\gamma + \sum_{n=1}^{x-1} \frac1n$$

for integers $x$. The constant $\gamma$ is the Euler-Mascheroni constant so $\psi(4) = 1 + 1/2 + 1/3 - \gamma$. Thus

$$L = 9 - \gamma.$$


Just for your curiosity.

$$\lim_\limits{x\to 3}\left(\frac{\left(x!-2x\right)}{x-3}\right)=\lim_\limits{y\to 0}\left(\frac{\Gamma (y+4)-2 (y+3)}{y}\right)$$ $$\Gamma (y+4)=6+(11-6 \gamma ) y+\frac{1}{2} \left(12-22 \gamma +6 \gamma ^2+\pi ^2\right) y^2+O\left(y^3\right)$$ $$\frac{\Gamma (y+4)-2 (y+3)}{y}=(9-6 \gamma )+\frac{1}{2} \left(12-22 \gamma +6 \gamma ^2+\pi ^2\right) y+O\left(y^2\right)$$ which shows the limit and how it is approached.

Using this for $x=\pi$ would give $6.32750$ while the exact value would be $6.39085$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.