1

$\lim_{x\to\infty }\frac{q^{x}}{x!}$ I need to find a limit of this sequence and prove it using the definition.

I know that we can have several variants of the limit of $q^{x}$: if $1>q>-1$ the limit is $0$, for $q=1$ the limit is $1$, for $q>1$ the limit is $\infty $, and for $q\leq -1$ it doesn't exist.

The limit of factorial $x!$ should be $\infty $, but I'm not sure and I don't know how to prove it.

Can you help me please with the whole fraction.

I don't know how to find a limit if we have something like $\frac{0}{\infty }$ or $\frac{1}{\infty }$ or $\frac{\infty}{\infty }$.

Martin Sleziak
  • 56,060
Green Banana
  • 329

2 Answers2

3

For each natural $x$ put $a_x=\frac{q^x}{x!}$. Let $N>|2q|$ be any natural number. Then $$|a_{N+1}|=\left|a_N\frac {q}{N+1}\right|<\frac {|a_N|}2\mbox{ , }|a_{N+2}|=\left|a_{N+1}\frac {q}{N+2}\right|<\frac {|a_{N+1}|}2<\frac {|a_N|}4,\dots,$$ $$|a_{N+k}|<\frac {|a_N|}{2^k},\dots$$ so the sequence $a_x$ tends to zero for any $q$.

Alex Ravsky
  • 106,166
1

By ratio test

$$\lim_{x\to+\infty}\frac{\frac{q^{x+1}}{(x+1)!}}{\frac{q^x}{x!}}=$$

$$\lim_{x\to+\infty}\frac{q}{x+1}=0$$

thus

the series $\sum \frac{q^x}{x!} $ converges and its general term

$\frac{q^x}{x!}$ goes to $0$.

hamam_Abdallah
  • 63,719