TIFR CSE 2010 | Part A | Question: 7

Question

The limit of $\dfrac{10^{n}}{n!}$ as $n \to \infty$ is.

• A. $0$
• B. $1$
• C. $e$
• D. $10$
• E. $\infty$

Comments

I think this question needs a kinda change, because $n!$ is not defined for non integers, so i think it's has to be nth like nth general term of sequence.

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@Lakshmi Narayana404 Either you take $n$ as positive integer or a positive real, answer will remain the same.

In case of positive integers, you get the limit of a sequence as zero and in case of positive reals, limit of a function as zero.

And also you can extend the factorial for positive reals using the Gamma Function.

Just check with your gate calculator whether you get a value for $0.5!$ or not and compare it with $\frac{\sqrt{\pi}}{2}.$  

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Right now i don't know much about gamma functions, but I just remember a glimpse of it as it's extension of factorial function for non negative $x$, so for this question, as limit of function is zero ( $f(n) = a_n$ $ \forall n\geq x_0$} so it implies the limit of sequence also zero.

Answer 1

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Answer 2

Asymptotic growth of N! is greater than 10^n

so at infinity it will be ZERO..