TITLE: Find (a) $\lim_{n\to\infty}(n^{1/n^2})$ (b) $\lim_{n\to\infty}(n!^{1/n^2})$
QUESTION [1 upvotes]: Find $$(a) \lim_{n\to\infty}(n^{1/n^2})$$ $$(b) \lim_{n\to\infty}(n!^{1/n^2})$$ 
for $(a)$
$1\leq n^{1/n^2}\leq n^{1/n} $
$\lim(1) _{n \to \infty}= \lim_{n \to \infty} (n^{1/n}) = 1$
Therefore by Squeeze theorem required limit for (a) is 1
for $(b)$
$1\leq n!^{1/n^2} < n!^{1/n}$
Since $n!\geq 1$ therefore, $\lim_{n\to\infty}(n!^{1/n}) = 1$ 
hence, by Squeeze theorem again required limit for $(b) = 1$
Is this correct ?
Can anyone suggest quicker methods to solve such question?
EDIT:
Apparently,
 $\lim_{n\to\infty}(n!^{1/n}) \to \infty$
My problem here is,
we know $lim_{n\to\infty}(n^{1/n}) = 1$ where $n \in N$
since, $n!\in N$ Why isn't the same true for $lim_{n\to \infty}((n!)^{1/n})$?

REPLY [0 votes]: (b) we can use sandwitch theorem as follows:
$1=\lim_{n\to\infty} (n^2)^{\frac{1}{n^2}}\leq (n!)^{\frac{1}{n^2}}\leq\lim_{n\to\infty} (n)^{\frac{1}{n}}=1$
remembering the most fundamental limit : $\lim_{n\to\infty} n^{\frac{1}{n}}=1$
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