Limit calculation $\lim_{n\to\infty}((\frac94)^n+(1+\frac1n)^{n^2})^{\frac1n}$
Here's what I did:
$\lim_{n\to\infty}((\frac94)^n+(1+\frac1n)^{n^2})^{\frac1n}\\
=(\lim(\frac94)^n+\lim((1+\frac1n)^{n})^n)^{\frac1n}\\
=(\lim(\frac94)^n+\lim e^n)^{\frac1n}\\$
Any hints on how to continue?
PS: no logs/integration/derivation because we haven't covered it. 
 A: $$
\begin{align}
&\lim_{n\to\infty}\left(\left(\frac94\right)^n+\left(1+\frac1n\right)^{n^2}\right)^{1/n}\tag{1}\\
&=\lim_{n\to\infty}\left(1+\frac1n\right)^n\lim_{n\to\infty}\left(\left(\frac94\left(1+\frac1n\right)^{-n}\right)^n+1\right)^{1/n}\tag{2}\\
&=e\lim_{n\to\infty}\left(\left(\frac94e^{-1}\right)^n+1\right)^{1/n}\tag{3}\\[9pt]
&=e\lim_{n\to\infty}\left(0+1\right)^{1/n}\tag{4}\\[18pt]
&=e\tag{5}
\end{align}
$$
Explanation:
$(1)$: original expression
$(2)$: bring a factor of $\left(1+\frac1n\right)^n$ outside the parentheses
$(3)$: evaluate $\lim\limits_{n\to\infty}\left(1+\frac1n\right)^n=e$
$(4)$: evaluate $\lim\limits_{n\to\infty}\left(\frac94e^{-1}\right)^n=0$
$(5)$: evaluate $\lim\limits_{n\to\infty}1^{1/n}=1$
A: We know that the sequence $(a_n)$ defined by
$$
a_n=\left(1+\frac1n\right)^n
$$
converges and its limit is $e$. Notice that
$$
\left[\left(\frac94\right)^n+\left(1+\frac1n\right)^{n^2}\right]^{1/n}=\left[\left(\frac94\right)^n+a_n^n\right]^{1/n}=a_n\left[1+\left(\frac{9}{4a_n}\right)^n\right]^{1/n}.
$$
Since
$$
\lim_{n\to\infty}\frac{9}{4a_n}=\frac{9}{4e}<1,
$$
it follows that
$$
\lim_{n\to\infty}\left[\left(\frac94\right)^n+\left(1+\frac1n\right)^{n^2}\right]^{1/n}=\lim_{n\to\infty}a_n\left[1+\left(\frac{9}{4a_n}\right)^n\right]^{1/n}=e(1+0)^0=e.
$$
A: The limit is indeed $e$. 
Let $a(n) = \frac94$ and $b(n) = (1+\frac1n)^n$ for $n \ge 1$. 
Since $\lim b(n) = e > 9/4 = 2.25$ ( versus $2.71728\ldots$ ) there is an $N$ such that if $n \ge N$ then $b(n) > 9/4$. Thus $x(n)=
\frac{a(n)}{b(n)} < 1$ for $n \ge N$. 
Let $S(n) =$ the original expression, then:
$S(n) = (a(n)^n + b(n)^n)^{\frac1n} = b(n)(1+(\frac{a(n)}{b(n)})^n)^{\frac1n} = b(n)r(n)$ 
whereas $r(n) = (1+(\frac{a(n)}{b(n)})^n)^{\frac1n}$ . 
We now estimate $r(n)$. It is easy to see that $1 < r(n) < 2^{\frac1n}$ for $n \ge N$. 
This means $\lim r(n) = 1$ and so $ \lim S(n) = e$ as claimed.
A: For positive $n$, we have that $0 < \left(1 + \frac{1}{n}\right)^{n^2} = \left(\left(1 + \frac{1}{n}\right)^n\right)^n < e^n$.  Thus, we can write
$$\left(1 +  \frac{1}{n}\right)^{n^2} < \left(\frac{9}{4}\right)^n + \left(1 + \frac{1}{n}\right)^{n^2} < \left(\frac{9}{4}\right)^n + e^n < 2 e^n$$
$$\left(\left(1 +  \frac{1}{n}\right)^{n^2}\right)^{1/n} < \left(\left(\frac{9}{4}\right)^n + \left(1 + \frac{1}{n}\right)^{n^2}\right)^{1/n} < \left(2e^n\right)^{1/n}.$$
We can use the squeeze theorem on the last inequality to obtain the limit.
