Calculate the limit using useful limits $$\lim_{x\to \infty }\left(\frac {4^n + n}{4^n + 8n}\right)^{\huge\frac{6\cdot4^n-2}{n}}$$
This is the problem.
Step 1) I know I should get a useful limit form=> $$\lim_{x\to \infty }( \frac {4^n + n}{4^n + 8n})^\frac {6\times4^n-2}{n}  =\lim_{x\to \infty }(1+ \frac {4^n + n}{4^n + 8n}-1)^\frac {6\times4^n-2}{n}  = \lim_{x\to \infty }(1- \frac {7n}{4^n + 8n})^\frac {6\times4^n-2}{n}   $$ And I am stuck here, I know that I should change the power of the Limit in order to get the useful limit but I don't know how
 A: The "useful limit" to consider here is
$$e^x = \lim_{n\to\infty} \left(1 + \frac xn\right)^n$$
Continuing from your last limit expression, write
$$\begin{align*}
\left(1 - \frac{7n}{4^n + 8n}\right)^{\frac{6\cdot4^n - 2}n} &= \left(1 - \frac1{\frac{4^n+8n}{7n}}\right)^{42 \left(\frac{4^n + 8n}{7n} - \frac87\right) - \frac2n} \\[1ex]
&= \left(\left(1 - \frac1m\right)^m\right)^{42} \cdot \left(1-\frac1m\right)^{-48-\frac2n}
\end{align*}$$
where $m=\frac{4^n+8n}{7n}$. As $n\to\infty$, we have $m\to\infty$. Can you finish from here?
A: As a general tactic for working with awkward exponents in limits, take logarithms of everything, then expand the logarithm into a power series, and aim for something like a rational function. This will make the function far more amenable to algebraic rearrangements. For all positive real $x$, we may use the bounds  $1-\frac1x\le \ln x \le x-1$, but as they converge at $x=1$, the bounds are particularly useful there. In the given function, as $n\to\infty$, the base becomes $\displaystyle\frac{4^{n}+n}{4^{n}+8n}\to 1$, so a series expansion into $\ln x \to x-1$ should be appropriate.
$$\begin{align}
y&=\left(\frac{4^{n}+n}{4^{n}+8n}\right)^{\left(\frac{6\cdot4^{n}-2}{n}\right)}
\\
\ln y &= \left(\frac{6\cdot4^{n}-2}{n}\right)\ln\left(\frac{4^{n}+n}{4^{n}+8n}\right)
\\
&\to\left(\frac{6\cdot4^{n}-2}{n}\right)\left(\frac{4^{n}+n}{4^{n}+8n}-1\right)
\\
&=-7\left(\frac{6\cdot4^{n}-2}{4^{n}+8n}\right)
\\
&\to -7 \left(\frac{6\cdot4^{n}}{4^{n}}\right)
\end{align}
$$
I think you can take it from here.
