How to calculate $\lim\limits_{ x \to \infty } \frac {7^{2x} +7^{-2x} }{ 3(7^{2x} - 7^{-2x})}$ $$\lim\limits_{ x \to \infty } \frac {7^{2x} +7^{-2x} }{ 3(7^{2x} - 7^{-2x})}$$
someone please help i’m not sure how to compute this. i’ve tried to do it this way:$$\lim\limits_{ x \to \infty } \frac {7^{2x} +7^{-2x} }{ 3(7^{2x} - 7^{-2x})} = \frac {1-1/7^{4x} }{1+ 1/21^{4x}} $$
$$\lim\limits_{ x \to \infty } \frac {7^{2x} +7^{-2x} }{ 3(7^{2x} - 7^{-2x})}
= \dfrac {\lim\limits_{ x \to \infty } 1-1/7^{4x} }{ \lim\limits_{ x \to \infty } 1+ 1/21^{4x} }$$
 but that’s defiantly wrong
 A: $$\begin{align}
\frac13 \frac{7^{2x} + 7^{-2x}}{7^{2x} - 7^{-2x}}  
&= \frac13 \frac{7^{2x}(7^{2x} + 7^{-2x})}{7^{2x}(7^{2x} - 7^{-2x})} \\
&= \frac13 \frac{7^{4x} + 1}{7^{4x} - 1} \\
&= \frac13 \frac{(7^{4x} -1) + 2}{7^{4x} - 1} \\
&= \frac13 \Big(1 + \frac{2}{7^{4x} - 1}\Big) \\
&= \frac13 + \frac{2}{3(7^{4x} - 1)}
\end{align}$$
Now, what happens if $x$ goes to infinity?
A: Put the constant outside of the limit:
$$
\begin {align}
l=&\lim\limits_{ x \to \infty }\color{blue}{\frac 1 3} \frac {7^{2x} +7^{-2x} }{ (7^{2x} - 7^{-2x})} \\
l=&\color{blue}{ \frac 1 3}\lim\limits_{ x \to \infty } \frac {7^{2x} +7^{-2x} }{ 7^{2x} - 7^{-2x}} \\
l =& \frac 1 3\lim\limits_{ x \to \infty } \frac {\color{blue}{7^{2x}}( 1+ 7^{-4x}) }{ \color{blue}{7^{2x}} (1 - 7^{-4x})} \\
l =& \frac 1 3\lim\limits_{ x \to \infty } \frac {\color{blue}{} 1+ 7^{-4x} }{ \color{blue}{} 1 - 7^{-4x}} \\
l=&\frac  1 3
\end{align}
$$
A: Remember that
$$\coth x = \frac{e^{x}+e^{-x}}{e^{x}-e^{-x}}, \quad\implies\quad \frac{7^{2 x}+7^{-2 x}}{7^{2 x}-7^{-2 x}} = \frac{e^{(2\ln 7)x}+e^{-(2\ln 7)x}}{e^{(2\ln 7) x}-e^{-(2\ln 7) x}}=\coth((2\ln7)x). $$
Knowing the behavior of $\coth$ at infinity, that is
$$ \lim_{x\to\infty} \coth x = \lim_{x\to\infty} \frac{1}{\tanh x} = 1, $$
you immediately get $L = 1/3$.
A: In that limes, $x\to\infty$ is equivalent to $7^{2x}\to\infty$, hence:
$$\lim\limits_{x \to \infty} \frac {7^{2x} +7^{-2x} }{ 3(7^{2x} - 7^{-2x})}
= \lim\limits_{z \to \infty} \frac {z +z^{-1}}{ 3(z - z^{-1})}
= \frac13\lim\limits_{z \to \infty} \frac {z^2+1}{z^2 - 1}
= \frac13
$$
