<s>[INST] 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
 [/INST] $$\begin{align}
\frac13 \frac{7^{2x} + 7^{-2x}}{7^{2x} - 7^{-2x}}  
&= \frac13 \frac{7^{2x}(7^{2x} + 7^{-2x})}{7^{2x}(7^{2x} - 7^{-2x})} \\
&= \frac13 \frac{7^{4x} + 1}{7^{4x} - 1} \\
&= \frac13 \frac{(7^{4x} -1) + 2}{7^{4x} - 1} \\
&= \frac13 \Big(1 + \frac{2}{7^{4x} - 1}\Big) \\
&= \frac13 + \frac{2}{3(7^{4x} - 1)}
\end{align}$$
Now, what happens if $x$ goes to infinity?
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