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

Put the constant outside of the limit:
$$
\begin {align}
l=&\lim\limits_{ x \to \infty }\color{blue}{\frac 1 3} \frac {7^{2x} +7^{-2x} }{ (7^{2x} - 7^{-2x})} \\
l=&\color{blue}{ \frac 1 3}\lim\limits_{ x \to \infty } \frac {7^{2x} +7^{-2x} }{ 7^{2x} - 7^{-2x}} \\
l =& \frac 1 3\lim\limits_{ x \to \infty } \frac {\color{blue}{7^{2x}}( 1+ 7^{-4x}) }{ \color{blue}{7^{2x}} (1 - 7^{-4x})} \\
l =& \frac 1 3\lim\limits_{ x \to \infty } \frac {\color{blue}{} 1+ 7^{-4x} }{ \color{blue}{} 1 - 7^{-4x}} \\
l=&\frac  1 3
\end{align}
$$
