How to calculate $\lim\limits_{ x \to \infty } \frac {7^{2x} +7^{-2x} }{ 3(7^{2x} - 7^{-2x})}$

Question

$$\lim\limits_{ x \to \infty } \frac {7^{2x} +7^{-2x} }{ 3(7^{2x} - 7^{-2x})}$$ 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

Answer 1

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

Answer 2

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

Answer 3

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 $L = 1/3$.

Answer 4

In that limes, $x\to\infty$ is equivalent to $7^{2x}\to\infty$, hence: $$\lim\limits_{x \to \infty} \frac {7^{2x} +7^{-2x} }{ 3(7^{2x} - 7^{-2x})} = \lim\limits_{z \to \infty} \frac {z +z^{-1}}{ 3(z - z^{-1})} = \frac13\lim\limits_{z \to \infty} \frac {z^2+1}{z^2 - 1} = \frac13 $$