# 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

• – Shaun Feb 12 at 23:37
• $\lim x→∞ \frac{7^2x +7^-2x}{3(7^2x - 7^-2x)}$ sorry if this doesn’t work i’m on a phone don’t have a laptop – dolphin Feb 12 at 23:48
• Write $\lim_{x\to\infty} \frac{7^{2x} +7^{-2x}}{3(7^{2x }- 7^{-2x})}$ for $\lim_{x\to\infty} \frac{7^{2x} +7^{-2x}}{3(7^{2x }- 7^{-2x})}$. – Shaun Feb 12 at 23:51
• What's with all these people with limit questions on their phone? – Gae. S. Feb 12 at 23:51
• Also, please edit your question accordingly. – Shaun Feb 12 at 23:52

\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?

• wow this is really confusing – dolphin Feb 12 at 23:50
• You got it? :)) – Azif00 Feb 12 at 23:51
• why is the answer not lim𝑥→∞[1/3+2⋅7^−2𝑥/3(7^2𝑥−7^−2𝑥)] – dolphin Feb 13 at 0:02
• I edited my answer. But yeah, it is the same. – Azif00 Feb 13 at 0:12

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}

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 $$L = 1/3$$.

• +1 for the original answer. – Aryadeva Feb 13 at 4:57
• @Isham Thank you! – giobrach Feb 13 at 8:55

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