# Find the limits. lim_{xrightarrow-infty}frac{2^x+4^x}{5^x-2^x}

Find the limits.
$\underset{x\to -\mathrm{\infty }}{lim}\frac{{2}^{x}+{4}^{x}}{{5}^{x}-{2}^{x}}$
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krolaniaN
Consider the limit
$\underset{x\to -\mathrm{\infty }}{lim}\frac{{2}^{x}+{4}^{x}}{{5}^{x}-{2}^{x}}$
Solve:
$\underset{x\to -\mathrm{\infty }}{lim}\frac{{2}^{x}+{4}^{x}}{{5}^{x}-{2}^{x}}$
Take ${2}^{x}$ common from numerator and denominator
$=\underset{x\to -\mathrm{\infty }}{lim}\frac{{2}^{x}\left(1+\frac{{4}^{x}}{{2}^{x}}\right)}{{2}^{x}\left(\frac{{5}^{x}}{{2}^{x}}-1\right)}$
$=\underset{x\to -\mathrm{\infty }}{lim}\frac{{2}^{x}\left(1+\left(\frac{4}{2}{\right)}^{x}\right)}{{2}^{x}\left(\left(\frac{5}{2}{\right)}^{x}-1\right)}$
$=\underset{x\to -\mathrm{\infty }}{lim}\frac{\left(1+{2}^{x}\right)}{\left(\left(2.5{\right)}^{x}-1\right)}$
$=\frac{\underset{x\to -\mathrm{\infty }}{lim}\left(1+{2}^{x}\right)}{\underset{x\to -\mathrm{\infty }}{lim}\left(\left(2.5{\right)}^{x}-1\right)}$
$=\frac{\left(1+{2}^{-\mathrm{\infty }}\right)}{\left({2.5}^{-\mathrm{\infty }}-1\right)}$
$=\frac{\left(1+0\right)}{\left(0-1\right)}$
$=-1$
Jeffrey Jordon