End Behavior Activity

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Finding the Pot of Gold at the End of the Rainbow, aka Finding the Limit as $x$ Goes to $\infty$ or $-\infty$

You goal is to find the pot of gold at the end of the creatively shaped rainbow. A function will be provided to let you know of the shape of the rainbow and the pot of gold will either be at the left or right "end" of the rainbow. To accurately find the pot of gold you must choose the end behavior, that is find $\displaystyle \lim_{x\to \infty} f(x)$ or $\displaystyle \lim_{x\to -\infty} f(x)$ . If the limit exists and is finite, select the correct choice and the value of the constant, using a decimal rather than a fraction if necessary, to find the pot of gold at the end of the rainbow.

$f(x) = \frac{2x-3}{5x+1}, \displaystyle \lim_{x\to \infty} f(x)$

$f(x) = \frac{4-x^2}{4+x}, \displaystyle \lim_{x\to \infty} f(x)$

$f(x) = \frac{3}{x}+\sin(x), \displaystyle \lim_{x\to -\infty} f(x)$

$f(x) = \frac{x^2-2x+3}{x^3+x^2-1}, \displaystyle \lim_{x\to -\infty} f(x)$

$f(x) = \frac{6x-1}{\sqrt{9x^2+2}}, \displaystyle \lim_{x\to -\infty} f(x)$

$f(x) = \frac{3e^x+e^{-x}}{2e^x-e^{-x}}, \displaystyle \lim_{x\to \infty} f(x)$

$f(x) = \frac{e^x+e^{-2x}}{5e^x-6e^{-x}}, \displaystyle \lim_{x\to -\infty} f(x)$

lim =
$\infty$
$-\infty$
Doesn't Exist and Not Infinite

$gold.png$ $elf.png$