What is $lim_(->-oo) f(x) = absx/(16-x^2)$ ?

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Answer 1 · 2015-11-21T17:23:58

$lim_(x rarr -oo)|x|/(16-x^2) = 0$

Since we're dealing with $-oo$ we can say $|x| = -x$ , so

$lim_(x rarr -oo)|x|/(16-x^2) = lim_(x rarr -oo)-x/(16-x^2)$

Now, with that out of the way, factor out an $x$ on the denominator
$lim_(x rarr -oo)-x/(16-x^2) = lim_(x rarr -oo)x/(x(16/x - x))$

Cancel the two $x$ s otu

$lim_(x rarr -oo) = 1/(16/x - x)$

Now, because that $16/x$ will become a $0$ in the limit, which will make the limit as a whole become a $0$ , or

$lim_(x rarr -oo) 1/(0 - x) = lim_(x rarr -oo) 1/x = 0$

What is lim_(->-oo) f(x) = absx/(16-x^2) lim_(->-oo) f(x) = absx/(16-x^2) ?