How do you find the limit $(4x)/(sqrt(2x^2+1))$ as $x->oo$ ?

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Answer 1 · 2018-04-13T22:56:01

We have that

$(4x)/(sqrt(2x^2+1))=(4x)/(sqrt(x^2(2+1/x^2)))=(4x)/(absx*(sqrt(2+1/x^2)))$

If x goes to $+oo$ the absolute value becomes positive hence $absx=x$
hence

$lim_(x->+oo) (4x)/(absx*(sqrt(2+1/x^2)))= lim_(x->+oo) (4x)/(x*(sqrt(2+1/x^2)))=4/sqrt2=2*sqrt2$

If x goes to $-oo$ the absolute value becomes negative hence $absx=-x$
hence

$lim_(x->+oo) (4x)/(absx*(sqrt(2+1/x^2)))= lim_(x->+oo) (4x)/(-x*(sqrt(2+1/x^2)))=-4/sqrt2=-2*sqrt2$

How do you find the limit (4x)/(sqrt(2x^2+1))(4x)/(sqrt(2x^2+1)) as x->oox->oo?