How do you determine the limit of $(x+2)/(3x^2)$ as x approaches 0-?

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Answer 1 · 2018-05-18T18:29:05

$lim_(x->0-)(x+2)/(3x^2)=oo$

I would do it this way:
Let $x=1/y$ with $x<0$ ,

$(x+2)/(3x^2)$ = $(1/y+2)/(3/y^2)$

= $(y+2y^2)/(3)$ (I multiplied numerator and denominator with $y^2$ to get rid of the fractions in the numerator and denominator)

We see that $lim_(y->-oo)(y+2y^2)/(3)=oo$ since $lim_(y->-oo)2/3y^2=oo$

Therefore $lim_(x->0-)(x+2)/(3x^2)=oo$

This makes sense, since as $x->0-$ the numerator will be closer and closer to 2, i.e.

$(x+2)/(3x^2)->2/(3x^2)-> 2/0$ which has $oo$ as it's limit.

How do you determine the limit of (x+2)/(3x^2)(x+2)/(3x^2) as x approaches 0-?