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Divide both numerator and denominator by x:
`x/(x^2+3) = 1/(x+3/x)` .
When `x -> oo,` `3/x -> 0` and the limit is `1/(oo + 0) = 1/oo ` = 0.
Plug in ` ` `oo` everywhere for x
Since you have ` ` `oo/oo` , you can use L^Hopital's Rule and differentiate the numerator and denominator independently.
1 divided by a really big number is simply 0.
the `lim_(x->oo)x/(x^2+3)=0 `
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