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Since this is a `(-oo)/(-oo)` form
Thus, applying L'Hospital's rule we get
`lim_(x->-oo)(5x^2)/(x+3) = lim_(x->-oo) (10x) = -oo`
Plug in` -oo` everywhere you have x
Since you have , you can use L^Hopital's Rule and differentiate the numerator and denominator independently.
the `lim_(x->-oo)(5x^2)/(x+3)=-oo `
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