# What is `lim_(x->oo)(x+7)/(x^2-8)`

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The limit `lim_(x->oo)(x+7)/(x^2 - 8)` has to be determined.

Substituting x = `oo` , gives the indeterminate form `oo/oo` . Using l'Hopital's rule substitute the numerator and denominator of the expression by their derivatives. This gives:

`lim_(x->oo) 1/(2x)`

Substitute `x = oo`

= `1/(2*oo)`

= `1/oo`

= 0

**The limit **`lim_(x->oo)(x+7)/(x^2 - 8) = 0`