# use limit laws to find the limit n-1/n^2+1 as x approaches negative infinity

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Find `lim_(x->-oo) (n-1)/(n^2+1)`

`lim_(x->-oo)(n-1)/(n^2+1)`

`=lim_(x->-oo)(n-1)/(n^2+1)*(1/n^2)/(1/n^2)` Multiply by "1"

`=lim_(n->-oo)(1/n-1/n^2)/(1+1/n^2)` Algebra

`=(lim_(n->-oo)(1/n-1/n^2))/(lim_(n->-oo)(1+1/n^2))` Limit of quotient is the quotient of limits

`=(lim_(n->-oo)1/n-lim_(n->-oo)1/n^2)/(lim_(n->-oo)1+lim_(n->-oo)1/n^2)` Limit of sum/difference is sum/diff of limits

`=(0-0)/(1+0)` `lim_(n->+-oo)1/n^k=0` for `k>1`

=0

**Thus the limit is zero.**

L'Hospital's rule can of course be used, but the instructions were to use the properties of limits.

You need to evaluate the following limit such that:

`lim_(n->-oo) (n - 1)/(n^2 + 1)`

Substituting `-oo` for n yields:

`lim_(n->-oo) (n - 1)/(n^2 + 1) = (-oo-1)/(oo+1) = -oo/oo`

Since the result is indeterminate -`oo/oo` , you may use l'Hospital's theorem such that:

`lim_(n->-oo) (n - 1)/(n^2 + 1) = lim_(n->-oo) ((n - 1)')/((n^2 + 1)')`

`lim_(n->-oo) (n - 1)/(n^2 + 1) = lim_(n->-oo) 1/(2n)`

Substituting -`oo` for n yields:

`lim_(n->-oo) 1/(2n) = 1/(2*(-oo)) = 1/(-oo) = 0`

**Hence, evaluating the given limit, using l'Hospital's theorem, yields `lim_(n->-oo) (n - 1)/(n^2 + 1) = 0.` **