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

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...

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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.

Approved by eNotes Editorial Team

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.`

Approved by eNotes Editorial Team