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

sciencesolve | Teacher | (Level 3) Educator Emeritus

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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.`

embizze | High School Teacher | (Level 1) Educator Emeritus

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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.