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

Multiply numerator and denominator by `1/n^2` :

`lim_(n->oo)(n^2-1)/(n^2+1)=lim_(n->oo)(1-1/n^2)/(1+1/n^2)` Now `lim(A/B)=(limA)/(limB)`

`=(lim_(n->oo)(1-1/n^2))/(lim_(n->oo)(1+1/n^2))` And `lim(A+B)=limA+limB`

`=(lim_(n->oo)1-lim_(n->oo)1/n^2)/(lim_(n->oo)1+lim_(n->oo)1/n^2)` `lim_(n->oo)1/n^2=0` ;`lim_(n->oo)c=c`

`=(1-0)/(1+0)`

`=1`

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

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This agrees with the fact from Algebra that the limit as x increases without bound of `f(x)=(p(x))/(q(x))` where `p(x) "and" q(x)` are polynomials is determined by the degrees of p(x) and q(x). Since they are equal, the limit is the quotient of the leading coefficients, or 1/1=1

The graph:

The limit `lim_(n->oo) (n^2 - 1)/(n^2 + 1)` has to be determined.

Substituting `n = oo` , gives `n^2 - 1 = n^2 + 1 = oo` .

As the indeterminate form `oo/oo` is obtained when `oo` is substituted for n, l'Hopital's rule can be applied and the numerator and denominator substituted with their derivatives.

This gives:

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

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