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

### 3 Answers | Add Yours

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

= `lim_(n-gtoo) [((n^2)(1/n-1/n^2))/((n^2)(1+1/n^2)]`

= `lim_(n-gtoo)` `[(1/n-1/n^2)/(1+1/n^2)]`

= `(1/oo-1/(oo)^2)/(1+1/(oo)^2)`

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

= 0

**Therefore` lim_(n-gtoo)` `(n-1)/(n^2+1) ` = 0**

When `n-gt-oo` same as above happens except the sign of `oo`

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

= `(1/-oo-1/(-oo)^2)/(1+1/(-oo)^2)` `

= (0-0)/(1-0)

**= 0**

**Therefore` lim_(n-gt-oo)` `(n-1)/(n^2+1) ` = 0**

The limits `lim_(n->oo) (n - 1)/(n^2 + 1)` and `lim_(n-> -oo) (n - 1)/(n^2 + 1)` have to be determined.

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

It is seen that as n tends to `oo` , (n^2 + 1) tends to `oo` faster than n - 1 as the power of n is higher in the denominator and `1/oo = 0` . As a result `lim_(n->oo) (n - 1)/(n^2 + 1) = 0`

Similarly for `lim_(n-> -oo) (n - 1)/(n^2 + 1)` , as n tends to `-oo` , n - 1 takes on a larger negative value and n^2 + 1 takes on a larger positive value but as the denominator has a higher power it tends to `oo` at a faster rate and `1/oo = 0` .

**This gives `lim_(n->oo) (n - 1)/(n^2 + 1) = 0` and **`lim_(n-> -oo) (n - 1)/(n^2 + 1) = 0`

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

As n tends to `oo` , so does n^2. If we substitute n = `oo` , we get the indeterminate form `oo/oo` .

This allows the use of l'Hopital's rule and the numerator and denominator can be substituted with their derivatives.

The limit is now: `lim_(n->oo) 1/(2*n)`

Substitute n = `oo` , this gives `1/oo` which is equal to 0

Similarly to find `lim_(n-> -oo) (n-1)/(n^2+1)` , as n tends to `-oo` , n^2 tends to `oo` . If we substitute n = `oo` , we get the indeterminate form `-oo/oo` .

This allows the use of l'Hopital's rule and the numerator and denominator can be substituted with their derivatives.

The limit is now:

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

Substituting n = `-oo` gives the result 0

The limit is 0 in both cases as -0 = 0.