# find limit of y=(5n^2)/(n^2+2)

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### 1 Answer

y=`(5n^2)/(n^2+2)`

=` (5*n^2)/[n^2(1+2/n^2)]`

= `5/(1+2/n^2)` -----(1)

To y to be a maximum, term `(1+1/n^2)` to be minimum.

To `(1+1/n^2)` to be a minimum 2/n^2 to be minimum.

To `2/n^2` to be minimum `n^2` to be maximum.

The max value n can take is a value that `n^2` tends infinity.

So from (1) **the max value of y is y = `lim_(x->oo)` `5/(1+2/n^2)`= 5**

To y to be a minimum, term (1+1/n^2) to be maximum.

To `(1+1/n^2)` to be a maximum `2/n^2` to be maximum.

To `2/n^2` to be maximum `n^2` to be minimum.

Since `n^2gt=0` the minimum value n can take is a value that `n^2` tends 0.

So the **minimum value of y = `lim_(x->0)``5/(1+2/n^2)` = 5/`oo` = 0**

so limits of the function are;

`y in (0,5)`

`n in (-oo,0)uu(0,oo)`