Given

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

as `x->oo` then we get `x^2/sqrt(x^2+1)=oo/oo`

since it is of the form `oo/oo` , we can use the L 'Hopital rule

so upon applying the L 'Hopital rule we get the solution as follows,

For the given general equation L 'Hopital rule is as follows

`lim_(x->a) f(x)/g(x)...

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Given

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

as `x->oo` then we get `x^2/sqrt(x^2+1)=oo/oo`

since it is of the form `oo/oo` , we can use the L 'Hopital rule

so upon applying the L 'Hopital rule we get the solution as follows,

For the given general equation L 'Hopital rule is as follows

`lim_(x->a) f(x)/g(x) is = 0/0` or `(+-oo)/(+-oo)` then by using the L'Hopital Rule we get the solution with the below form.

`lim_(x->a) (f'(x))/(g'(x))`

so , now evaluating

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

=`lim_(x->oo)((x^2)')/((sqrt(x^2+1))')`

First let us solve `(sqrt(x^2+1))' `

=> `d/dx (sqrt(x^2+1)) `

let `u=x^2+1 `

so,

`d/dx (sqrt(x^2+1)) `

=`d/dx (sqrt(u))`

= `d/(du) sqrt(u) * d/dx (u) ` [as `d/dx f(u) = d/(du) f(u) (du)/dx` ]

= `[(1/2)u^((1/2)-1) ]*(d/dx (x^2+1))`

= `[(1/2)u^(-1/2)]*(2x)`

=`[1/(2sqrt(x^2 +1))]*(2x)`

=`x/sqrt(x^2+1)`

so now the below limit can be given as

=`lim_(x->oo)((x^2)')/((sqrt(x^2+1))')`

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

=`lim_(x->oo) (2sqrt(x^2+1))`

Now on substituting the value of `x =oo` we get

=` (2sqrt((oo)^2+1))`

= `oo`