lim_(x->oo)x^2/sqrt(x^2+1) Evaluate the limit, using L’Hôpital’s Rule if necessary.

 
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