# Calculus FInd the Limit by using L'Hospital's Rule. lim x--> infinity (x^2 +2)^(1/2))/(2x^2+1)^(1/2)) You need to substitute `oo`  for x in equation under limit to check if the limit is indeterminate such that:

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

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

Since the limit is indeterminate, you may use l'Hospital's theorem such that:

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

`lim_(x-gtoo) ((sqrt(x^2+2))')/((sqrt(2x^2+1))') = lim_(x-gtoo)...

You need to substitute `oo`  for x in equation under limit to check if the limit is indeterminate such that:

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

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

Since the limit is indeterminate, you may use l'Hospital's theorem such that:

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

`lim_(x-gtoo) ((sqrt(x^2+2))')/((sqrt(2x^2+1))') = lim_(x-gtoo) ((2x)/(2sqrt(x^2+2)))/((4x)/(2sqrt(2x^2+1)))`

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

`lim_(x-gtoo) (sqrt(2x^2+1))/(2sqrt(x^2+2)) = lim_(x-gtoo) (xsqrt(2+1/x^2))/(2xsqrt(1+2/x^2))`

`lim_(x-gtoo) (sqrt(2+1/x^2))/(2sqrt(1+2/x^2)) = sqrt2/2`

Hence, evaluating the limit to the function yields `lim_(x-gtoo) (sqrt(x^2+2))/(sqrt(2x^2+1)) = sqrt2/2.`

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