Calculus

FInd the Limit by using L'Hospital's Rule.

lim x--> infinity (x^2 +2)^(1/2))/(2x^2+1)^(1/2))

Expert Answers

An illustration of the letter 'A' in a speech bubbles

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.`

See eNotes Ad-Free

Start your 48-hour free trial to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Get 48 Hours Free Access
Approved by eNotes Editorial