find the limit x→ ∞, [1 - sqrt{ x/(x+1) } / 2 +sqrt{ (4x +1) /(x+2) }

[hndalama]

limit x→ ∞, [1 - sqrt{ x/(x+1) }] / [2 +sqrt{ (4x +1) /(x+2) }]

in both sqrt symbols, the numerator and denominator have ∞/∞ so I use Lhopitals rule to determine the limit at ∞, in the numerator the sqrt symbol is 1, so the numerator simplifies to 0. The denominator simplifies to 6 but since the numerator is 0 my answer is the limit is 0. The correct answer is actually 2/7. How do they solve for that and what is wrong with my method?

 

[Nico55]

Hi, I think that given answer 2/7 is wrong (stated in your book, given by your teacher?). When I saw the limit of your function I tried it out and got 0 too. I see you used l'Hopital, I used the technique that under the square root, the +1 in 'x+1' is rejectable as we are working at infinity here, so we get sqrt(x/x), so sqrt(1)=1, so you automatically get 0 for the limit as you don't get 0 in the denominator. Also, if you draw the function, you will see the x-axis is the asymptote of the function, not y=2/7, so my guess is the answer they gave you is false.

 

[stapel]

limit x→ ∞, [1 - sqrt{ x/(x+1) }] / [2 +sqrt{ (4x +1) /(x+2) }]

in both sqrt symbols, the numerator and denominator have ∞/∞....

How are you getting that 1 - 1 (in the numerator) and 2 + 2 (in the denominator) are equalling "infinity"?

Please show all of your work. Thank you!