How to solve this calculus (limit) problem?

Question

what is the limit of (2x^2+3)^(1/2) - (2x^2-5)^(1/2) as x-->infinity

that is: the limit of the squareroot of (2x^2+3) minus the squareroot of (2x^2-5), as x approaches infinity.

Can you show me the steps algebraicly? I had it on a test yesterday and could only solve it by graphing it.

Also how do you algabraicly solve limits as x--> infinity when it is not a rational function?

Answer 1

well dear, Remember always,the important thing in limit is ,you should simplify the function first.
(2x^2+3)^(1/2) - (2x^2-5)^(1/2) =

√(2x^2+3) - √(2x^2-5) =
now ;
lim √(2x^2+3) - √(2x^2-5) = lim(√(2x^2+3))^2 - (√(2x^2-5))^2
x –> ∞
now you can remove ' √' so we have
lim 2x^2 + 3 - 2x^2 + 5 = 3 + 5 = 8
x –> ∞


Good Luck.

Answer 2

Divide by x^2 and try to eliminate all terms with x in the Nr.
If x is in the Dr. it goes to 0.

Answer 3

Multiply and divide by:

sqrt(2x^2 +3) + sqrt(2x^2 - 5)

At nominator use the identity:

(a - b)(a + b) = a^2 - b^2

and you get nominator = 8

If x --> inf. then the denominator --> infinity.

So the limit is zero

Answer 4

Lim ((2x^2+3)^(1/2) - (2x^2-5)^(1/2))
X -> ∞

[(2x^2+3)^(1/2) - (2x^2-5)^(1/2)][(2x^2+3)^(1/2) + (2x^2-5)^(1/2)]
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
[(2x^2+3)^(1/2) + (2x^2-5)^(1/2)]


(2x^2+3) - (2x^2-5)
- - - - - - - - - - - - - - - - - - - - - - - - -
[(2x^2+3)^(1/2) + (2x^2-5)^(1/2)]


8
- - - - - - - - - - - - - - - - - - - - - - - -
[(2x^2+3)^(1/2) + (2x^2-5)^(1/2)]


= 8/∞

= 0

Answer 5

Look at each expression under each radical. As x goes to infinity they are the same but with a different constant added (one being negative). The effect of said constant in each becomes diminishingly small to the value of each radical. If equal numbers are subtracted from each other, what do you get? (Romans didn't have it.).... :-)

Answer 6

notice that: lim (x-->2) [2^x + 2^(3 - x) - 6]/{?[2^(-x)] - 2^(a million - x)} = lim (x-->2) {2^x + 8[2^(-x)] - 6}/{2^(-x/2) - 2[2^(-x)]} = lim (x-->2) (2^x + 8/2^x - 6)/[2^(-x/2) - 2/2^x]. Then, if we replace u = 2^x, we see that u --> 4 as x --> 2. The shrink now turns into: lim (x-->2) (2^x + 8/2^x - 6)/[2^(-x/2) - 2/2^x] = lim (u-->4) (u + 8/u - 6)/(a million/?u - 2/u). via multiplying the numerator and denominator via u^(3/2): lim (u-->4) (u + 8/u - 6)/(a million/?u - 2/u) = lim (u-->4) [u^(5/2) + 8?u - 6u^(3/2)]/(u - 2?u) = lim (u-->4) [?u(?u + 2)(?u - 2)(u - 2)]/[?u(?u - 2)], via factoring the numerator = lim (u-->4) [(?u + 2)(u - 2)] = (2 + 2)(4 - 2) = 8. i desire this enables!

Answer 7

i dont speak english