lim as x approaches a to (1/the square root of x+2) - (1/the square root of a+2) all divided by x-a?

Question

There is no math tutor available for my summer calculus class. I'm very appreciative of any help.

Answer 1

lim x-> a of [(x+2)^(-1/2) - (a+2)^(-1/2)]/(x-a) is of the form f(x)/g(x) and in the specified limit, both f(x) and g(x) -> 0. We can therefore try to use l'Hopital's rule (see source) to evaluate the limit of f(x)/g(x)

Take the derivative of the numerator and denominator of your expression and ratio them:

{(-1/2)*(x+2)^(-3/2)}/1

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

Evaluate this as x-> a:

= -1/[2*(a+2)^3/2]

This is the desired limit.

Answer 2

Wow.. another calculus question.. like I said in another question I haven't taken a calc class in years, so this is just a stab at it, doesn't mean I'm right.. always like a challenge but this is borderline nerd, lol..

Someone please correct me if I'm wrong again..

*****

Assuming you're taking the square root of (x+2) and the square root of (a+2):

The limit as you approach a from the right hand side (ie. x is greater than a):

Let's say a = 1 for arguement sake. Then if x is approaching 1 from the right hand side (ie. x > a=1) then the fraction you're gonna get from 1 / (square root (x+2)) is going to be smaller than 1 / (square root (a+2)). So the whole numerator is going to negative infinity.

For the denominator, as x approaches a from the right hand side, this number is always going to be positive since x is larger than a, so the denominator goes to positive infinity in this case.

Since you have a negative over a positive, the whole limit is going to negative infinity.

You can try it coming from the left hand side of a as well, and you'll find that the numerator is going to positive infinity while the denominator is going to negative infinity, meaning that the whole limit is going to negative infinity by either case.

(ouch, my head - and I don't even know if I'm right.. gonna go do something fun now.)

Answer 3

Hint: "Rationalize"
That is, in your big compound fraction, multiply both the numerator and denominator by the following:
1/sqroot(x+2) + 1/sqroot(a+2)
(This formula is called the "conjugate" of your original numerator, if you recall your algebra terminology.)
Then simplify the numerator by multiplying it out - and like magic, the square roots will all cancel out. You will get (x-a) in the numerator (along with other stuff), which will cancel the (x-a) in the denominator, and then you can take the limit just by plugging in x = a.

You have to do the algebra carefully. It's easy to make a mistake because it's so complicated.

Answer 4

HF Shaw got it right. First, plug in x = a to see if L'Hopital's applies. You get [1/sqrt(a+2) - 1/sqrt(a+2)] / (a - a) = 0/0 (except, of course, if a = -2).

Your numerator has two terms; HF Shaw has the derivative of the first term, and since the second term is a constant g(a), its derivative is zero.

The derivative of your denominator, x - a, is 1.

So by applying L'Hopital's rule, you get HF Shaw's derivative, divided by 1. Then let x approach a in the limit, and you get your answer.

Answer 5

This is the difference quotient that you use to compute the derivative of the function f(x) = 1/sqrt(x+2) = (x+2)^(-1/2)

f'(x) = (-1/2)(x+2)^(-3/2)
f'(a) = (-1/2)(a+2)^(-3/2)

However, you probably have to actually compute that value.
I will post that in a minute.

Simplify the numerator:
1/sqrt(x+2) - 1/sqrt(a+2) =
sqrt(a+2) - sqrt(x+2) /sqrt(x+2)sqrt(a+2)

Multiply top and bottom by (sqrt(a+2) + sqrt(x+2))
(sqrt(a+2) + sqrt(x+2))(sqrt(a+2) - sqrt(x+2)) /sqrt(x+2)sqrt(a+2) (sqrt(a+2) + sqrt(x+2)) =

((a+2) - (x+2))/ /sqrt(x+2)sqrt(a+2) (sqrt(a+2) + sqrt(x+2))
(a - x)/ sqrt(x+2)sqrt(a+2) (sqrt(a+2) + sqrt(x+2))

Now divide by (x-a) and take the limit
(a - x)/ sqrt(x+2)sqrt(a+2) (sqrt(a+2) + sqrt(x+2))/(x-a) =

-1/ sqrt(x+2)sqrt(a+2) (sqrt(a+2) + sqrt(x+2))

lim x->a -1/ sqrt(x+2)sqrt(a+2) (sqrt(a+2) + sqrt(x+2)) =

-1/sqrt(a+2)sqrt(a+2) (sqrt(a+2) + sqrt(x+2)) =
(-1/2)sqrt(a+2)^3 = (-1/2)(a+2)^(-3/2)

Answer 6

Look up L'Hopital's rule in your calculus book. I think this may be a case where it can be applied to find the limit.

Answer 7

I DON'T TUTOR!

...unless you are really HOT!!

THEN I'M WILLING TO TEACH YOU EVERYTHING!

=)