calculus limit?

Question

Hi, can someone help me figure this out:

Let f(x) = √(x-1)

compute:

lim h->0 (f[h+2] - f[2]) / (h)

It wants the fraction simplified as much as possible and then limit taken. It also says to rationalize the numerator. Finally, if answer is not a teminating decimal, give the exact answer in rational radical form.

Thanks in advance...

Answer 1

The first thing to do for a problem like this is to computer f(h+2), & f(2). f(h+2)= √(h+1), f(2)=1

So you have lim h->0 (√(h+1) - 1) / h

First simplify the fraction. by multiplying the numerator & denominator by the conjugate (That is multiplying the top & bottom of the fraction by ( √(h+1) + 1 )

you get: h / ( h √(h+1) + h)

Then you can factor out h on the top & bottom. (See fraction below!)

h (1)
__________

h (√(h+1) + 1)
, & h cancels.

Then you have lim h->0 1/ (√(h+1) + 1) , which has a value of 1/2, when you plug in 0 for h.

So lim h->0 = 1/2

Enjoy!!

Answer 2

You have to rationalize a numerator that looks like √(h+2-1) - √(2-1) which simplifies to √(h+1) - 1

To rationalize a denominator like that, you would multiply by the conjugate, √(h+1) + 1, in numerator and denominator, then simplify. Same thing works for rationalizing a numerator.

I thought doing it in my head that a √2 would stick around in the denominator. But as Pascal shows you, it doesn't (so you do have a terminating decimal).

Answer 3

[h→0]lim (f(2+h) - f(2))/h
Evaluate the function:
[h→0]lim (√(1+h) - √1)/h
Multiply both the numerator and denominator by (√(1+h) + 1):
[h→0]lim (√(1+h) - 1)(√(1+h) + 1)/(h(√(1+h) + 1))
Multiply out the numerator:
[h→0]lim (1+h-1)/(h(√(1+h) + 1))
Simplify:
[h→0]lim (h)/(h(√(1+h) + 1))
Cancel the h:
[h→0]lim 1/(√(1+h) + 1)
Evaluate the limit:
1/(√1 + 1)
Simplify:
1/2