is there an easier way to solve besides rationalizing...like taking the deriv...?

Question

lim root (x^2+9) -3 / (t^2)
t->0

Answer 1

The problem isn't clearly stated. I assume you're dealing with a fraction where the numerator is(t^2+9)^(1/2) - 3, and the denominator is t^2. This does indeed go to 0/0 as t approaches zero, so you should use l'hopital's rule: if f(t) and g(t) both approach zero, then lim f(t)/g(t) = lim f'(t)/g'(t).

The derivative of the denominator is 2t, which still goes to 0.
Using the chain rule, the derivative of the numerator is
(1/2)*(t^2+9)^(-1/2)*2t, which also goes to zero.
So we apply l'hopital again.

The second derivative of the denominator is 2
To take the second derivative of the numerator, substitute u=t^2. Then du = 2t, so our first derivative turns into:
(1/2)*(u+9)^(-1/2) du
which gives us a second derivative of (1/2)*(-1/2)*(u+9)^(-3/2)
substituting t^2 in for u and simplifying,
(-1/4)*(t^2+9)^(-3/2)
but t is going to zero, so this turns into:
(-1/4)*9^(-3/2)
(-1/4)*(1/27)
(-1/108)

so assuming I did everything right here,
the limit is (-1/108)/2, or -1/216

In any case, when you have a limit that goes to 0/0 or infinity/infinity, l'hopital's rule is the thing to use.

Answer 2

No. L'Hopital's rule only applies when the limit reduces to 0/0 or inf/inf.

In this case, t only exists in the denominator. The limit doesn't exist since it approaches -inf from the right side and +inf from the left side.

If the x is supposed to be t, then the limit still doesn't exist. you wind up with:

(t - 3 - 3i) / t^2

The numerator approaches -3 - 3i while the denominator approaches 0.

Answer 3

I think so, but you may as well take the derivative and rationalize just to make sure you are right. That's what I would do.

Answer 4

I'm not getting your question, however you can use the l'hospital rule. That is by differentiating the numerator and denorminator separetely and substituting the limit given.