calculus limit problem?

Question

i need help finding the limit as x approaches 2 for the function 1/(cuberoot(x^2 -4))

sorry, thats the clearest way i can write the problem. in words its 1 divided by the cube root of the quantity x squared minus 4.

i understand most of how limits work, but im real stuck on how to go about this one. could someone at least get me started?

thanks so much

maybe im not as familiar with things as i thought. what does l'hopitals rule mean?

Answer 1

You people are a load of hacks.

The limit of a function raised to the p power is equivalent to the pth power of the limit of the function for non-negative p. It's a simple property of limits.

Secondly, when you look at f(x) = 1/ (x^2 - 4) the function itself has a vertical assymptote at x= 2. It approaches positive infinity from the right side and negative infinity from the left. This is true even if you ignore the last paragragh since p =1/3 changes nothing in this argument.

Since the limit approaches 2 different values from different sides, the limit does not exist. QED

Answer 2

The question is "how does the function y=1/(x^2-4)^(1/3) behave if x is very very close to 2?"

When x is very very close to 2, x^2-4 will be a number with very small magnitude, but it may be postive (if x is slightly larger than 2) or negative (if x is slightly less than 2). 1 divided by a number very small in magnitude will be very large, but the sign will depend on the sign of the denominator.

In this case
\lim_{x \to 0+}1/(x^2-4)^(1/3)=+infinity
and
\lim_{x \to 0-}1/(x^2-4)^(1/3)=-infinity.

Because the left and right hand limits are different, for this problem the limit does not exist so you might write
"\lim_{x \to 0}1/(x^2-4)^(1/3) does not exist because the left and right limits are not equal."

Remember to be careful with the definition of limit. Some teachers would want you to write \lim_{x \to 0} 1/x^2=+ infinity because the function y=1/x^2 is increasing without bound in a positive sense as x approaches zero. Other teachers would want you to say that the limit does not exist because infinity is not a real number. Be sure that you understand the context of the question! :)

L'Hospital's rule says that if f(x)/g(x) is an indeterminate form of type 0/0 or infinity/infinity if x=a, then
\lim_{x \to a}f(x)/g(x)=
\lim_{x \to a}f'(x)/g'(x).

For this problem, you do not have an indeterminate form (because of the 1 in the numerator) so L'Hospital's rule does not apply in this case. Wait until you've had more calculus before you use L'Hospital's rule! :)

Answer 3

Replace x^2 -4 by Y.
As X appraoches 2, x^2 -4 approaches 0.Then, Y approaches 0 too.
Rewrite the limit using the new substitution.

Cuberoot(x^2 - 4) = cuberootY
thus the function becomes 1/cuberoot(Y).
The limit of the function of 1/cuberoot(Y) is the same as the cube root of the limit as y approaches 0 of the function 1/(Y) (Power rule).
When Y approaches 0 from the left the limit is negative infinity and when Y approaches 0 from the right the limit is positive infinity. Hence, the limit DNE.

Answer 4

Okay,

When evaluated at x=2, the limit as 1/(cuberoot(x^2 -4)) = 1/0

=> the limit does not exist

You only need to evaluate further when the evaluated function is "0/0".

Note [where "a" is in (-infinity, 0) U (0, infinity)...a non-zero number]:
Case 1. 0/a, the limit is 0.
Case 2. a/0, the limit DNE.
Case 3. "0/0", indeterminate, the function needs to be further evaluated.

The answer to your question is: the limit as x approaches 2 of 1/(cuberoot(x^2 -4)) DNE.

Hope this helps and good luck!

Answer 5

Don't worry... you don't need l'Hopital.

If you substitute x = 2 you find the fraction 1/0. The limit is therefore + or - infinity, depending on what direction you come from.

(L'Hopital is used when both numerator and denominator go to zero. That is not the case here.)

Answer 6

You can't use L'Hopital because the denominator is to a non-whole number power (sorry can't say it much clearer). Therefore, if you take the derivative, you won't get anything easier to work with and will still be stuck with the x^2. I'd give the actual answer some thought but I'm at work

Answer 7

Check L'hopital

Answer 8

1/(x *2 -4) ** 1/3 is infinite as numerator is finite and non zero and denominator tends to be zero