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Another limit using l'hopitals

by magnifik
Tags: lhopitals, limit
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magnifik
#1
Jan17-10, 12:54 PM
P: 361
1. The problem statement, all variables and given/known data
limit as x goes to infinity of (1/x^2) - (cscx)^2


2. Relevant equations



3. The attempt at a solution
I made it so the denominator is x^2, but then it would 1-inf/inf which isn't indeterminate. i need help setting it up so it would be in indeterminate form. thanks.
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Dick
#2
Jan17-10, 02:30 PM
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You probably mean lim x->0, right? Just make a common denominator and combine those two terms into a single fraction. It's probably easier to write 1/sin(x)^2 instead of csc(x)^2, but it will still take several derivatives before you get a nonindeterminant answer from l'Hopital.
magnifik
#3
Jan17-10, 03:28 PM
P: 361
nope, the question states lim x-> inf

Dick
#4
Jan17-10, 03:31 PM
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Another limit using l'hopitals

Quote Quote by magnifik View Post
nope, the question states lim x-> inf
Then tell me about the limiting behavior of 1/x^2 and csc(x)^2 as x->inf. Is that expression really indeterminant?
magnifik
#5
Jan17-10, 03:33 PM
P: 361
that's what my original problem was
Dick
#6
Jan17-10, 03:39 PM
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Quote Quote by magnifik View Post
that's what my original problem was
Sketch a graph of each one. The limiting behavior should be visually obvious.
Altabeh
#7
Jan17-10, 03:56 PM
P: 665
Quote Quote by magnifik View Post
1. The problem statement, all variables and given/known data
limit as x goes to infinity of (1/x^2) - (cscx)^2


2. Relevant equations



3. The attempt at a solution
I made it so the denominator is x^2, but then it would 1-inf/inf which isn't indeterminate. i need help setting it up so it would be in indeterminate form. thanks.
The limit does not exist by any means. If needed, a proof can be given.

AB


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