# Another limit using l'hopitals

by magnifik
Tags: lhopitals, limit
 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.
 Sci Advisor HW Helper Thanks P: 25,235 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.
 P: 361 nope, the question states lim x-> inf
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P: 25,235
Another limit using l'hopitals

 Quote by magnifik 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?
 P: 361 that's what my original problem was
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P: 25,235
 Quote by magnifik that's what my original problem was
Sketch a graph of each one. The limiting behavior should be visually obvious.
P: 665
 Quote by magnifik 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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