For what values of k is lim x -> INF coshkx/sin2x finite?
(Enter your answer as an interval or list of intervals, for example, (-infinity,3], (5,7). Enter none if no values of k will work.)
Please help..
To determine the values of k for which the limit of cosh(kx)/sin^2(x) as x approaches infinity is finite, we can follow these steps:
Step 1: Simplify the expression.
cosh(kx) can be written as (e^(kx) + e^(-kx))/2, and sin^2(x) can be written as (1 - cos^2(x)).
Substituting these into the expression, we get:
(e^(kx) + e^(-kx))/(2(1 - cos^2(x)))
Simplifying further, we have:
(e^(kx) + e^(-kx))/(2 - 2cos^2(x))
Step 2: Determine the behavior of the expression as x approaches infinity.
Notice that as x approaches infinity, e^(kx) will also approach infinity if k > 0, and e^(-kx) will approach zero if k > 0. Similarly, if k < 0, e^(kx) will approach zero, and e^(-kx) will approach infinity. This means that the expression will not be finite for any values of k that are positive or negative.
Step 3: Analyze the cases when k = 0.
When k = 0, the expression becomes:
(e^(0) + e^(0))/(2 - 2cos^2(x))
= 2/(2 - 2cos^2(x))
= 1/(1 - cos^2(x))
We notice that this expression is equivalent to 1/sin^2(x).
Step 4: Determine the behavior of the expression as x approaches infinity when k = 0.
As x approaches infinity, sin(x) oscillates between -1 and 1, and sin^2(x) will always be positive. Therefore, 1/sin^2(x) will also be positive and finite as x approaches infinity.
Step 5: Final answer.
From step 4, we conclude that the limit of cosh(kx)/sin^2(x) as x approaches infinity will be finite if k = 0.
Therefore, the values of k for which the limit is finite are k = 0.