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 given limit is finite, let's first write the expression in a more manageable form.
Using the identity cosh(x) = (e^x + e^(-x))/2 and sin(2x) = 2sin(x)cos(x), we can rewrite the expression:
lim x -> INF cosh(kx)/sin(2x)
= lim x -> INF [(e^(kx) + e^(-kx))/2] / [2sin(x)cos(x)]
= lim x -> INF [e^(kx) + e^(-kx)] / [4sin(x)cos(x)]
Now, let's focus on the numerator. We have e^(kx) and e^(-kx), both of which increase exponentially as x approaches infinity.
If k is positive, then e^(kx) will grow without bound as x approaches infinity. Similarly, if k is negative, then e^(-kx) will grow without bound as x approaches infinity.
In either case, the numerator will approach infinity faster than the denominator, and the limit will be infinite.
Hence, for a finite limit, k must be equal to zero. When k = 0, the expression becomes:
lim x -> INF [e^(0) + e^(0)] / [4sin(x)cos(x)]
= lim x -> INF [2] / [4sin(x)cos(x)]
= 1/[2sin(x)cos(x)]
Now, let's look at the denominator. The function sin(x)cos(x) will oscillate between positive and negative values as x approaches infinity, resulting in no specific limiting behavior.
Therefore, for all values of k, except for k = 0, the given limit will be infinite.
In conclusion, the values of k for which the given limit is finite are k = 0.
Answer: [0]