For what values of k is lim x -> INF coshkx/sin2x finite?
For any non-zero value of k, cosh(kx)->∞ as x->∞
For k=0, cosh(kx)->1 as x->∞
However, the numerator being finite does not make the limit existent because the denominator is undefined as x->∞ because sin(2x) as x->∞ is undefined. (It oscillates).
For what values of k is lim as x approaches infinity of (coshkx)/(sinh3x) finite?
Enter your answer as an interval or list of intervals, for example, (-infinity,3], (5,7).
To find the values of k for which the limit of (cosh(kx))/sin(2x) as x approaches infinity is finite, we can analyze the behavior of the function at infinity.
The function cosh(kx) grows exponentially as x gets larger, while sin(2x) oscillates between -1 and 1 as x increases.
For the limit to be finite, we need to balance the growth of cosh(kx) with the oscillation of sin(2x). This can be done if the exponent k is small enough that the growth of cosh(kx) does not dominate the oscillation of sin(2x).
Since sin(2x) oscillates between -1 and 1, we can say that for the limit to be finite, the numerator cosh(kx) should not grow faster than 1/x. This means that kx should not have a greater exponent than x.
Thus, k <= 1 in order for the limit to be finite.
To determine the values of k for which the given limit is finite, we can start by simplifying the expression first.
The limit is given by:
lim(x -> ∞) cosh(kx) / sin^2(x)
First, let's recall the definitions of hyperbolic cosine (cosh) and sine (sinh):
cosh(x) = (e^x + e^(-x)) / 2
sinh(x) = (e^x - e^(-x)) / 2
Using these definitions, we can rewrite the expression:
lim(x -> ∞) [ (e^(kx) + e^(-kx)) / 2 ] / sin^2(x)
Now, we can simplify further by dividing numerator and denominator by e^(kx):
lim(x -> ∞) [ (1 + e^(-2kx)) / (2e^(kx)) ] / sin^2(x)
As x approaches infinity, the term e^(-2kx) goes to 0 since it's an exponential function with a negative exponent. Therefore, we can ignore this term in the expression:
lim(x -> ∞) (1 / 2e^(kx)) / sin^2(x)
Next, let's focus on the sin^2(x) term. We know that sin(x) oscillates between -1 and 1, so when we square it, it becomes positive and always less than or equal to 1. Therefore, the denominator sin^2(x) is always bounded.
For the limit to be finite, we need the numerator to also be bounded as x approaches infinity. This means that the exponential function in the numerator, 1 / 2e^(kx), must approach 0 as x goes to infinity. In other words, we need e^(kx) to approach infinity.
Since e^(kx) is an exponential function, it will approach infinity as long as kx is positive. Therefore, we need k to be positive for the limit to be finite.
In summary, for the given limit lim(x -> ∞) cosh(kx) / sin^2(x) to be finite, the value of k must be positive (k > 0).