lim x → 0 [e^-nx+e^nx-2cos(nx/2)-kx^2]/(sinx-tanx) exists and finite , then possible values of k and n are

To determine the values of \( k \) and \( n \) for which the limit \[ \lim_{x \to 0} \frac{e^{-nx} + e^{nx} - 2\cos\left(\frac{nx}{2}\right) - kx^2}{\sin x - \tan x} \] exists and is finite, we need to analyze both the numerator and the denominator as \( x \) approaches 0. Let's break this down step by step.

Analyzing the Denominator

The denominator is \( \sin x - \tan x \). We can use Taylor series expansions around \( x = 0 \) for both functions:

• \( \sin x = x - \frac{x^3}{6} + O(x^5) \)
• \( \tan x = x + \frac{x^3}{3} + O(x^5) \)

Subtracting these gives:

\[ \sin x - \tan x = \left(x - \frac{x^3}{6}\right) - \left(x + \frac{x^3}{3}\right) = -\frac{x^3}{6} - \frac{x^3}{3} = -\frac{x^3}{2} + O(x^5) \]

Examining the Numerator

Next, we look at the numerator \( e^{-nx} + e^{nx} - 2\cos\left(\frac{nx}{2}\right) - kx^2 \). Again, we can use Taylor expansions:

• \( e^{-nx} = 1 - nx + \frac{(nx)^2}{2} + O(x^3) \)
• \( e^{nx} = 1 + nx + \frac{(nx)^2}{2} + O(x^3) \)
• \( \cos\left(\frac{nx}{2}\right) = 1 - \frac{(nx/2)^2}{2} + O(x^4) = 1 - \frac{n^2x^2}{8} + O(x^4) \)

Combining these expansions, we have:

\[ e^{-nx} + e^{nx} = 2 + \frac{(nx)^2}{2} - nx + nx + O(x^3) = 2 + \frac{n^2x^2}{2} + O(x^3) \]

Now substituting into the numerator:

\[ e^{-nx} + e^{nx} - 2\cos\left(\frac{nx}{2}\right) - kx^2 = \left(2 + \frac{n^2x^2}{2} + O(x^3)\right) - \left(2 - \frac{n^2x^2}{8} + O(x^4)\right) - kx^2 \]

This simplifies to:

\[ \frac{n^2x^2}{2} + \frac{n^2x^2}{8} - kx^2 + O(x^3) = \left(\frac{4n^2}{8} + \frac{n^2}{8} - k\right)x^2 + O(x^3) = \left(\frac{5n^2}{8} - k\right)x^2 + O(x^3) \]

Setting Up the Limit

Now, we can substitute back into the limit:

\[ \lim_{x \to 0} \frac{\left(\frac{5n^2}{8} - k\right)x^2 + O(x^3)}{-\frac{x^3}{2} + O(x^5)} \]

To ensure this limit exists and is finite, the coefficient of \( x^2 \) in the numerator must be zero, because the denominator behaves like \( -\frac{x^3}{2} \). Therefore, we set:

\[ \frac{5n^2}{8} - k = 0 \implies k = \frac{5n^2}{8} \]

Conclusion on Values of k and n

Thus, for the limit to exist and be finite, \( k \) must be expressed in terms of \( n \) as follows:

\[ k = \frac{5n^2}{8} \]

In summary, the possible values of \( k \) and \( n \) depend on each other, with \( k \) being determined by the square of \( n \). This relationship allows us to find a range of values for \( n \) and subsequently \( k \) based on the context of the problem.