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

To determine the values of \( k \) and \( n \) such that 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 the Taylor series expansions for small \( x \) to simplify this expression:

• For \( \sin x \), the expansion is \( x - \frac{x^3}{6} + O(x^5) \).
• For \( \tan x \), the expansion is \( x + \frac{x^3}{3} + O(x^5) \).

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:

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

Combining these, we have:

\[ e^{-nx} + e^{nx} = 2 + \frac{n^2x^2}{2} + O(x^3). \] Thus, the numerator becomes: \[ 2 + \frac{n^2x^2}{2} - 2\left(1 - \frac{n^2x^2}{8}\right) - kx^2 = \frac{n^2x^2}{2} - 2 + \frac{n^2x^2}{4} - kx^2. \] Combining the terms gives: \[ \left(\frac{n^2}{2} + \frac{n^2}{4} - k\right)x^2. \]

Setting Up the Limit

Now we can rewrite the limit as follows:

\[ \lim_{x \to 0} \frac{\left(\frac{n^2}{2} + \frac{n^2}{4} - k\right)x^2}{-\frac{x^3}{2}} = \lim_{x \to 0} \frac{-2\left(\frac{n^2}{2} + \frac{n^2}{4} - k\right)}{x} = \text{finite}. \] For this limit to exist and be finite, the coefficient of \( x^2 \) in the numerator must equal zero: \[ \frac{n^2}{2} + \frac{n^2}{4} - k = 0. \]

Solving for \( k \) and \( n \)

Combining the terms gives:

\[ \frac{3n^2}{4} - k = 0 \implies k = \frac{3n^2}{4}. \] Now, we need to ensure that the limit is finite. The limit will be finite if \( n \) is a non-zero constant. Thus, we can choose any positive value for \( n \) and compute \( k \) accordingly. For example, if we let \( n = 2 \), then: \[ k = \frac{3(2^2)}{4} = \frac{12}{4} = 3. \]

Final Values

In summary, the values of \( k \) and \( n \) that ensure the limit exists and is finite are:

• Any non-zero \( n \)
• Corresponding \( k = \frac{3n^2}{4} \)

This approach ensures that both the numerator and denominator behave appropriately as \( x \) approaches 0, leading to a well-defined limit. If you have any further questions or need clarification on any part of this process, feel free to ask!