Question

\(\lim_{x \to -1} \left( \frac{x^4 + x^2 + x + 1}{x^2 - x + 1} \right)^{\frac{1 - \cos(x+1)}{(x+1)^2}}\) is equal to

• A. 1
• B. \(\sqrt{\frac{2}{3}}\)
• C. \(\sqrt{\frac{3}{2}}\)
• D. \(e^{1/2}\)

Hint:

For \(1^\infty\) forms, use \(e^{\lim (f(x)-1)g(x)}\). But here base is not 1, so direct substitution works.

Answer 1

The Correct Option is B

To solve the given problem, we need to evaluate the limit:

\(\lim_{x \to -1} \left( \frac{x^4 + x^2 + x + 1}{x^2 - x + 1} \right)^{\frac{1 - \cos(x+1)}{(x+1)^2}}\)

First, let's analyze the inside of the limit:

1. Evaluate the expression inside the limit: \(\frac{x^4 + x^2 + x + 1}{x^2 - x + 1}\) at \(x = -1\).
   • Numerator: \(x^4 + x^2 + x + 1 = (-1)^4 + (-1)^2 + (-1) + 1 = 1 + 1 - 1 + 1 = 2\).
   • Denominator: \(x^2 - x + 1 = (-1)^2 - (-1) + 1 = 1 + 1 + 1 = 3\).

   The expression simplifies to \(\frac{2}{3}\) when \(x = -1\).

2. Next, examine the exponent: \(\frac{1 - \cos(x+1)}{(x+1)^2}\) as \(x \to -1\).
   • Let \(y = x + 1\). Thus, as \(x \to -1\), \(y \to 0\).
   • The expression becomes \(\frac{1 - \cos y}{y^2}\).
   • Using the Taylor expansion for \(\cos y\) around \(y = 0\), we have: \(\cos y \approx 1 - \frac{y^2}{2} + \frac{y^4}{24} + \ldots\), leading to \(1 - \cos y \approx \frac{y^2}{2}\).
   • Therefore, \(\frac{1 - \cos y}{y^2} \approx \frac{\frac{y^2}{2}}{y^2} = \frac{1}{2}\).
3. Substitute back into the original limit:

   \(\lim_{x \to -1} \left( \frac{2}{3} \right)^{\frac{1}{2}} = \left( \sqrt{\frac{2}{3}} \right)\).

Thus, the value of the limit is \(\sqrt{\frac{2}{3}}\), which is the correct answer.