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 limit problem, we need to find:

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

Let's break it down step-by-step:

1. First, consider the base of the exponent, \(\frac{x^4 + x^2 + x + 1}{x^2 - x + 1}\). Substitute \(x = -1\):
   • When \(x = -1\), the numerator becomes:
     • \((-1)^4 + (-1)^2 + (-1) + 1 = 1 + 1 - 1 + 1 = 2\)
   • The denominator becomes:
     • \((-1)^2 - (-1) + 1 = 1 + 1 + 1 = 3\)
   • Thus, the base is \(\frac{2}{3}\).
2. Now consider the exponent, \(\frac{1 - \cos(x+1)}{(x+1)^2}\). As \(x \to -1\), \(x+1\) approaches 0, and we use the standard limit:
   • \(\lim_{y \to 0} \frac{1 - \cos y}{y^2} = \frac{1}{2}\)
3. So, the limit becomes:
   • \(\left(\frac{2}{3}\right)^{\frac{1}{2}}\) = \(\sqrt{\frac{2}{3}}\)

Therefore, the correct answer is \(\sqrt{\frac{2}{3}}\).