Use continuity to evaluate the limit: $\underset{x\to 1}{lim}\frac{ln(7-x^2)}{1+x}$

• A. $\frac{ln\left(8\right)}{2}$
• B. $ln\left(6\right)$
• C. $\frac{ln\left(6\right)}{2}$
• D. $0$

Answer: (C)

Verified step by step guidance

1. Step 1: Understand the problem. The goal is to evaluate the limit \( \lim_{x \to 1} \frac{\ln(7 - x^2)}{1 + x} \) using the concept of continuity. Continuity allows us to substitute the value of \( x \) directly into the function if the function is continuous at that point.
2. Step 2: Check the continuity of the function \( \ln(7 - x^2) \). The natural logarithm function \( \ln(u) \) is continuous for \( u > 0 \). Therefore, we need \( 7 - x^2 > 0 \) at \( x = 1 \). Substituting \( x = 1 \), we find \( 7 - 1^2 = 6 > 0 \), so \( \ln(7 - x^2) \) is continuous at \( x = 1 \).
3. Step 3: Check the continuity of the denominator \( 1 + x \). The expression \( 1 + x \) is a polynomial, which is continuous everywhere. Substituting \( x = 1 \), we find \( 1 + 1 = 2 \), so the denominator is continuous at \( x = 1 \).
4. Step 4: Substitute \( x = 1 \) directly into the function \( \frac{\ln(7 - x^2)}{1 + x} \). This gives \( \frac{\ln(7 - 1^2)}{1 + 1} = \frac{\ln(6)}{2} \).
5. Step 5: Conclude that the limit is \( \frac{\ln(6)}{2} \), as the function is continuous at \( x = 1 \). Continuity allows us to evaluate the limit by direct substitution.