Questions: Evaluate the limit as x approaches 1 of (x^2 - 1) / ln(x). Determine whether the function f(x) = 1 / (x^2 - 1) is continuous at x=1. If not, classify the discontinuity.

Solution Steps

1. To evaluate the limit \(\lim _{x \rightarrow 1} \frac{x^{2}-1}{\ln (x)}\), we can apply L'Hôpital's Rule since the limit is in the indeterminate form \(\frac{0}{0}\). This involves differentiating the numerator and the denominator separately and then taking the limit again.
2. To determine the continuity of the function \(f(x)=\frac{1}{x^{2}-1}\) at \(x=1\), we need to check if the function is defined at \(x=1\), and if the limit of the function as \(x\) approaches 1 from both sides exists and equals the function value at \(x=1\).

To evaluate the limit \(\lim _{x \rightarrow 1} \frac{x^{2}-1}{\ln (x)}\), we notice that both the numerator and the denominator approach 0 as \(x\) approaches 1, resulting in an indeterminate form \(\frac{0}{0}\). Therefore, we apply L'Hôpital's Rule, which involves differentiating the numerator and the denominator separately:

• The derivative of the numerator \(x^2 - 1\) is \(2x\).
• The derivative of the denominator \(\ln(x)\) is \(\frac{1}{x}\).

Applying L'Hôpital's Rule, the limit becomes: \[ \lim _{x \rightarrow 1} \frac{2x}{\frac{1}{x}} = \lim _{x \rightarrow 1} 2x^2 = 2 \times 1^2 = 2 \]

To determine if the function \(f(x)=\frac{1}{x^{2}-1}\) is continuous at \(x=1\), we check the following conditions:

1. Defined at \(x=1\): The function \(f(x)\) is not defined at \(x=1\) because the denominator \(x^2 - 1\) becomes 0, leading to division by zero.

2. Limit from the Left and Right:

   • The limit of \(f(x)\) as \(x\) approaches 1 from the left is \(-\infty\).
   • The limit of \(f(x)\) as \(x\) approaches 1 from the right is \(+\infty\).

Since the function is not defined at \(x=1\) and the left and right limits are not equal, the function is not continuous at \(x=1\). The discontinuity is an infinite discontinuity.

Final Answer