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The given function is:

\[
f(x) = \frac{2x - \sqrt{x + 5}}{x^2 - 1}
\]

The problem asks to calculate the left-hand limit and the right-hand limit of \( f(x) \) at \( x = 1 \) and \( x = -1 \).

### Step 1: Factorize the denominator
The denominator can be factored as:
\[
x^2 - 1 = (x - 1)(x + 1)
\]

### Step 2: Limit at \( x = 1 \)
We are calculating:
\[
\lim_{x \to 1^-} f(x) \quad \text{(left-hand limit)} \quad \text{and} \quad \lim_{x \to 1^+} f(x) \quad \text{(right-hand limit)}
\]
Substitute \( x = 1 \) directly into the function:

\[
f(1) = \frac{2(1) - \sqrt{1 + 5}}{1^2 - 1} = \frac{2 - \sqrt{6}}{0}
\]

Since the denominator becomes zero and the numerator is nonzero, the limits involve division by zero, which tends to infinity. We need to analyze whether it is positive or negative infinity depending on the sign of the numerator and the denominator on either side of \( x = 1 \).

- For \( x \to 1^- \) (values slightly less than 1), \( x - 1 \) is negative and \( x + 1 \) is positive. Thus, the denominator is negative, and since \( 2x - \sqrt{x+5} \) is positive near 1, we conclude:
  \[
  \lim_{x \to 1^-} f(x) = -\infty
  \]

- For \( x \to 1^+ \) (values slightly greater than 1), both \( x - 1 \) and \( x + 1 \) are positive, leading to a positive denominator. Thus:
  \[
  \lim_{x \to 1^+} f(x) = +\infty
  \]

### Step 3: Limit at \( x = -1 \)
Now calculate the limits at \( x = -1 \):
\[
\lim_{x \to -1^-} f(x) \quad \text{(left-hand limit)} \quad \text{and} \quad \lim_{x \to -1^+} f(x) \quad \text{(right-hand limit)}
\]

Substitute \( x = -1 \) into the function:
\[
f(-1) = \frac{2(-1) - \sqrt{-1 + 5}}{(-1)^2 - 1} = \frac{-2 - \sqrt{4}}{0} = \frac{-2 - 2}{0}
\]
Again, the denominator becomes zero. As \( x \to -1 \), we analyze the sign of the numerator and denominator:

- For \( x \to -1^- \), \( x + 1 \) is negative, making the denominator negative. The numerator \( 2x - \sqrt{x + 5} \) is negative near \( x = -1 \), so:
  \[
  \lim_{x \to -1^-} f(x) = +\infty
  \]

- For \( x \to -1^+ \), \( x + 1 \) is positive, leading to a positive denominator. Thus:
  \[
  \lim_{x \to -1^+} f(x) = -\infty
  \]

### Final Answer:
- \( \lim_{x \to 1^-} f(x) = -\infty \)
- \( \lim_{x \to 1^+} f(x) = +\infty \)
- \( \lim_{x \to -1^-} f(x) = +\infty \)
- \( \lim_{x \to -1^+} f(x) = -\infty \)

Would you like further details or clarification on any steps?

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Here are 5 related questions to expand your understanding:
1. How do you evaluate limits involving radicals in the numerator?
2. What happens when both the numerator and denominator approach 0 at a limit point?
3. How does factoring help simplify limit problems involving rational functions?
4. What is the significance of analyzing left-hand and right-hand limits separately?
5. Can you explain the concept of vertical asymptotes in rational functions?

**Tip:** When dealing with limits leading to \(\frac{a}{0}\) (where \( a \neq 0 \)), the sign of the numerator and denominator is crucial to determine if the limit tends to \( +\infty \) or \( -\infty \).

Calculer la limite à gauche et la limite à droite de f(x) en 1 et en -1 pour la fonction f(x) = (2x - sqrt(x + 5)) / (x^2 - 1).

Explore how to calculate the left-hand and right-hand limits of a rational function involving radicals at x = 1 and x = -1. This problem explains key concepts of limits, vertical asymptotes, and radicals in advanced algebra. Suitable for high school students in Grades 11-12.

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