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solve this exercice without using hopital rule

Let's solve this exercise step-by-step.

### Given Function
\[
f(x) = \frac{\sqrt{8x + 1}}{2(x - 1)} + \frac{3|x|}{1 - x^2}
\]

### 1) Determining the Domain of \( f \), \( D_f \)

To find the domain \( D_f \), we need to identify the values of \( x \) for which the function \( f(x) \) is defined. This involves checking the validity of each part of the function:

#### a) \( \frac{\sqrt{8x + 1}}{2(x - 1)} \)

- **Square Root Condition**: The expression under the square root, \( 8x + 1 \), must be non-negative:
  \[
  8x + 1 \geq 0 \implies x \geq -\frac{1}{8}
  \]

- **Denominator Condition**: The denominator \( 2(x - 1) \neq 0 \), which implies:
  \[
  x \neq 1
  \]

#### b) \( \frac{3|x|}{1 - x^2} \)

- **Denominator Condition**: The expression \( 1 - x^2 \neq 0 \):
  \[
  1 - x^2 \neq 0 \implies x \neq \pm 1
  \]

Thus, combining all conditions:

1. \( x \geq -\frac{1}{8} \)
2. \( x \neq 1 \)
3. \( x \neq -1 \)

The domain \( D_f \) is:
\[
D_f = \left[ -\frac{1}{8}, 1 \right) \cup (1, +\infty)
\]

### 2) Determining if \( f \) Admits a Continuation by Continuity at \( x = 1 \)

To determine if \( f(x) \) can be extended by continuity at \( x = 1 \), we must check the limit of \( f(x) \) as \( x \to 1 \).

#### Computing the Limit \( \lim_{x \to 1} f(x) \)

Let's find the limit of each term separately:

\[
\lim_{x \to 1} \frac{\sqrt{8x + 1}}{2(x - 1)}
\]
and
\[
\lim_{x \to 1} \frac{3|x|}{1 - x^2}.
\]

##### 1) First Term: \( \lim_{x \to 1} \frac{\sqrt{8x + 1}}{2(x - 1)} \)

Notice that as \( x \to 1 \):
- The numerator \( \sqrt{8x + 1} \to \sqrt{9} = 3 \).
- The denominator \( 2(x - 1) \to 0 \).

So, this term involves a limit of the form \( \frac{3}{0} \), which tends to infinity. Hence:

\[
\lim_{x \to 1} \frac{\sqrt{8x + 1}}{2(x - 1)} = +\infty.
\]

##### 2) Second Term: \( \lim_{x \to 1} \frac{3|x|}{1 - x^2} \)

As \( x \to 1 \):
- The numerator \( |x| \to 1 \), so \( 3|x| \to 3 \).
- The denominator \( 1 - x^2 \to 0 \).

Depending on whether \( x \to 1^+ \) or \( x \to 1^- \), the denominator will approach 0 from different sides:

\[
\lim_{x \to 1^+} \frac{3|x|}{1 - x^2} = -\infty, \quad \lim_{x \to 1^-} \frac{3|x|}{1 - x^2} = +\infty.
\]

#### Conclusion for Continuation by Continuity

Since both terms do not converge to a finite value as \( x \to 1 \) and instead tend towards infinity (with differing signs for the second term), the function \( f(x) \) **does not admit a continuation by continuity at \( x = 1 \)**.

### Summary

1. **Domain**: \( D_f = \left[ -\frac{1}{8}, 1 \right) \cup (1, +\infty) \).
2. **No continuation by continuity at \( x = 1 \)**.

Would you like further details or have any questions?

### Additional Questions:
1. How would you handle discontinuities for other points in \( D_f \)?
2. What is the limit behavior of \( f(x) \) as \( x \to -\frac{1}{8} \)?
3. Can \( f(x) \) have any other points of discontinuity besides \( x = 1 \)?
4. How would you express \( f(x) \) in terms of piecewise functions?
5. How does \( f(x) \) behave as \( x \to \infty \)?

### Tip:
When dealing with functions that have absolute values and square roots, always consider both the domain constraints and the behavior of the function around critical points.

On considère la fonction f définie par f(x) = \frac{\sqrt{8x + 1}}{2(x - 1)} + \frac{3|x|}{1 - x^2}. 1) Déterminer D_f, le domaine de définition de f. 2) f admet-elle un prolongement par continuité en 1 ?

In this exercise, we solve a function involving square roots and absolute values, determining its domain and whether it admits continuity at x = 1. This problem covers advanced calculus concepts suitable for university-level students.

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