Given the function $f(x) = \frac{x^2 + 2x + 1}{x-1}$, we are asked to: 1) Find real numbers $a$, $b$, and $c$ such that $f(x) = ax + b + \frac{c}{x-1}$. 2) Find the limits of $f$ at the boundaries of its domain. 3) Determine the table of variations of $f$.

1. Problem Description

Given the function $f(x) = \frac{x^2 + 2x + 1}{x-1}$, we are asked to:

1) Find real numbers $a$, $b$, and $c$ such that $f(x) = ax + b + \frac{c}{x-1}$.

2) Find the limits of $f$ at the boundaries of its domain.

3) Determine the table of variations of $f$.

2. Solution Steps

1) To find $a$, $b$, and $c$, we can perform polynomial long division on $\frac{x^2 + 2x + 1}{x-1}$:

$x^2 + 2x + 1 = (x-1)(x+3) + 4$.

Thus, $\frac{x^2 + 2x + 1}{x-1} = \frac{(x-1)(x+3) + 4}{x-1} = x+3 + \frac{4}{x-1}$.

Therefore, $a=1$, $b=3$, and $c=4$.

2) The domain of $f$ is $D_f = \mathbb{R} \setminus \{1\} = (-\infty, 1) \cup (1, \infty)$. We need to find the limits of $f$ as $x \to -\infty$, $x \to 1^-$, $x \to 1^+$, and $x \to \infty$.

$\lim_{x \to -\infty} f(x) = \lim_{x \to -\infty} \left(x+3 + \frac{4}{x-1}\right) = -\infty + 3 + 0 = -\infty$.

$\lim_{x \to \infty} f(x) = \lim_{x \to \infty} \left(x+3 + \frac{4}{x-1}\right) = \infty + 3 + 0 = \infty$.

$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} \left(x+3 + \frac{4}{x-1}\right) = 1+3 + \frac{4}{1^- - 1} = 4 + \frac{4}{0^-} = 4 - \infty = -\infty$.

$\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} \left(x+3 + \frac{4}{x-1}\right) = 1+3 + \frac{4}{1^+ - 1} = 4 + \frac{4}{0^+} = 4 + \infty = \infty$.

3) To determine the table of variations of $f$, we need to find the derivative of $f(x)$.

$f(x) = x+3 + \frac{4}{x-1} = x+3 + 4(x-1)^{-1}$.

$f'(x) = 1 - 4(x-1)^{-2} = 1 - \frac{4}{(x-1)^2} = \frac{(x-1)^2 - 4}{(x-1)^2} = \frac{x^2 - 2x + 1 - 4}{(x-1)^2} = \frac{x^2 - 2x - 3}{(x-1)^2} = \frac{(x-3)(x+1)}{(x-1)^2}$.

$f'(x) = 0$ when $x = -1$ or $x = 3$.

The denominator $(x-1)^2$ is always positive (except at $x=1$).

The sign of $f'(x)$ depends on the sign of $(x-3)(x+1)$.

- If $x < -1$, then $x-3 < 0$ and $x+1 < 0$, so $f'(x) > 0$.

- If $-1 < x < 1$, then $x-3 < 0$ and $x+1 > 0$, so $f'(x) < 0$.

- If $1 < x < 3$, then $x-3 < 0$ and $x+1 > 0$, so $f'(x) < 0$.

- If $x > 3$, then $x-3 > 0$ and $x+1 > 0$, so $f'(x) > 0$.

$f(-1) = \frac{(-1)^2 + 2(-1) + 1}{-1-1} = \frac{1-2+1}{-2} = \frac{0}{-2} = 0$.

$f(3) = \frac{3^2 + 2(3) + 1}{3-1} = \frac{9+6+1}{2} = \frac{16}{2} = 8$.

Table of variations:

x | -inf -1 1 3 +inf

---------------------------------

f'(x) | + 0 - - 0 +

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f(x) | -inf inc 0 dec -inf inf dec 8 inc +inf

3. Final Answer

1) $a=1$, $b=3$, $c=4$.

2) $\lim_{x \to -\infty} f(x) = -\infty$, $\lim_{x \to \infty} f(x) = \infty$, $\lim_{x \to 1^-} f(x) = -\infty$, $\lim_{x \to 1^+} f(x) = \infty$.

3) Table of variations:

x | -inf -1 1 3 +inf

---------------------------------

f'(x) | + 0 - - 0 +

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f(x) | -inf inc 0 dec -inf inf dec 8 inc +inf