• $\lim\limits_{x \to 1} \dfrac{x^2 - 1}{x - 1}$

$$\begin{align*} \lim\limits_{x \to 1} \dfrac{x^2 - 1}{x - 1} &=\lim\limits_{x \to 1} \dfrac{(x - 1)(x + 1)}{x - 1}\\ &=\lim\limits_{x \to 1} (x + 1) \\&= 1 + 1 = 2 \end{align*}$$

Alors

$$\boxed{ \lim\limits_{x \to 1} \dfrac{x^2 - 1}{x - 1}=2 }$$

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• $\lim\limits_{x \to 2} \dfrac{x - 2}{x^2 - 3x + 2}$

on a

$$\begin{align*} x^2 - 3x + 2 &=a(x-x_1)(x-x_2)\\ \end{align*}$$

avec : $x^2 - 3x + 2=ax^2+bx+c$

et donc $a=1$ , $x_1=2$

et $x_1.x_2=\frac ca=\frac21=2$

Donc $x_2=\frac1x_1=1$

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

Ainsi,

$$\begin{align*} \lim\limits_{x \to 2}\dfrac{x - 2}{x^2 - 3x + 2} &=\lim\limits_{x \to 2} \dfrac{x - 2}{(x - 1)(x - 2)} \\&=\lim\limits_{x \to 2}\dfrac{1}{x - 1} \\&=\frac{1}{2-1}=1 \end{align*}$$

alors :

$$\boxed{ \lim\limits_{x \to 2} \dfrac{x - 2}{x^2 - 3x + 2}=1 }$$

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• $\lim\limits_{x \to 1} \dfrac{\sqrt{x} - 1}{x - 1}$

$$\begin{align*} \dfrac{\sqrt{x} - 1}{x - 1} &= \dfrac{(\sqrt{x} - 1)(\sqrt{x} + 1)}{(x - 1)(\sqrt{x} + 1)}\\ &= \dfrac{x - 1}{(x - 1)(\sqrt{x} + 1)} \\ &= \dfrac{1}{\sqrt{x} + 1} \text{ pour } x \ne 1 \end{align*}$$

On a

$$\lim\limits_{x \to 1} \dfrac{1}{\sqrt{x} + 1} = \dfrac{1}{\sqrt{1} + 1} = \dfrac{1}{2}$$

Alors :

$$\boxed{ \lim\limits_{x \to 1} \dfrac{\sqrt{x} - 1}{x - 1}=\dfrac12 }$$

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• $\lim\limits_{x \to 0} \dfrac{\sqrt{x + 1} - 1}{x^2 - x}$

$$\begin{align*} \dfrac{\sqrt{x + 1} - 1}{x^2 - x} &= \dfrac{(\sqrt{x + 1} - 1)(\sqrt{x + 1} + 1)}{(x^2 - x)(\sqrt{x + 1} + 1)} \\~\\ &= \dfrac{x + 1 - 1}{(x^2 - x)(\sqrt{x + 1} + 1)} \\~\\ &= \dfrac{x}{(x^2 - x)(\sqrt{x + 1} + 1)} \\~\\ &=\dfrac{x}{x(x - 1)(\sqrt{x + 1} + 1)} \\~\\ &=\dfrac{1}{(x - 1)(\sqrt{x + 1} + 1)} \end{align*}$$

On a :

$$\lim\limits_{x \to 0} \dfrac{1}{(x - 1)(\sqrt{x + 1} + 1)} = \dfrac{1}{(0 - 1)(\sqrt{0 + 1} + 1)} = -\dfrac{1}{2}$$

Alors :

$$\boxed{ \lim\limits_{x \to 0} \dfrac{\sqrt{x + 1} - 1}{x^2 - x} =-\dfrac{1}{2} }$$