$\text{1)}$ 

`\lim_{x \to -3} \frac{x + \sqrt{2x + 15}}{-2x^2 - 5x + 3}`

`= \lim_{x \to -3} \frac{x + \sqrt{2x + 15}}{-(2x-1)(x+3)}`

`= \lim_{x \to -3} \frac{(x + \sqrt{2x + 15})(x - \sqrt{2x + 15})}{-(2x-1)(x+3)(x - \sqrt{2x + 15})}`

`= \lim_{x \to -3} \frac{x^2 - 2x - 15}{-(2x-1)(x+3)(x - \sqrt{2x + 15})}`

`= \lim_{x \to -3} \frac{(x + 3)(x - 5)}{-(2x-1)(x+3)(x - \sqrt{2x + 15})}`

`= \lim_{x \to -3} \frac{x - 5}{-(2x - 1)(x - \sqrt{2x + 15})}`

`= \frac{-3 - 5}{-(2(-3) - 1)(-3 - \sqrt{2(-3) + 15})}`

`= \frac{-8}{7 \cdot (-6)}`

`= \frac{4}{21}`

$\text{2)}$

Ta có: `u_{n+1} - u_n = \frac{1}{2^n}` với $n \in \mathbb{N}^*$

`\Rightarrow u_n - u_1 = \sum_{i=1}^{n-1} (u_{i+1} - u_i) = \sum_{i=1}^{n-1} \frac{1}{2^i}`

`u_n - 1 = \frac{1}{2} + \frac{1}{2^2} + . . . + \frac{1}{2^{n-1}}`

VP là tổng $n-1$ số hạng đầu của cấp số nhân có `a_1 = \frac{1}{2}` và `q = \frac{1}{2}`:

`u_n - 1 = \frac{1}{2} \cdot \frac{1 - (\frac{1}{2})^{n-1}}{1 - \frac{1}{2}}`

`u_n - 1 = 1 - \frac{1}{2^{n-1}}`

`u_n = 2 - \frac{1}{2^{n-1}}`

`\Rightarrow \lim(u_n - 2) = \lim( 2 - \frac{1}{2^{n-1}} - 2) = \lim( -\frac{1}{2^{n-1}}) = 0`