Giải giúp em ạ

1. $\lim_{x\rightarrow-1}\frac{x^2+x+1}{2x^5+3}$ Thay $x = -1$ vào: \[ \lim_{x\rightarrow-1}\frac{x^2+x+1}{2x^5+3} = \frac{(-1)^2 + (-1) + 1}{2(-1)^5 + 3} = \frac{1 - 1 + 1}{-2 + 3} = \frac{1}{1} = 1 \] 2. $\lim_{x\rightarrow3}\frac{x-3}{x^2+2x-15}$ Ta thấy $x^2 + 2x - 15 = (x - 3)(x + 5)$, do đó: \[ \lim_{x\rightarrow3}\frac{x-3}{(x-3)(x+5)} = \lim_{x\rightarrow3}\frac{1}{x+5} = \frac{1}{3+5} = \frac{1}{8} \] 3. $\lim_{x\rightarrow2}\frac{x^2-5x+6}{x^2-8x+15}$ Ta thấy $x^2 - 5x + 6 = (x - 2)(x - 3)$ và $x^2 - 8x + 15 = (x - 3)(x - 5)$, do đó: \[ \lim_{x\rightarrow2}\frac{(x-2)(x-3)}{(x-3)(x-5)} = \lim_{x\rightarrow2}\frac{x-2}{x-5} = \frac{2-2}{2-5} = \frac{0}{-3} = 0 \] 4. $\lim_{x\rightarrow1}\frac{x^2-1}{x^2+2x-3}$ Ta thấy $x^2 - 1 = (x - 1)(x + 1)$ và $x^2 + 2x - 3 = (x - 1)(x + 3)$, do đó: \[ \lim_{x\rightarrow1}\frac{(x-1)(x+1)}{(x-1)(x+3)} = \lim_{x\rightarrow1}\frac{x+1}{x+3} = \frac{1+1}{1+3} = \frac{2}{4} = \frac{1}{2} \] 5. $\lim_{x\rightarrow\frac12}\frac{8x^2-1}{6x^2-5x+1}$ Thay $x = \frac{1}{2}$ vào: \[ \lim_{x\rightarrow\frac12}\frac{8x^2-1}{6x^2-5x+1} = \frac{8\left(\frac{1}{2}\right)^2 - 1}{6\left(\frac{1}{2}\right)^2 - 5\left(\frac{1}{2}\right) + 1} = \frac{8 \cdot \frac{1}{4} - 1}{6 \cdot \frac{1}{4} - \frac{5}{2} + 1} = \frac{2 - 1}{\frac{3}{2} - \frac{5}{2} + 1} = \frac{1}{0} = \text{không tồn tại} \] 6. $\lim_{x\rightarrow2}\frac{x^2-3x+2}{(x-2)^2}$ Ta thấy $x^2 - 3x + 2 = (x - 2)(x - 1)$, do đó: \[ \lim_{x\rightarrow2}\frac{(x-2)(x-1)}{(x-2)^2} = \lim_{x\rightarrow2}\frac{x-1}{x-2} = \frac{2-1}{2-2} = \frac{1}{0} = \text{không tồn tại} \] 7. $\lim_{x\rightarrow1}\frac{\sqrt{x+3}-2}{x-1}$ Nhân cả tử và mẫu với $\sqrt{x+3} + 2$: \[ \lim_{x\rightarrow1}\frac{(\sqrt{x+3}-2)(\sqrt{x+3}+2)}{(x-1)(\sqrt{x+3}+2)} = \lim_{x\rightarrow1}\frac{x+3-4}{(x-1)(\sqrt{x+3}+2)} = \lim_{x\rightarrow1}\frac{x-1}{(x-1)(\sqrt{x+3}+2)} = \lim_{x\rightarrow1}\frac{1}{\sqrt{x+3}+2} = \frac{1}{\sqrt{1+3}+2} = \frac{1}{4} \] 8. $\lim_{x\rightarrow1}\frac{x-\sqrt{2x-1}}{x^2-12x+11}$ Nhân cả tử và mẫu với $x + \sqrt{2x-1}$: \[ \lim_{x\rightarrow1}\frac{(x-\sqrt{2x-1})(x+\sqrt{2x-1})}{(x^2-12x+11)(x+\sqrt{2x-1})} = \lim_{x\rightarrow1}\frac{x^2-(2x-1)}{(x^2-12x+11)(x+\sqrt{2x-1})} = \lim_{x\rightarrow1}\frac{x^2-2x+1}{(x^2-12x+11)(x+\sqrt{2x-1})} = \lim_{x\rightarrow1}\frac{(x-1)^2}{(x-1)(x-11)(x+\sqrt{2x-1})} = \lim_{x\rightarrow1}\frac{x-1}{(x-11)(x+\sqrt{2x-1})} = \frac{1-1}{(1-11)(1+\sqrt{2-1})} = \frac{0}{-10 \cdot 2} = 0 \] 9. $\lim_{x\rightarrow0}\frac{\sqrt{x^2+1}-1}{x^2-3x}$ Nhân cả tử và mẫu với $\sqrt{x^2+1} + 1$: \[ \lim_{x\rightarrow0}\frac{(\sqrt{x^2+1}-1)(\sqrt{x^2+1}+1)}{(x^2-3x)(\sqrt{x^2+1}+1)} = \lim_{x\rightarrow0}\frac{x^2+1-1}{(x^2-3x)(\sqrt{x^2+1}+1)} = \lim_{x\rightarrow0}\frac{x^2}{(x^2-3x)(\sqrt{x^2+1}+1)} = \lim_{x\rightarrow0}\frac{x}{(x-3)(\sqrt{x^2+1}+1)} = \frac{0}{(0-3)(\sqrt{0+1}+1)} = \frac{0}{-3 \cdot 2} = 0 \] 10. $\lim_{x\rightarrow3}\frac{x^2-4x+3}{\sqrt{2x^2-9}-x}$ Nhân cả tử và mẫu với $\sqrt{2x^2-9} + x$: \[ \lim_{x\rightarrow3}\frac{(x^2-4x+3)(\sqrt{2x^2-9}+x)}{(\sqrt{2x^2-9}-x)(\sqrt{2x^2-9}+x)} = \lim_{x\rightarrow3}\frac{(x-3)(x-1)(\sqrt{2x^2-9}+x)}{2x^2-9-x^2} = \lim_{x\rightarrow3}\frac{(x-3)(x-1)(\sqrt{2x^2-9}+x)}{x^2-9} = \lim_{x\rightarrow3}\frac{(x-3)(x-1)(\sqrt{2x^2-9}+x)}{(x-3)(x+3)} = \lim_{x\rightarrow3}\frac{(x-1)(\sqrt{2x^2-9}+x)}{x+3} = \frac{(3-1)(\sqrt{2 \cdot 3^2 - 9} + 3)}{3+3} = \frac{2(\sqrt{18-9} + 3)}{6} = \frac{2(3 + 3)}{6} = \frac{2 \cdot 6}{6} = 2 \] 11. $\lim_{x\rightarrow-2}\frac{2x^2-5x+2}{\sqrt{x^2+5}-3}$ Nhân cả tử và mẫu với $\sqrt{x^2+5} + 3$: \[ \lim_{x\rightarrow-2}\frac{(2x^2-5x+2)(\sqrt{x^2+5}+3)}{(\sqrt{x^2+5}-3)(\sqrt{x^2+5}+3)} = \lim_{x\rightarrow-2}\frac{(2x^2-5x+2)(\sqrt{x^2+5}+3)}{x^2+5-9} = \lim_{x\rightarrow-2}\frac{(2x^2-5x+2)(\sqrt{x^2+5}+3)}{x^2-4} = \lim_{x\rightarrow-2}\frac{(2x^2-5x+2)(\sqrt{x^2+5}+3)}{(x-2)(x+2)} = \lim_{x\rightarrow-2}\frac{(2x^2-5x+2)(\sqrt{x^2+5}+3)}{(x-2)(x+2)} = \frac{(2(-2)^2-5(-2)+2)(\sqrt{(-2)^2+5}+3)}{(-2-2)(-2+2)} = \frac{(8+10+2)(\sqrt{4+5}+3)}{(-4)(0)} = \frac{20 \cdot 6}{0} = \text{không tồn tại} \] 12. $\lim_{x\rightarrow4}\frac{5-\sqrt{x^2+9}}{4x^2-19x+12}$ Nhân cả tử và mẫu với $5 + \sqrt{x^2+9}$: \[ \lim_{x\rightarrow4}\frac{(5-\sqrt{x^2+9})(5+\sqrt{x^2+9})}{(4x^2-19x+12)(5+\sqrt{x^2+9})} = \lim_{x\rightarrow4}\frac{25-(x^2+9)}{(4x^2-19x+12)(5+\sqrt{x^2+9})} = \lim_{x\rightarrow4}\frac{16-x^2}{(4x^2-19x+12)(5+\sqrt{x^2+9})} = \lim_{x\rightarrow4}\frac{(4-x)(4+x)}{(4x^2-19x+12)(5+\sqrt{x^2+9})} = \lim_{x\rightarrow4}\frac{(4-x)(4+x)}{(4x-3)(x-4)(5+\sqrt{x^2+9})} = \lim_{x\rightarrow4}\frac{-(x-4)(4+x)}{(4x-3)(x-4)(5+\sqrt{x^2+9})} = \lim_{x\rightarrow4}\frac{-(4+x)}{(4x-3)(5+\sqrt{x^2+9})} = \frac{-(4+4)}{(4 \cdot 4 - 3)(5 + \sqrt{4^2 + 9})} = \frac{-8}{(16-3)(5 + \sqrt{16+9})} = \frac{-8}{13 \cdot 8} = \frac{-1}{13} \]