Toán 11 giải nhanh giúp vs tính lim $\frac{ \sqrt[2]{x+3} + x^{2} + x – 4}{x-1}$ x →1

Ta có

$\underset{x \to 1}{\lim} \dfrac{\sqrt{x+3} + x^2 + x – 4}{x-1} = \underset{x \to 1}{\lim}  \dfrac{\sqrt{x+3} – 2 + x^2 + x – 2}{x-1}$

$= \underset{x \to 1}{\lim} \dfrac{\frac{x+3-4}{\sqrt{x+3} + 2} + (x-1)(x+2)}{x-1}$

$= \underset{x \to 1}{\lim} \left( \dfrac{1}{\sqrt{x+3} + 2} + x + 2 \right)$

$= \dfrac{1}{2 + 2} + 1 + 2 = \dfrac{13}{4}$

Vậy

$\underset{x \to 1}{\lim} \dfrac{\sqrt{x+3} + x^2 + x – 4}{x-1} = \dfrac{13}{4}$.