ĐGNL ĐHQG Hà Nội - Tư duy định lượng - Các dạng vô định của giới hạn

Bước 1: Tính$$\underset{x\to 2}{\lim }f\left(x\right)$$

Đặt$$g\left(x\right)=\frac{f\left(x\right)-16}{x-2}$$ta có:$$f\left(x\right)=\left(x-2\right)g\left(x\right)+16$$

$$\Rightarrow \underset{x\to 2}{\lim }f\left(x\right)=\underset{x\to 2}{\lim }\left[\left(x-2\right)g\left(x\right)+16\right]=16$$

Bước 2:

Ta có:

$$\begin{array}{l}\underset{x\to 2}{\lim }\frac{\sqrt{2f(x)-16}-4}{x^2+x-6}\\ =\underset{x\to 2}{\lim }\frac{2f(x)-16-16}{(x^2+x-6)\left(\sqrt{2f(x)-16}+4\right)}\\ =\underset{x\to 2}{\lim }\frac{2f(x)-32}{(x-2)(x+3)\left(\sqrt{2f(x)-16}+4\right)}\\ =\underset{x\to 2}{\lim }\frac{f(x)-16}{x-2}.\underset{x\to 2}{\lim }\frac{2}{(x+3)\left(\sqrt{2f(x)-16}+4\right)}\\ =12.\frac{2}{5.\left(\sqrt{2.16-16}+4\right)}=\frac{3}{5}\end{array}$$

$$\begin{array}{l}=>a=3;b=5\\ \Rightarrow a^2+b^2=34\end{array}$$

$$\begin{array}{l}L=\underset{x\to x_0}{\lim }\frac{\sqrt{f\left(x\right)+2}-f\left(x\right)}{f\left(x\right)-2}\\ =\underset{x\to x_0}{\lim }\frac{f\left(x\right)+2-f^2\left(x\right)}{f\left(x\right)-2}.\frac{1}{\sqrt{f\left(x\right)+2}+f\left(x\right)}\\ =\underset{x\to x_0}{\lim }\frac{-\left[f\left(x\right)+1\right]\left[f\left(x\right)-2\right]}{f\left(x\right)-2}.\frac{1}{\sqrt{f\left(x\right)+2}+f\left(x\right)}\\ =-\frac{3}{4}\end{array}$$

Bước 1: Tính$$\underset{x\to 4}{\lim }f\left(x\right)$$

Đặt$$\frac{f\left(x\right)-2018}{x-4}=g\left(x\right)\Rightarrow f\left(x\right)=\left(x-4\right)g\left(x\right)+2018$$

$$\Rightarrow \underset{x\to 4}{\lim }f\left(x\right)=2018$$

Bước 2: Nhân cả tử và mẫu với$$\sqrt{x}+2$$. Tính L

$$\begin{array}{l}L=\underset{x\to 4}{\lim }\frac{1009[f(x)-2018]}{\left(\sqrt{x}-2\right)\left[\sqrt{2019f(x)+2019}+2019\right]}\\ =\underset{x\to 4}{\lim }\frac{1009[f(x)-2018](\surd x+2)}{(x-4)\left[\sqrt{2019f(x)+2019}+2019\right]}\\ =1009.\underset{x\to 4}{\lim }\frac{f(x)-2018}{x-4}.\frac{\sqrt{x}+2}{\sqrt{2019f(x)+2019}+2019}\\ =1009.2019\frac{\sqrt{2018}+2}{\sqrt{2019.2018+2019}+2019}\\ =\mathrm{1009.2019.}\frac{\sqrt{4}+2}{2019+2019}=2018\end{array}$$