Hoạt động 2 trang 67 SGK Toán 11 Cánh diều Tập 1 - CD

a) Giả sử (x_n) là dãy số bất kì thỏa mãn limx_n = 1. Khi đó ta có:

$\lim f\left(x_n\right)=\lim \left(x_n^2-1\right)=\lim x_n^2$-1 = 1-1 = 0.

$\Rightarrow$limf(x) = 0.

limg(x_n) = lim(x_n+1) = limx_n+1 = 2

$\Rightarrow$limg(x) = 2.

b) Ta có: f(x) + g(x) = x² – 1 + x + 1 = x² + x

(x_n) là dãy số bất kì thỏa mãn limx_n = 1. Khi đó ta có:

$\lim \left(f\left(x_n\right)+g\left(x_n\right)\right)=\lim \left(x_n^2+x_n\right)=\lim x_n^2+\lim x_n=1^2+1=2$.

$\Rightarrow \underset{x\to 1}{\lim }\left(f\left(x\right)+g\left(x\right)\right)=2$.

Ta lại có: $\underset{x\to 1}{\lim }f\left(x\right)+\underset{x\to 1}{\lim }g\left(\mathrm{x }\right)$= 0 + 2 = 2.

Vậy $\underset{x\to 1}{\lim }\left(f\left(x\right)+g\left(x\right)\right)=\underset{x\to 1}{\lim }f\left(x\right)+\underset{x\to 1}{\lim }g\left(\mathrm{x }\right)$= 2.

c) Ta có: f(x) – g(x) = x² – 1 – x – 1 = x² – x – 2

(x_n) là dãy số bất kì thỏa mãn limx_n = 1. Khi đó ta có:

$\lim \left(f\left(x_n\right)-g\left(x_n\right)\right)=\lim \left(x_n^2-x_n-2\right)$

$=\lim x_n^2-\lim x_n-2=1^2-1-2=-2$

$\Rightarrow \underset{x\to 1}{\lim }\left(f\left(x\right)-g\left(x\right)\right)=-2$.

Ta lại có: $\underset{x\to 1}{\lim }f\left(x\right)-\underset{x\to 1}{\lim }g\left(x\right)$ = 0-2 = -2.

Vậy $\underset{x\to 1}{\lim }\left(f\left(x\right)-g\left(x\right)\right)=\underset{x\to 1}{\lim }f\left(x\right)-\underset{x\to 1}{\lim }g\left(x\right)$= -2.

d) Ta có: f(x).g(x) = (x² – 1)(x + 1) = x³ + x² – x – 1

(x_n) là dãy số bất kì thỏa mãn limx_n = 1. Khi đó ta có:

$\lim \left(f\left(x_n\right).g\left(x_n\right)\right)=\lim \left(x_n^3+x_n^2-x_n-1\right)$

$=\lim x_n^3+\lim x_n^2-\lim x_n$-1 = 1³+1²-1-1 = 0

$\Rightarrow \underset{x\to 1}{\lim }\left(f\left(x\right).g\left(x\right)\right)$=0.

Ta lại có: $\underset{x\to 1}{\lim }f\left(x\right).\underset{x\to 1}{\lim }g\left(x\right)$ = 0.2 = 0.

Vậy $\underset{x\to 1}{\lim }\left(f\left(x\right).g\left(x\right)\right)=\underset{x\to 1}{\lim }f\left(x\right).\underset{x\to 1}{\lim }g\left(x\right)$.

e) Ta có: $\frac{f\left(x\right)}{g\left(x\right)}=\frac{x^2-1}{x+1}$

(x_n) là dãy số bất kì thỏa mãn limx_n = 1. Khi đó ta có:

$\lim \frac{f\left(x_n\right)}{g\left(x_n\right)}=\lim \frac{x_n^2-1}{x_n+1}=\lim \frac{\left(x_n-1\right)\left(x_n+1\right)}{x_n+1}=\lim \left(x_n-1\right)$ = 0.

$\Rightarrow \underset{x\to 1}{\lim }\frac{f\left(x\right)}{g\left(x\right)}$ = 0.

Ta lại có: $\frac{\lim f\left(x\right)}{\lim g\left(x\right)}=\frac{0}{2}$ = 0

Vậy $\underset{x\to 1}{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\frac{\underset{x\to 1}{\lim }f\left(x\right)}{\underset{x\to 1}{\lim }g\left(x\right)}$.

-- Mod Toán 11 HỌC247