Question

Cho hàm số \( y = f(x) \) xác định trên \( \mathbb{R} \) và \(\lim_{x \to 3} \frac{f(x) - f(3)}{x-3} = 10\).

Giới hạn \(\lim_{x \to 1} \frac{f(2x^2+1) - f(3)}{x-1} = \underline{(1)}\).

Answer 1

Đặt \(t=2x^2+1;x\rightarrow1\Rightarrow t\rightarrow3\)

\(\lim\limits_{x\rightarrow1}\dfrac{f\left(2x^2+1\right)-f\left(3\right)}{x-1}=\lim\limits_{x\rightarrow1}\left[\dfrac{f\left(2x^2+1\right)-f\left(3\right)}{2x^2-2}.\dfrac{2x^2-2}{x-1}\right]\)

\(=\lim\limits_{t\rightarrow3}\dfrac{f\left(t\right)-f\left(3\right)}{t-3}.\lim\limits_{x\rightarrow1}\dfrac{2\left(x^2-1\right)}{x-1}=\lim\limits_{t\rightarrow3}\dfrac{f\left(t\right)-f\left(3\right)}{t-3}.\lim\limits_{x\rightarrow1}2\left(x+1\right)\)

\(=10.2\left(1+1\right)=40\)

Vậy \(\left(1\right)\) là \(40\)