Biết rằng lim x chạy đến 1 (f(x) - 5)/(x - 1) = 2 và

* Hướng dẫn giải

Đáp án: \(\frac{{17}}{6}\)

Giải chi tiết:

Đặt \(h\left( x \right) = \frac{{f\left( x \right) - 5}}{{x - 1}} \Leftrightarrow f\left( x \right) = \left( {x - 1} \right)h\left( x \right) + 5\)

\( \Rightarrow \mathop {\lim }\limits_{x \to 1} f\left( x \right) = 5\).

Đặt \(k\left( x \right) = \frac{{g\left( x \right) - 1}}{{x - 1}} \Rightarrow g\left( x \right) = \left( {x - 1} \right)k\left( x \right) + 1\).

\( \Rightarrow \mathop {\lim }\limits_{x \to 1} g\left( x \right) = 1\).

Ta có: $L=\underset{x\to 1}{lim}\frac{\sqrt{f\left(x\right).g\left(x\right)+4}\mathrm{ }-3}{x-1}$

$L=\underset{x\to 1}{lim}\frac{f\left(x\right).g\left(x\right)+4-9}{(x-1)[\sqrt{f\left(x\right).g\left(x\right)+4}\mathrm{ }+3]}$

$L=\underset{x\to 1}{lim}\frac{f\left(x\right).g\left(x\right)-5}{(x-1)[\sqrt{f\left(x\right).g\left(x\right)+4}\mathrm{ }+3]}$

$L=\underset{x\to 1}{lim}\frac{f\left(x\right)[g\left(x\right)-1]+f\left(x\right)-5}{(x-1)[\sqrt{f\left(x\right).g\left(x\right)+4}\mathrm{ }+3]}$

$L=\underset{x\to 1}{lim}\frac{f\left(x\right)[g\left(x\right)-1]+[f\left(x\right)-5]}{(x-1)[\sqrt{f\left(x\right).g\left(x\right)+4}\mathrm{ }+3]}$

$L=\underset{x\to 1}{lim}\frac{g\left(x\right)-1}{x-1}.\frac{f\left(x\right)}{\sqrt{f\left(x\right).g\left(x\right)+4}\mathrm{ }+3}+\underset{x\to 1}{lim}\frac{f\left(x\right)-5}{x-1}.\frac{1}{\sqrt{f\left(x\right).g\left(x\right)+4}\mathrm{ }+3}$

$L=3.\frac{5}{\sqrt{5.1+4}\mathrm{ }+3}+2.\frac{1}{\sqrt{5.1+4}\mathrm{ }+3}$

\(L = \frac{{15}}{6} + \frac{2}{6} = \frac{{17}}{6}\).