Giải thích các bước giải:
$\lim_{x\to-1}\dfrac{\sqrt{13x^2+2x+5}+ax+b}{x^3+2x^2+x}=c$
$\to\lim_{x\to-1}\dfrac{\sqrt{13x^2+2x+5}+ax+b}{x(x+1)^2}=c$
Đặt $f(x)=\sqrt{13x^2+2x+5}+ax+b$
$\to$Để triệt tiêu $(x+1)^2\to f(x)=0$ và $f'(x)=0$ tại $x=-1$
$\to f(x)=0\to \sqrt{13(-1)^2+2(-1)+5}+a(-1)+b=0$
$\to f(x)=0\to 4-a+b=0\to b=a-4$
Lại có :
$f'(x)=0$ có nghiệm tại $x=-1$
$\to \dfrac{13x+1}{\sqrt{13x^2+2x+5}}+a=0$ có nghiệm tại $x=-1$
$\to \dfrac{13(-1)+1}{\sqrt{13(-1)^2+2(-1)+5}}+a=0$
$\to a=3\to b=-1$
$\to\lim_{x\to-1}\dfrac{\sqrt{13x^2+2x+5}+3x-1}{x^3+2x^2+x}=c$
$\to\lim_{x\to-1}\dfrac{\sqrt{13x^2+2x+5}-4+3(x+1)}{x(x+1)^2}=c$
$\to\lim_{x\to-1}\dfrac{\dfrac{13x^2+2x+5-4^2}{\sqrt{13x^2+2x+5}+4}+3(x+1)}{x(x+1)^2}=c$
$\to\lim_{x\to-1}\dfrac{\dfrac{13x^2+2x-11}{\sqrt{13x^2+2x+5}+4}+3(x+1)}{x(x+1)^2}=c$
$\to\lim_{x\to-1}\dfrac{\dfrac{(x+1)(13x-11)}{\sqrt{13x^2+2x+5}+4}+3(x+1)}{x(x+1)^2}=c$
$\to\lim_{x\to-1}\dfrac{\dfrac{13x-11}{\sqrt{13x^2+2x+5}+4}+3}{x(x+1)}=c$
$\to\lim_{x\to-1}\dfrac{13x-11+3(\sqrt{13x^2+2x+5}+4)}{x(x+1)(\sqrt{13x^2+2x+5}+4)}=c$
$\to\lim_{x\to-1}\dfrac{13x+1+3\sqrt{13x^2+2x+5}}{x(x+1)(\sqrt{13x^2+2x+5}+4)}=c$
$\to\lim_{x\to-1}\dfrac{13x+13+3(\sqrt{13x^2+2x+5}-4)}{x(x+1)(\sqrt{13x^2+2x+5}+4)}=c$
$\to\lim_{x\to-1}\dfrac{13(x+1)+3\dfrac{13x^2+2x+5-4^2}{\sqrt{13x^2+2x+5}+4}}{x(x+1)(\sqrt{13x^2+2x+5}+4)}=c$
$\to\lim_{x\to-1}\dfrac{13(x+1)+3\dfrac{(13x-11)(x+1)}{\sqrt{13x^2+2x+5}+4}}{x(x+1)(\sqrt{13x^2+2x+5}+4)}=c$
$\to\lim_{x\to-1}\dfrac{13+3\dfrac{13x-11}{\sqrt{13x^2+2x+5}+4}}{x(\sqrt{13x^2+2x+5}+4)}=c$
$\to\dfrac{13+3\dfrac{13(-1)-11}{\sqrt{13(-1)^2+2(-1)+5}+4}}{(-1)(\sqrt{13(-1)^2+2(-1)+5}+4)}=c$
$\to c=-\dfrac12$
