Tính giới hạn của các hàm số sau: 
a) $\mathop {\lim }\limits_{x \to 2+} \frac{5x+1}{x-2}$
b)$\mathop {\lim }\limits_{x \to -\infty } \frac{-3x^{3}+2}{2x+1}$ 
c)$\mathop {\lim }\limits_{x \to -\infty }\frac{5x^{3}-x^{2}+1}{3x^{2}+x}$
$a/\lim_{x\to 2^+}\frac{5x+1}{x-2}=+\infty$ do $\lim_{x\to 2^+}(x-2)=0;\lim_{x\to2^+}(5x+1)=11;x-2>0 \forall x>2$
$b/\lim_{x\to-\infty}\frac{-3x^3+2}{2x+1}=\lim_{x\to-\infty}x^2\frac{-3+\frac{2}{x^3}}{2+\frac{1}{x}}=-\infty$
$c/\lim_{x\to-\infty}\frac{5x^3-x^2+1}{3x^2+x}=\lim_{x\to-\infty}x\frac{5-\frac{1}{x}+\frac{1}{x^3}}{3+\frac{1}{x}}=+\infty$
thì cũng là nhấn V và vote à :P –  ♥♥♥ Panda Sơkiu Panda Mập ♥♥♥ 12-05-14 12:56 PM
ều. coi lại đi nha... câu này tui cũng có lâu rồi à. từ hồi mấy người chưa vào cái page này nữa ấy. Mà 2 câu khác hẳn –  ♂Vitamin_Tờ♫ 12-05-14 12:21 PM
cai cau cmt y chang anh Tan van cmt :)) –  ♥♥♥ Panda Sơkiu Panda Mập ♥♥♥ 12-05-14 12:07 PM
Ấn V và vote up nếu đúng, lần sau mình sẵn sàng giúp. Tks –  ♂Vitamin_Tờ♫ 12-05-14 08:15 AM

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