Giải thích các bước giải:
Câu 1:
a.Ta có:
$\lim 2n^5+7n^2-n$
$=\lim n^5(2+\dfrac7{n^3}-\dfrac1{n^4})$
$=+\infty(2+0-0)$
$=+\infty$
b.Ta có:
$\lim\dfrac{-n^4+n^3}{4-6n^2+n^4}$
$=\lim\dfrac{-1+\dfrac1n}{\dfrac4{n^4}-\dfrac6{n^2}+1}$
$=\dfrac{-1+0}{0-0+1}$
$=-1$
Câu 2:
a.Ta có:
$\lim_{x\to-4}\dfrac{x^2+3x-4}{2x+8}$
$=\lim_{x\to-4}\dfrac{(x+4)(x-1)}{2(x+4)}$
$=\lim_{x\to-4}\dfrac{x-1}{2}$
$=\dfrac{-4-1}{2}$
$=-\dfrac52$
b.Ta có:
$\lim_{x\to 2^-}\dfrac{7-3x^2}{2x-4}$
Ta có: $\lim_{x\to 2^-}7-3x^2=7-3\cdot (-2)^2=-5$
$\lim_{x\to 2^-}2x-4=2\cdot 2-4=0$ và $2x-4<0$ vì $x<2$
$\to \lim_{x\to 2^-}\dfrac{7-3x^2}{2x-4}=+\infty$
c.Ta có:
$\lim_{x\to -1}\dfrac{2x+1+\sqrt{3x+4}}{2x+2}$
$=\lim_{x\to -1}\dfrac{2(x+1)+\sqrt{3x+4}-1}{2(x+1)}$
$=\lim_{x\to -1}\dfrac{2(x+1)+\dfrac{3x+4-1}{\sqrt{3x+4}+1}}{2(x+1)}$
$=\lim_{x\to -1}\dfrac{2(x+1)+\dfrac{3x+3}{\sqrt{3x+4}+1}}{2(x+1)}$
$=\lim_{x\to -1}\dfrac{2(x+1)+\dfrac{3(x+1)}{\sqrt{3x+4}+1}}{2(x+1)}$
$=\lim_{x\to -1}\dfrac{2+\dfrac{3}{\sqrt{3x+4}+1}}{2}$
$=\dfrac{2+\dfrac{3}{\sqrt{3\cdot (-1)+4}+1}}{2}$
$=\dfrac74$
