Lời giải:
\(\lim\limits_{x\to 0}\frac{2-\sqrt{4-x}}{x}=\lim\limits_{x\to 0}\frac{4-(4-x)}{x(2+\sqrt{4-x})}=\lim\limits_{x\to 0}\frac{x}{x(2+\sqrt{4-x})}=\lim\limits_{x\to 0}\frac{1}{2+\sqrt{4-x}}=\frac{1}{4}\)
\(\lim\limits_{x\to -2}\frac{x^3+8}{x^2-4}=\lim\limits_{x\to -2}\frac{(x+2)(x^2-2x+4)}{(x-2)(x+2)}=\lim\limits_{x\to -2}\frac{x^2-2x+4}{x-2}=-3\)
Tìm các giới hạn sau :
a, lim\(\dfrac{2x^2+x-6}{x^3+8}\) khi x→-2
b, lim\(\dfrac{x^4-x^2-72}{x^2-2x-3}\) khi x→3
c, lim\(\dfrac{x^5+1}{x^3+1}\) khi x→-1
d, lim \(\left(\dfrac{2}{x^2-1}-\dfrac{1}{x-1}\right)\) khi x→1
1. lim (\(\dfrac{1}{x}+\dfrac{1}{x^2}\) ) khi x->0
2 lim \(\dfrac{x^3+8}{X+2}\) khi x->-2
\(\lim\limits_{x\rightarrow1^+}\dfrac{\sqrt{x^3-x^2}}{\sqrt{x-1}+1-x}\)
\(\lim\limits_{x\rightarrow3}\dfrac{\sqrt{x^2+x}-2\sqrt{3}}{x-3}\)
\(\lim\limits_{x\rightarrow-2}\dfrac{x^4+8x}{x^3+2x^2+x+2}\)
\(\lim\limits_{x\rightarrow+\infty}\dfrac{x\sqrt{x^2+1}+2x+1}{\sqrt[3]{2x^3+x+1}+x}\)
\(\lim\limits_{x\rightarrow1}\dfrac{\sqrt{2x^2-x+1}-\sqrt[3]{2x+3}}{3x^2-2}\)
\(\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt{4x^2+x}+\sqrt[3]{8x^3+x-1}}{\sqrt[4]{x^4+3}}\)
1) \(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{1+4x}.\sqrt[3]{1+6x}.\sqrt[4]{1+8x}-1}{x}\)
2)\(\lim\limits_{x\rightarrow1}\dfrac{\sqrt[3]{1+7x}-x^3+3x-4}{x-1}\)
3) \(\lim\limits_{x\rightarrow-\infty}\dfrac{x^3-x^2+1}{2x^2+3x-1}\)
4) \(\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt{x}+\sqrt[3]{x}+\sqrt[4]{x}}{\sqrt{4x+1}}\)
5) \(\lim\limits_{x\rightarrow-\infty}\dfrac{x+\sqrt{x^2+2}}{\sqrt[3]{8x^3+x^2+1}}\)
6) \(\lim\limits_{x\rightarrow-\infty}\dfrac{\sqrt{4x^2+3x-7}}{\sqrt[3]{27x^3+5x^2+x-4}}\)
1) \(\lim\limits_{x\rightarrow0}\dfrac{2\sqrt{1+x}-\sqrt[3]{8-x}}{x}\)
2)\(\lim\limits_{x\rightarrow1}\dfrac{\sqrt[3]{x+7}-\sqrt{x+3}}{x^2-3x+2}\)
3)\(\lim\limits_{x\rightarrow1}\dfrac{\sqrt[3]{x^2+7}-\sqrt{5-x^2}}{x^2-1}\)
4)\(\lim\limits_{x\rightarrow-2}\dfrac{\sqrt{x+11}-\sqrt[3]{8x+43}}{2x^2+3x-2}\)
5) \(\lim\limits_{x\rightarrow0}\dfrac{\sqrt[n]{1+ax}-\sqrt[m]{1+bx}}{x}\)
6)\(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{1+4x}.\sqrt[3]{1+6x}-1}{x}\)
\(\lim\limits_{x\rightarrow1^+}\dfrac{x^2-x+1}{x^2-1}\)
\(\lim\limits_{x\rightarrow0}\dfrac{2\sqrt{1+x}-\sqrt[3]{8-x}}{x}\)
\(\lim\limits_{x\rightarrow3}\dfrac{\sqrt{x+6}-3}{\sqrt{2x-2}-2}\)
\(\lim\limits_{x\rightarrow3}\dfrac{\sqrt{2x+3}-x}{x^2-4x+3}\)
\(\lim\limits_{x\rightarrow0}\dfrac{\sqrt[3]{x+1}-1}{\sqrt[4]{2x+1}-1}\)
\(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{1+4x}-\sqrt[3]{1+6x}}{x^2}\)
cho lim \(\dfrac{f\left(x\right)-5}{x-1}=4\) khi x->1 , lim \(\dfrac{g\left(x\right)-1}{x-1}=5\) khi x->1
tinh lim \(\dfrac{\sqrt{f\left(x\right)\times g\left(x\right)+4}-1}{x-1}\)khi x->1
