Nó ko phải dạng vô định thì bạn cứ thay vô và tính toán bình thường
\(\lim\limits_{x\rightarrow0}\dfrac{1-\dfrac{1}{x}}{1+\dfrac{1}{x}}=\lim\limits_{x\rightarrow0}\dfrac{x-1}{x+1}=\dfrac{0-1}{0+1}=-1\)
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
a) lim \(\dfrac{2x^3-5x-4}{\left(x+1\right)^2}\) khi x tiến đến -1.
b) lim (x3 + 2x2\(\sqrt{x}\) - 1) khi x tiến đến dương vô cùng.
Giúp mình với ạ.
1. lim (\(\dfrac{1}{x}+\dfrac{1}{x^2}\) ) khi x->0
2 lim \(\dfrac{x^3+8}{X+2}\) khi 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
\(\lim\limits_{x\rightarrow2}\left(\dfrac{1}{x-2}-\dfrac{12}{x^3-8}\right)\)
\(\lim\limits_{x\rightarrow2}\left(\dfrac{1}{x^2-3x+2}+\dfrac{1}{x^2-5x+6}\right)\)
\(\lim\limits_{x\rightarrow+\infty}x\left(\sqrt{x^2+1}-x\right)\)
\(\lim\limits_{x\rightarrow+\infty}\left(\sqrt{x^2+1}-\sqrt[3]{x^3-1}\right)\)
\(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{x^2+1}-1}{\sqrt{x^2+16}-4}\)
Tính các giới hạn sau:
a) \(\lim\limits_{x\rightarrow1^+}\dfrac{x^3+x+1}{x-1}\)
b) \(\lim\limits_{x\rightarrow-1^+}\dfrac{3x+2}{x+1}\)
c) \(\lim\limits_{x\rightarrow2^-}\dfrac{x-15}{x-2}\)
\(\lim\limits_{x\rightarrow3}\dfrac{\sqrt{2x+3}-x}{x^2-4x+3}\)
\(\lim\limits_{x\rightarrow0}\dfrac{\sqrt[3]{x+1}-1}{\sqrt[4]{2x+1}-1}\)
\(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{1+4x}-\sqrt[3]{1+6x}}{x^2}\)
Tính các giới hạn sau:
1. \(\lim\limits_{x\rightarrow a}\dfrac{x^2-\left(a+1\right)x+a}{x^3-a^3}\)
2. \(\lim\limits_{x\rightarrow1}\left(\dfrac{1}{1-x}-\dfrac{3}{1-x^3}\right)\)
3. \(\lim\limits_{h\rightarrow0}\dfrac{\left(x+h\right)^3-x^3}{h}\)
\(\lim\limits_{x\rightarrow1^+}\dfrac{x^2-x+1}{x^2-1}\)
\(\lim\limits_{x\rightarrow0}\dfrac{2\sqrt{1+x}-\sqrt[3]{8-x}}{x}\)
\(\lim\limits_{x\rightarrow3}\dfrac{\sqrt{x+6}-3}{\sqrt{2x-2}-2}\)
