Đặt \(t=\sqrt[7]{2-x}\rightarrow x=2-t^7.x\rightarrow1\Rightarrow t\rightarrow1\)
\(\Rightarrow lim_{t\rightarrow1}\dfrac{t-1}{-\left(t^7-1\right)}=lim_{x\rightarrow1}\dfrac{t-1}{-\left(t-1\right)\left(t^6+t^5+t^4+t^3+t^2+t+1\right)}\)\(=lim_{t\rightarrow1}\dfrac{-1}{t^6+t^5+t^4+t^3+t^2+t+1}=-\dfrac{1}{7}\)
\(lim_{x->1}\frac{\sqrt[3]{8x+11}-\sqrt{x+7}}{x^2-3x+2}\)
\(\lim_{x\to -\infty} ((2x+1)^2+4\sqrt{x^2+4}\sqrt[3]{x^3+3x^2})\)
Tính giới hạn
a, \(Lim_{n->+\infty}\frac{1+sin\left(n\right)+2^{n+2}}{2-2n+2^n}\)
b,\(Lim_{x->0}\frac{e^x-1-xcos\left(x\right)}{x\left(e^{2x}-1\right)}\)
c,\(Lim_{n->+\infty}\sqrt[2n]{8^n+9^n}\)
d,\(Lim_{x->0}\frac{\ln\left(1+x\right)-xe^3}{x\tan\left(2x\right)}\)
Tính giới hạn
a, \(Lim_{n->+\infty}\frac{1+sin\left(n\right)+2^{n+2}}{2-2n+2^n}\)
b,\(Lim_{x->0}\frac{e^x-1-xcos\left(x\right)}{x\left(e^{2x}-1\right)}\)
c,\(Lim_{n->+\infty}\sqrt[2n]{8^n+9^n}\)
d,\(Lim_{x->0}\frac{\ln\left(1+x\right)-xe^3}{x\tan\left(2x\right)}\)
\(Lim_{x\to3}\)\(\frac{2 - \sqrt(x+1)\sqrt[3](x-2)}{2- \sqrt(x-2)\sqrt[3](x+5)}\)
\(lim_{x\rightarrow1^-}\frac{\sqrt{x^2-x+3}}{2\left|x\right|-1}\)
\(\lim_{x \to 0}\left ( \frac{e^x}{e^x-1}-\frac{1}{x} \right )\)
\(lim_{x\rightarrow2^-}\frac{x^2-4}{\sqrt{\left(x^2+1\right)\left(2-x\right)}}\)
\(lim_{x\rightarrow0}\dfrac{\sqrt{x^3+1}-1}{x^2+x}\)