Question

\(\lim_{x \to 0}\left ( \frac{e^x}{e^x-1}-\frac{1}{x} \right )\)

Answer 1

2 cách:

C1: Xài VCB tương đương khi x ->0

\(\lim\limits_{x\rightarrow0}\left(\dfrac{e^x}{e^x-1}-\dfrac{1}{x}\right)=\lim\limits_{x\rightarrow0}\left(\dfrac{e^x-1+1}{e^x-1}-\dfrac{1}{x}\right)=\lim\limits_{x\rightarrow0}\left(\dfrac{x+1}{x}-\dfrac{1}{x}\right)=\lim\limits_{x\rightarrow0}\dfrac{x}{x}=1\)

C2: Xài L'Hospital

\(=\lim\limits_{x\rightarrow0}\dfrac{e^x.x-e^x+1}{x.e^x-x}=\lim\limits_{x\rightarrow0}\dfrac{e^x.x+e^x-e^x}{e^x.x+e^x-1}=\lim\limits_{x\rightarrow0}\dfrac{e^x.x}{e^x.x+e^x-1}=\lim\limits_{x\rightarrow0}\dfrac{e^x.x+e^x}{e^x.x+2e^x-1}=1\)