问题1. 给定\(\delta>0\)，定义函数\(f: (-\delta,\delta)\backslash \{0\}\rightarrow \mathbb{R}\)，它满足

\[\lim_{x\to 0}\Big(f(x)+\frac{1}{|f(x)|}\Big)=0. \]

求证：\(\lim_{x\to 0}f(x)\)存在，且值为\(-1\).

\(\color{black}{证.}\) 定义\(B_r:=(-r,r)\backslash \{0\},\forall r>0\)，且\(g(x):=f(x)+\frac{1}{|f(x)|},x\in B_{\delta}\).

假设\(\color{blue}{\forall r>0,\exists a\in B_r,f(a)>0}\)，则

\[g(a)=|f(a)|+\frac{1}{|f(a)|}\geqslant 2. \]

这与\(\lim g(x)=0\)矛盾！因此，\(\color{blue}{\exists r>0,\forall x\in B_r,f(x)\leqslant 0.}\)

当\(x\in B_r\)时，根据\(g\)的定义，解得

\[0\geqslant f(x)=\frac{g(x)-\sqrt{g(x)^2+4}}{2}. \]

利用\(\lim g(x)=0\)及\(\frac{x-\sqrt{x^2+4}}{2}\)的连续性，得到\(\lim f(x)=-1\).\(\qquad\vartriangleleft\)