Question

f(x) is defined everywhere except at x=0. Given that lim x->0 (f(x)+1/f(x))=2, show that lim x->0 exists and equals 2.

Answer 1

First note that: $$\left(f(x)+\frac{1}{f(x)}\right)^2=f(x)^2+2f(x)\cdot\frac{1}{f(x)}+\frac{1}{f(x)^2}$$$$=f(x)^2+\frac{1}{f(x)^2}+2.$$ What can you conlcude about $$\lim_{x\rightarrow 0}f(x)^2+\frac{1}{f(x)^2}=?$$ from this information?

Answer 2

After that, note that:$$\left(f(x)-\frac{1}{f(x)}\right)^2=f(x)^2-2f(x)\cdot\frac{1}{f(x)^2}+\frac{1}{f(x)^2}$$ $$f(x)^2+\frac{1}{f(x)^2}-2$$ What can we conclude about: $$\lim_{x\rightarrow 0}\left(f(x)-\frac{1}{f(x)}\right)^2=?$$

Answer 3

Use the information about $$\lim_{x\rightarrow 0}\left(f(x)-\frac{1}{f(x)}\right)^2$$to deduce what $$\lim_{x\rightarrow 0}f(x)-\frac{1}{f(x)}=?$$is (use the fact that the function g(x)=x^2 is continuous).

Answer 4

Then finally, once you have $$\lim_{x\rightarrow 0}f(x)-\frac{1}{f(x)}=L,$$and: $$\lim_{x\rightarrow 0}f(x)+\frac{1}{f(x)}=2,$$ can you cleverly put these together to deduce what the limit of f(x) is as x goes to 0?