Question: Theorem 71. Let f be a function that is defined on the real numbers, andlet c be in a,b. Further assume f(c)>0. Suppose that for every naturalnumber n that there exists a point an in the open interval (c-1n,c+1n)such that f(an)≤0. Then f is not continuous at x=c1n isn't special. What's important is that it converges to zero.

Theorem $71.$ Let $fbe$ a function that $is$ defined $on$ the real numbers, and

let $cbeina,b.$ Further assume $f(c)>0.$ Suppose that for every natural

number $n$ that there exists a point $a_{n}in$ the open interval $(c-\frac{1}{n},c+\frac{1}{n})$

such that $f(a_n)\le 0.$ Then $fis$ not continuous $atx=c\frac{1}{n}$ isn't special. What's important $is$ that $it$ converges $to$ zero.