Question 6.11
3 months ago
We're given that f(x)=1xf(x)=\frac{1}{x}f(x)=x1 is continuous. Suppose ggg is continuous at aaa and that g(a)≠0g(a)\ne 0g(a)=0. Using Theorem 2, f∘gf\circ gf∘g also is continuous at aaa.
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Q 6.11