i am reading Thomas' calcullus. In chapter 2, continuity section i need to understand a theorem.The Theorem and prof are given bellow according to the book.

Theroem 10 - Limits of Continuous Function : if g is continuous at the point b and $\lim_{x\to c} f(x) = b$ then $\lim_{x\to c} g(f(x)) = g(lim_{x\to c} f(x))$

Prof: let ε > 0 be given, Since g is continuous at b, there exist a number δ1 such that |g(y) - g(b)| < ε whenever 0 < |y-b| < δ1 -----------(1)

Since $\lim_{x\to c} f(x) = b$ , there exist a δ > 0 such that |f(x) - b| < δ1 whenever 0 < |x-c| < δ -------------- (2)

If we let y =f(x), we then have that |y -b| < δ1 whenever 0 < |x-c| < δ

Which implies from the first statement that *|g(y) - g(b)| = |g(f(x)) - g(b)| < ε whenever 0 < |x-c| < δ.From the definition of the limit, this proves that
$\lim_{x\to c} g(f(x)) = g(b)$

My Question is, how equation (1) is written? And how and why δ1 is used in equation (2)? Can you graph g(x) and f(x) according to the above?

Best Regards ' sabbir


Equation 1 comes directly from the definition of continuity of $g$ at a point. Can you be more specific about which part you don't understand?

Also, $\delta_1$ does not appear in Equation 2. Are you asking how you combine the two equations to get the result? Allow me to explain it in a slightly different way:

From line 2, we know that if we choose $x$ such that $0<|x-c|<\delta$ then we will have $|f(x)-b|<\delta_1$. But this is just the hypothesis of the first line (ignoring the possibility that $|f(x)-b|=0$), so in this case we have that $|g(f(x))-g(b)|< \epsilon$. In the case where $|f(x)-b|=0$, it follows trivially that $|f(x)-b|<\epsilon$


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