f(x)= sqrt(x), g(x)=x^2+1; prove that h(x)=f(g(x)) is continuous at x=1

Let \( f(x) = \sqrt{x} \) and \( g(x) = x^2 + 1 \). We have to prove that the composite function \( h(x) = f(g(x)) \) is continuous at \( x = 1 \).

Solution:

For a function \( f(x) \) to be continuous at a point \( x = c \), the following three conditions must be satisfied:

1. \( f(c) \) is defined;
2. \( \lim_{x \to c} f(x) \) exists;
3. \( \lim_{x \to c} f(x) = f(c) \).

\[ h(x) = f(g(x)) = \sqrt{x^2 + 1}. \]

1) \( h(1) = \sqrt{1^2 + 1} = \sqrt{2} \). Therefore, \( h(1) \) is defined.

2)

\( \lim_{x \to 1} h(x) \\ = \lim_{x \to 1} \sqrt{x^2 + 1} \\ = \sqrt{1^2 + 1} = \sqrt{2} \\ \)

Therefore, \( \lim_{x \to 1} h(x) \) exists.

3) \( \lim_{x \to 1} h(x) = h(1) = \sqrt{2} \)

Since all 3 conditions are satisfied, the composite function \( h(x) = f(g(x)) \) is continuous at \( x = 1 \).

Please let me know in the comments if you find any error in this solution.