Show continuity using epsilon delta definition for piecewise function

Question

Using epsilon delta definition, show that $g$ is continuous on the whole of $\mathbb R$

$$g(x)=\cases{x^2 & \text{ if } x<1\\ \sqrt{x} & \text{ if } x≥1.}$$

Answer 1

We observe that $f$ is continuous for both $x>1$ and $x<1$ given that both the square root and the quadratic functions are continuous in their domains of definition. We need to test the continuity of $f$ at $x=1$.

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For right-side continuity at $1$, we have for $x>1$

$$|f(x)-f(1)|=|\sqrt{x}-1|=\frac{x-1}{\sqrt{x}+1}$$

Now, let's take $\delta=1$ and $0<x-1<\delta=1$. Then, $\frac{1}{\sqrt{x}+1}<1/2$ and

$$|f(x)-f(1)|<\frac12 (x-1)$$

Thus, for $x>1$given $\epsilon>0$, $|f(x)-f(1)|<\epsilon$ when $x-1=\min(1,\epsilon/2)$. Thus, $f$ is continuous from the right.

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For left-side continuity at $1$, we have for $x<1$

$$|f(x)-f(1)|=|x^2-1|=|x-1||x+1|$$

Now, let's take $\delta'=1$ and $-1=-\delta'<x-1<0$. Then, $|x+1|<2$ and

$$|f(x)-f(1)|<2 |x-1|$$

Thus, for $x<1$given $\epsilon>0$, $|f(x)-f(1)|<\epsilon$ when $|x-1|=\min(-1,2\epsilon)$. Thus, $f$ is continuous from the left.

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Since $f$ is both left and right continuous at $x=1$, then $f$ is continuous at $x=1$.