1. ## Functions question

The function f is defined by $\displaystyle f(x)=sinx\ for -\frac{\pi}{4}<x\leq\frac{\pi}{4}, and\ f(x+\frac{\pi}{2})=f(x)+\sqrt{2}$ $\displaystyle for\ all\ x \in \mathbb{R}$

$\displaystyle 1)\ Find\ an\ expression\ for\ f(x)\ for\ \frac{\pi}{4}<x\leq\frac{3\pi}{4}$

$\displaystyle 2)\ Prove\ that\ f\ is\ continuous\ at\ \frac{\pi}{4}.$ $\displaystyle That\ is,\ prove\ \lim_{x\rightarrow\frac{\pi}{4}}f(x)=f(\frac{\pi}{ 4})$

2. A possible solution is $\displaystyle f(x) = \frac{2\sqrt{2}x}{\pi}$

Since $\displaystyle \lim_{x \to \frac{\pi}{4}} \sin{x} = \sin\frac{\pi}{4} = \frac{1}{\sqrt{2}}, \lim_{x \to \frac{\pi}{4}} \frac{2\sqrt{2}x}{\pi} = \frac{1}{\sqrt{2}}$, f(x) is continuous at $\displaystyle \frac{\pi}{4}$

3. Originally Posted by nahduma
A possible solution is $\displaystyle f(x) = \frac{2\sqrt{2}x}{\pi}$

Since $\displaystyle \lim_{x \to \frac{\pi}{4}} \sin{x} = \sin\frac{\pi}{4} = \frac{1}{\sqrt{2}}, \lim_{x \to \frac{\pi}{4}} \frac{2\sqrt{2}x}{\pi} = \frac{1}{\sqrt{2}}$, f(x) is continuous at $\displaystyle \frac{\pi}{4}$
Shouldn't you consider the left and right hand limit?

4. Originally Posted by acevipa
Shouldn't you consider the left and right hand limit?
Exactly.
the right hand limit is the limit when $\displaystyle x>\frac{\pi}{4} ie. \lim_{x\to \frac{\pi}{4}} \frac{2\sqrt{2}x}{\pi}.$
the left hand limit is the limit when $\displaystyle x<\frac{\pi}{4} ie. \lim_{x\to\frac{\pi}{4}} \sin{x}.$

Hope this clears it up.