# Properties of functions.

1. Nov 19, 2007

### azatkgz

Among the following function,
-one is continuous everywhere,but not differentiable at 0
-one is differentiable,but derivative is discontinuous at 0
-one is differentiable and has continuous derivative

1)$$f(x) = \left\{\begin{array}{cc}\ln (1 + x^3)\sin\frac {1}{x},x > 0 \\ 0,x\leq0\end{array}\right$$

2)$$g(x) = \left\{\begin{array}{cc}\ln^2(1 + x)\sin\frac {1}{x},x > 0 \\ 0,x\leq0\end{array}\right$$

3)$$h(x) = \left\{\begin{array}{cc}\ln (1 + \frac {\sin x}{2}),x > 0 \\ x\leq0\end{array}\right$$

Solution:

1)$$\lim_{x\rightarrow 0}\frac {\ln (1 + x^3)\sin\frac {1}{x}}{x} = \lim_{x\rightarrow 0}\frac {(x^3 + 0{x^6})\sin\frac {1}{x}}{x} = 0$$

differentiable

$$f'(x) = \frac {3x^2\sin\frac {1}{x}}{1 + x^3} - \frac {\cos\frac {1}{x}\ln (1 + x^3)}{x^2}$$

2)$$\lim_{x\rightarrow 0}\frac {(x^2 + o(x^4))\sin\frac {1}{x}}{x} = 0$$ differentiable

$$g'(x) = \frac {2\ln (1 + x)\sin\frac {1}{x}}{1 + x} - \frac {\cos\frac {1}{x}\ln^2 (1 + x)}{x^2}$$

3)$$\lim_{x\rightarrow 0}\frac {(\frac {\sin x}{2} + 0(\sin^2x))\sin\frac {1}{x}}{x} = \lim_{x\rightarrow 0}\sin\frac {1}{x}$$ not differentiable

I couldn't find which function f'(x) or g'(x) is not differentiable at 0,which one continuous.

2. Nov 20, 2007

### Gib Z

This may not be the most analytical method, but try graphing these functions, it becomes easier to see.