So, this is the question:
Does the graph
g(x) = x sin (1/x), ≠ 0
g(x) = 0, x = 0
have a tangent at the origin? Give reasons for your answer.
I am really unsure on how to even begin to think about this problem.
Do you mean the tangent line? In order for a graph to have a tangent line at a point, it needs to be differentiable at that point. This means that the derivative will exist.
We could use derivative rules to find the derivative of that function and see whether it exists or not, but in this case its simpler to use the limit definition of derivative at a point.
$\displaystyle g'(0) = \lim_{x \to 0} \frac{g(x)-g(0)}{x - 0}$
$\displaystyle = \lim_{x \to 0} \frac{x \cdot \sin{\frac{1}{x}} - 0}{x} $
$\displaystyle = \lim_{x \to 0} \sin{\frac{1}{x}}$
Now to see that this limit is undefined, notice that the graph will oscillate infinitely faster and faster as x approaches 0. To see this, make the substitution x = 1/y. So as x approaches 0 from the right, y approaches infinity, since 1/0 gets larger and larger.
$\displaystyle = \lim_{y \to \infty} \sin(y)$
And this is undefined due to the oscillating nature of the sine function. Thus the limit does not exist, the derivative is undefined, and there is no tangent line.
In order to have a tangent at a point, the function needs to be differentiable at that point.
For a function to be differentiable at a point, it needs to be continuous and have a continuous derivative at that point.
At the point $\displaystyle \displaystyle \begin{align*} x = 0 \end{align*}$ we have $\displaystyle \displaystyle \begin{align*} g(0) = 0 \end{align*}$ and
$\displaystyle \displaystyle \begin{align*} \lim_{x \to 0} g(x) &= \lim_{x \to 0} x\sin{\left(\frac{1}{x}\right)} \\ &= \lim_{x \to 0}x \,\lim_{x \to 0}\sin{\left(\frac{1}{x}\right)} \\ &= 0 \cdot \textrm{ "Something between -1 and 1"} \\ &= 0 \end{align*}$
Since g(0) = 0 and the function approaches 0 as x goes to 0, the function is continuous. Now to show that it's differentiable, we need to show the function has a continuous derivative at that point.
If $\displaystyle \displaystyle \begin{align*} x \neq 0 \end{align*}$ we have $\displaystyle \displaystyle \begin{align*} g'(x) = \sin{\left(\frac{1}{x}\right)} - \frac{1}{x}\cos{\left(\frac{1}{x}\right)} \end{align*}$ and if $\displaystyle \displaystyle \begin{align*} x = 0 \end{align*}$ we have $\displaystyle \displaystyle \begin{align*} g'(x) = 0 \end{align*}$
$\displaystyle \displaystyle \begin{align*} \lim_{x \to 0}g'(x) = \lim_{x \to 0}\left[ \sin{\left(\frac{1}{x}\right)} - \frac{1}{x}\cos{\left(\frac{1}{x}\right)} \right] \end{align*}$
which does not exist, since as x gets smaller, the cosine value oscillates, and divding by 0 makes the value huge, but still oscillate. The sine value also oscillates.
Since the function does not have a continuous derivative, there can not be a tangent to the function at x = 0.
Why on earth would that hold? o.O sin(0) = 0, but sin'(0) = 1...
There is also a distinction between being differentiable and continuously differentiable. Consider for example the function
$\displaystyle f(x) \;=\; \begin{cases} x^2\sin (1/x) & \text{if }x \ne 0 \\ 0 & \text{if }x=0\end{cases}$
Has a derivative defined on all x but the derivative is not continuous.
The function is DEFINED to be the ZERO FUNCTION at x = 0. So its derivative AT THAT POINT is the derivative of the function AT THAT POINT.
The derivative of the zero function is 0.
But since the derivative everywhere else does NOT approach 0 as x approaches 0, the function is considered to NOT be differentiable at that point.
Your argument is fallacious. If this were true, you could "define" any particular point to a value, making the derivative there zero because the value is constant. You cannot use the derivative rules at the endpoint of a piecewise segment, nor a single point. You could just as easily say that this function:So its derivative AT THAT POINT is the derivative of the function AT THAT POINT.
$\displaystyle f(x) \;=\; \begin{cases} \sin(x) & \text{if }x \ne 0 \\ 0 & \text{if }x=0\end{cases}$
or even this one
$\displaystyle g(x) \;=\; \begin{cases} 1 & \text{if }x \ne 0 \\ 0 & \text{if }x=0\end{cases}$
Have a derivative of 0 at the origin, when we know this is false. f(x) is simply sin(x), equal AT ALL POINTS IN THE DOMAIN. Therefore the derivative is equal too, and f'(0) = 1, even though I defined f(0) = 0.
And g'(0) does not exist because the function isn't even continuous at 0, let alone differentiable. Yet the function is "defined to be the zero function" at x=0, is it not?
Unless I've been educated incorrectly, the derivative of a real-valued function has only one definition, and it is this limit, if and only if it exists, regardless of whether I chose to define the function piecewise or with 1 equation, as long as the values at each point are well-defined.
$\displaystyle g'(a) = \lim_{x \to a} \frac{g(x)-g(a)}{x - a} = \lim_{h \to 0} \frac{g(a+h)-g(a)}{h}$
The derivative rules are just theorems on how to find this limit.
If that doesn't convince you, consider
$\displaystyle h(x) = f(x) - \sin(x)$
Where f(x) is the function I defined above. Clearly, this function is zero at all points. However, since f(0) = 0 was piecewise defined, you are arguing that f'(0) = 0, and so , h'(0) = 0 - sin'(0) = -1, when the real calculation should be h'(0) = 1 - 1 = 0, a result that makes sense being this function actually is zero everywhere.
It is possible for a function to be differentiable at a point, but whose derivative is discontinuous, as long as the limit definition holds.But since the derivative everywhere else does NOT approach 0 as x approaches 0, the function is considered to NOT be differentiable at that point.
A function with a derivative defined for all x, but whose derivative is discontinuous.