# Thread: Calculate derivative using formal definition

1. ## Calculate derivative using formal definition

I'm really hoping someone can help me calculate the derivative (using the formal definition) for the following function:

Here is my attempted solutions but I know it is wrong:

Can someone tell me what I am doing incorrectly?

2. ## Re: Calculate derivative using formal definition

Originally Posted by otownsend
I'm really hoping someone can help me calculate the derivative (using the formal definition) for the following function:

Here is my attempted solutions but I know it is wrong:

Can someone tell me what I am doing incorrectly?
We can't tell you where you went wrong if you don't post the whole problem... What's h(t)?

-Dan

3. ## Re: Calculate derivative using formal definition

h(t) is in the first image amended to my post. Do you not see it?

4. ## Re: Calculate derivative using formal definition

Originally Posted by otownsend
I'm really hoping someone can help me calculate the derivative (using the formal definition) for the following function:

Here is my attempted solutions but I know it is wrong:
Can someone tell me what I am doing incorrectly?
Let's start from $\dfrac{k^2\sin\left(\tfrac{1}{k}\right)}{k}=k \cdot \sin\left(\tfrac{1}{k}\right)$

At this point we need to look at a graph, SEE HERE
Looking at that graph, what does the limit as $k\to 0$ look like?

How do you prove it? If you need hints here they are:
1) $-1\le\sin\left(\tfrac{1}{k}\right)\le 1$
2) If $k>0$ then $-k\le k\sin\left(\tfrac{1}{k}\right)\le k$ and we know $k\to 0^+$
This is the tricky one, what about $k\to 0^-~?$
3) Well $\sin$ is an odd function so $\sin\left(\tfrac{1}{-k}\right)=-\sin\left(\tfrac{1}{|k|}\right)$ as can be seen on the graph.
In other words: the limit as $k\to 0^+$ is equal to the limit as $k\to 0^-$

5. ## Re: Calculate derivative using formal definition

Are you saying that sin(1/-k) is the function from the left of 0 ?

6. ## Re: Calculate derivative using formal definition

Originally Posted by otownsend
Are you saying that sin(1/-k) is the function from the left of 0 ?
To be fair to myself, I did say that this was tricky.
Odd functions are just that.
$\sin(-t)=-\sin(t)$ or $\sin(t)=-\sin(-t)$

THUS $\displaystyle \large{\mathop {\lim }\limits_{t \to {0^ - }} \sin (t) = \mathop {\lim }\limits_{t \to {0^ - }} - \sin ( - t) = \mathop {\lim }\limits_{t \to {0^ + }} - \sin (t)}$