left and right hand limits

[rberg1897]

Let

​

​

According to the definition of derivative, to compute

, we need to compute the left-hand limit and the right-hand limit

I've solved similar problems, but I can't seem to get the left hand limit. Using lim x-->0+ (f(x)-f(a))/(x-a), I got the right hand limit to equal 4x, which is correct according to the program I'm using, but using the same formula for limx-->0-((-4x^2+7x)-0)/(x-0) equals -4x+7, which apparently is incorrect. I'm just not sure what I'm doing wrong.

 

[pka]

Let

According to the definition of derivative, to compute

, we need to compute the left-hand limit and the right-hand limit

You don't need to do much.
\(\displaystyle \displaystyle\lim _{x \to 0^ + } f(x) = - 4\;\& \,\lim _{x \to 0^ - } f(x) = 0\).
Therefore, \(\displaystyle f\) is not continuous at \(\displaystyle x=0\) so \(\displaystyle f'(0)\) cannot exist.

 

[HallsofIvy]

For x< 0, f(x)= -4x^2+ 7x while f(0)= -4. That is, f(x)- f(0)= -4x^2+ 7x+ 4. That does not go to 0 as x goes to 0 so the limit, as x goes to 0 from below does not exist. The fact that the numerator of the "difference quotient" does not go to 0 is due to the fact that f(x) is not continuous at x= 0.