Function f differentiable but not continuous

[elfin fable]

Let be f(x) =
{-(x^2)/2, if x < 3}
{-3x, if x >= 3}
That function isn't continuous in x = 3, but the limit
$\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$

Then, f is differentiable in 3, but not continuous. Where I am wrong?

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[compact bear]

Nicolás123_

[elfin fable]

It is assumed that if a function f is differentiable at x = c, then f is continuous at x = c. However, the function f defined piecewise by:
f(x) = (-x^2)/2 for x < 3
and
f(x) = -3x, for x >= 3
It is not continuous if x = 3 (because the limit does not exist at x = 3), but it is differentiable at x = 3 since the following limit exists:

$\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$

[compact bear]

Nicolás123_

[elfin fable]

Where i am wrong?

[faint anvil]

try computing this limit for f(x)=-3x

for h->o+

[elfin fable]

I have already do this, but i have been getting aproximatelly = -3

[fierce sun]

Just because these two functions have the same derivative at x=3 doesn't mean the difference quotient will converge. Try computing it for h less than zero where you have to use the top function. That limit does not exist.

[elfin fable]

I just realized my mistake... To make the limit on the left I was assuming f(3) = (-3^2)/2 and therefore the limit existed for me; I get it. Thank you

[minor tusk]

@elfin fable has given 1 rep to @fierce sun

[elfin fable]

+close