Question

If because x is differentiable at 2 doesn't necessarily make it continuous at 0, right?

Answer 1

oh yes it does

Answer 2

\[\text{differentiable}\rightarrow \text{continuous}\] but not the converse

Answer 3

But we don't know if its differentiable at 0..

Answer 4

the fact that
\[\lim-{h\to 0}\frac{f(2+h)-f(2)}{h}\] exists implies that
\[\lim_{h \to 0} f(2+h)-f(2)=0\] (otherwise the limit of the difference quotient would not exist, and this implies
\[\lim_{h\to 0}f(2+h)=f(2)\] means f is continuous at 2

Answer 5

differentiable at 2 means the limit
\[\lim_{h\to 0}\frac{f(2+h)-f(2)}{h}\]exists

Answer 6

Hmm, that makes sense but my teacher just told me its false

Answer 7

oh no i am sure what your teacher said is "just because a function is continuous does not mean it is differentiable"

Answer 8

in other words you can have a function continuous at a point, but not differentiable there.

however, if it is differentiable at a point, it is continuous there

Answer 9

It was a true or false question, stating if x is differentiable at 2, then it is continuous at 0.
he said the answer is false.

Answer 10

oh i am sorry. i did not read correctly. of course. if f is continuous at 2, we know nothing about it at 0. for all we know it doesn't even exists at 0

Answer 11

kind of a weird question, but it is my fault for not reading carefully

Answer 12

That's correct. For example, the function
f(x) = 1, x > 0
= 0, x =< 0

is differentiable at x=2, but not continuous at x=0.

Answer 13

Okay, thanks!

Answer 14

Perhaps you guys can help me with another question i was unsure of.. "Find all of the horizontal asymptotes for y=\[\sqrt{9x^+5}\]/x-1

Answer 15

\[\sqrt{9x^2+5}\]*