# Differentiability and continuity

• Apr 29th 2010, 08:39 PM
monster
Differentiability and continuity
Hace a problem that asks me to determine if the following function;

f(x) = $\displaystyle \frac{sin(x)}{x}$ if x<0

and $\displaystyle 1 + x^2$, if $\displaystyle x => 0$

is continuous at x = 0 and then if it's differentiable at x=0,

I figured to compute the limit as x -> 0 from x<0 and x>0 to determine continuity.

when i did lim(x->0) $\displaystyle \frac{sinx}{x}$ i used lhopitals to give 1/1 which isnt equal to f(0) = 0 so i figured the function wasn't continuous.
But i assume i have made a mistake here as when i went ahead and checked for differentiability i found the function to be diff at x=0. which doesn't make sense as if differentiable it must be continuous?

Cheers.
• Apr 29th 2010, 08:50 PM
NoobzUseRez
The function is continuous at 0. The calculus definition of continuity is as followed: A function $\displaystyle f$ is continuous at a point $\displaystyle c$ if both the $\displaystyle \displaystyle\lim_{x\to c}f(x)$ and the $\displaystyle f(c)$ exist and the $\displaystyle \lim_{x\to c}f(x)=f(c)$.
• Apr 29th 2010, 08:51 PM
shenanigans87
Just by looking at the equations you can see they are continuous. The only way $\displaystyle \frac{sin(x)}{x}$ is not continuous is if x can be 0 which you apparently ruled out by limiting the domain to x < 0
• Apr 29th 2010, 09:22 PM
monster
sorry, my notations at the start is confusing as i dont know how to do a split function, ill try to fix in original post,

suppose to be one functions f(x) that equals sinx/x for x<0 and 1+x^2 for x => 0
• Apr 29th 2010, 09:37 PM
monster
Quote:

Originally Posted by NoobzUseRez
The function is continuous at 0. The calculus definition of continuity is as followed: A function $\displaystyle f$ is continuous at a point $\displaystyle c$ if both the $\displaystyle \displaystyle\lim_{x\to c}f(x)$ and the $\displaystyle f(c)$ exist and the $\displaystyle \lim_{x\to c}f(x)=f(c)$.

so i just take the limit of the 1+ x^2 part of the function? there is no need to worry about behavior at x<0 ?
• Apr 29th 2010, 10:09 PM
NoobzUseRez
If you want to see if a limit exists, show that the limit from the left side equals the limit from the right side. So take the limit of $\displaystyle \frac{sin(x)}{x}$ from the left side, take the limit of $\displaystyle x^2 +1$ from the right, and see if they equal (which they will).
• Apr 30th 2010, 03:26 AM
HallsofIvy
Quote:

Originally Posted by monster
Hace a problem that asks me to determine if the following function;

f(x) = $\displaystyle \frac{sin(x)}{x}$ if x<0

and $\displaystyle 1 + x^2$, if $\displaystyle x => 0$

is continuous at x = 0 and then if it's differentiable at x=0,

I figured to compute the limit as x -> 0 from x<0 and x>0 to determine continuity.

when i did lim(x->0) $\displaystyle \frac{sinx}{x}$ i used lhopitals to give 1/1

Yes, that is correct.

Quote:

which isnt equal to f(0) = 0
No, $\displaystyle f(0)= 1+ 0^2= 1$

Quote:

so i figured the function wasn't continuous.
But i assume i have made a mistake here as when i went ahead and checked for differentiability i found the function to be diff at x=0. which doesn't make sense as if differentiable it must be continuous?