# Thread: continuity & differentiability, limits

1. ## continuity & differentiability, limits

Give an example of a function, if such function exists (by sketching OR writing it) that satisfies the conditions given in a), b), c) (each separately):

a) f(-2)=4, f is continuous at x=(-2) but has no derivative at this point

b) f is not defined at x=(-3) but has a limit at x=(-3) and is not continuous at x=(-3)

c) f is continuous at x=(-4) from the right and from the left but is not "continuous" at this point

2. Originally Posted by calculus_2718
Give an example of a function, if such function exists (by sketching OR writing it) that satisfies the conditions given in a), b), c) (each separately):

a) f(-2)=4, f is continuous at x=(-2) but has no derivative at this point

b) f is not defined at x=(-3) but has a limit at x=(-3) and is not continuous at x=(-3)

c) f is continuous at x=(-4) from the right and from the left but is not "continuous" at this point
I can give you examples for the questions a) and b) but not for c):

a) $\displaystyle f(x)=-|x+2|+4$ See the first graph.

b) $\displaystyle f(x) = e^{-\frac1{x+3}}$ See second graph. ( The exponent is $\displaystyle -\dfrac1{x+3}$

You can prove that $\displaystyle \lim_{x \to -3} f(x) = 0$

As I've mentioned above I can't think of an example which satisfies the third question.

3. Originally Posted by calculus_2718
Give an example of a function, if such function exists (by sketching OR writing it) that satisfies the conditions given in a), b), c) (each separately):

a) f(-2)=4, f is continuous at x=(-2) but has no derivative at this point

b) f is not defined at x=(-3) but has a limit at x=(-3) and is not continuous at x=(-3)

c) f is continuous at x=(-4) from the right and from the left but is not "continuous" at this point
Some examples
(i) $\displaystyle y = \sqrt{x+2} + 4$

(ii) $\displaystyle y = \frac{x^2-9}{x+3}$

So now the question is why?

4. Originally Posted by calculus_2718
Give an example of a function, if such function exists (by sketching OR writing it) that satisfies the conditions given in a), b), c) (each separately):

a) f(-2)=4, f is continuous at x=(-2) but has no derivative at this point
f(x)= |x+2|+ 4 works.

b) f is not defined at x=(-3) but has a limit at x=(-3) and is not continuous at x=(-3)
f(x)= 1 for all x except -3 and not defined at x= -3. If you prefer a "formula", $\displaystyle f(x)= \frac{x+3}{x+3}$ or [tex]f(x)= \frac{x^2- 9}{x+ 3} will work.

c) f is continuous at x=(-4) from the right and from the left but is not "continuous" at this point
This can't be done. "continuous at x= -4 from the right" means $\displaystyle \lim_{x\rightarrow -4^+} f(x)= f(-4)$ and "continuous at x= -4 from the left" means $\displaystyle \lim_{x\rightarrow -4^-} f(x)= f(-4)$. Those both say that f(-4) exists and together they say $\displaystyle \lim_{x\rightarrow -4} f(x)$ exists and is equal to f(-4), precisely the condition that f be continuous at x= -4.