Continuity & Discontinuity in limits

[qwerty159]

Homework Statement

a) Determine the points where the function f (x) = (x + 3) / (x^2 − 3x − 10) is discontinuous. Then define a new function g that is a a continuous extension of f .

b) Determine what value of the constant k makes the Piecewise function
{ (x − k)/ (k^2 + 1) , x < 0
f (x) = { x^2 + k , x ≥ 0

continuous on lR. Justify your answer.

c) True or False: if f and g are both continuous at 0, then f ◦ g is also continuous at 0. Justify your answer by proving it (if true) or giving a counterexample (if false).

Homework Equations

The Attempt at a Solution

a) it is discontinuous at 5 and -2.Factored the denominator.

b) lim x-> 0- = k
& lim x-> 0+ = k

c) lim x-> 0 f(x) = f(0) & lim x-> 0 g(x) = g(0) so, f ◦ g is lim x-> 0 f(g(x)) = 0. It is continuous.

I hope I get a reply..No one is helping on other sites either. Thanks.

 

[flyingpig]

a) Fix the discontinous points, give g any value at those points

b)

You have

$$f(x) = \begin{cases} \frac{x-k}{k^2 + 1} &\text{if } x < 0\\ x^2 + k &\text{if } x \geq0 \end{cases}$$

And you said that

$$\lim_{x\to0^-} \frac{x-k}{k^2 + 1} = \lim_{x\to0^+}x^2 + k$$, find that k

c) That seems right.

 

[qwerty159]

I still don't understand a) . I got b). How sure are you about c) ?

 

[HallsofIvy]

Homework Statement

a) Determine the points where the function f (x) = (x + 3) / (x^2 − 3x − 10) is discontinuous. Then define a new function g that is a a continuous extension of f .

b) Determine what value of the constant k makes the Piecewise function
{ (x − k)/ (k^2 + 1) , x < 0
f (x) = { x^2 + k , x ≥ 0

continuous on lR. Justify your answer.

c) True or False: if f and g are both continuous at 0, then f ◦ g is also continuous at 0. Justify your answer by proving it (if true) or giving a counterexample (if false).

Homework Equations

The Attempt at a Solution

a) it is discontinuous at 5 and -2.Factored the denominator.

b) lim x-> 0- = -k

This is wrong. Try again.

& lim x-> 0+ = k

c) lim x-> 0 f(x) = f(0) & lim x-> 0 g(x) = g(0) so, f ◦ g is lim x-> 0 f(g(x)) = 0.

No, you do not know that the limit is 0 nor do you know that you should be looking at limit of g(x) as x goes to 0 because you do not know that f(0)= 0.

It is continuous.

I hope I get a reply..No one is helping on other sites either. Thanks.

 

[qwerty159]

what is wrong ? Give me an explanation or something so i understand better.

 

[flyingpig]

I still don't understand a) . I got b). How sure are you about c) ?

Nope...I read this as what HallsofIvy pointed out. It says the f and g are continuous at x = 0, but f and g may not be 0.

 

[HallsofIvy]

Well, how about you tell how you got that
$$\lim_{x\to 0}\frac{x- k}{k^2+ 1}= -k$$!
What do you get if you just take x= 0?

 

[qwerty159]

Explanation..please

 

[qwerty159]

Well, how about you tell how you got that
$$\lim_{x\to 0}\frac{x- k}{k^2+ 1}= -k$$!
What do you get if you just take x= 0?

ok

-k = k
-k - k = 0
-2k = 0
k = 0/-2
k = 0

Explain the other ones please

 

[qwerty159]

Guys...Please help me.

 

[flyingpig]

ok

-k = k
-k - k = 0
-2k = 0
k = 0/-2
k = 0

Explain the other ones please

What is 0 - k?

 

[qwerty159]

-k ...what kinda question is that LOL. I know that.

 

[qwerty159]

Please help me quickly..it is getting late.

 

[qwerty159]

Can you guys please explain it to me a) and c) because i already got b) ..

 

[flyingpig]

-k ...what kinda question is that LOL. I know that.

$$\frac{0-k}{k^2 + 1}$$

 

[SammyS]

...

c) True or False: if f and g are both continuous at 0, then f ◦ g is also continuous at 0. Justify your answer by proving it (if true) or giving a counterexample (if false).
...

The Attempt at a Solution

...

c) lim x-> 0 f(x) = f(0) & lim x-> 0 g(x) = g(0) so, f ◦ g is lim x-> 0 f(g(x)) = 0. It is continuous.

...

What if g(0) = a, where a ≠ 0.

What do you know about f(a) ?