NEED HELP!!! Limits and Continuity!

[Kobe0824]

Hi im new to this forum and i desperately need help. Anyways, I have 7 problems that i need help with and i would really appreciate it if you could give a step-by-step solution. And its okay if you can only answer one of them and not all of them. Thanks.

1. F(x)=(1/x+1) and G(x)=x^2-5. Find values of discontinuity for F(G(x))
2. Find value of c so continuity exists when F(x)= x+3, x =<(less than or equal) to -1 and 2x-c, x > -1
3. F(x)=5/x-1, G(x)=x^4 Find F(G(x)) and find value of discontinuity
4. Lim h=>0 of Sqrt(x+h)-8 - sqrt(x-8)/h
5. Use Squeeze theorem to find Lim x=>a of F(x) when B-(x-a)=< (less than or equal) F(x) =< (less than or equal) B+(x-a)
6. x-2/x^2-4. Find discontinuity and either if its removable or nonremovable.
7. Graph F(x)= Double absolute (x-4) and find continuity and everything else related to this subject. I forgot what exactly the question asked but i think its this.

 

[galactus]

Hi im new to this forum and i desperately need help. Anyways, I have 7 problems that i need help with and i would really appreciate it if you could give a step-by-step solution. And its okay if you can only answer one of them and not all of them. Thanks.

Here's two of them to help with the others.

1.\(\displaystyle F(x)=\frac{1}{x+1}\) and \(\displaystyle G(x)=x^{2}-5\). Find values of discontinuity for F(G(x))

This is a composition. Sub g(x) into f(x) and get \(\displaystyle f(g(x))=\frac{1}{x^{2}-5+1}=\frac{1}{x^{2}-4}=\frac{1}{(x+2)(x-2)}\)

There can be no division by 0, so what values of x make the denominator equal 0?. These are the points of discontinuity.

There are vertical asymptotes at these values, thus discontinuous.

That's all there is to it. I will try to explain continuity in the next problem.

2. Find value of c so continuity exists when\(\displaystyle F(x)= \left\{ \begin{array}{rcl} x+3, \;\ x \leq -1\\ 2x-c, \;\ x > -1\end{array}\right\)

\(\displaystyle \lim_{x\to -1^{-}}(x+3)=2, \;\ \lim_{x\to -1^{+}}(2x-c)=-2-c, \;\ and \;\ f(-1)=2\)

Thus \(\displaystyle \lim_{x\to 1}f(x)=f(-1)=2 \;\ if \;\ c=-4\), so f is continuous if c=-4.

In simpler terms, for x+3, if x=-1 we get 2. So, what does c have to be so that we get 2 when we sub x=-1 into 2x-c?.

\(\displaystyle 2(-1)-c=2, \;\ c=-4.\).

Think of continuity has drawing a graph without lifting the pencil from the paper. If there is a gap, then it is discontinuous.

In order to be continuous, the line y=x+3 and the line y=2x-c have to intersect at x=-1. So, this happens if c=-4 and the other line becomes y=2x+4. See?. Here is a graph. See how the lines y=x+3 and y=2x+4 intersect at x=-1?.
Suppose instead that c=-3 and we got the line y=2x+3. I graphed this hypothetical line in blue on the graph. See how it does not intersect at (-1,2), but at (-1,1) and we have a gap. This is discontinuous. So, the line we found, y=2x+4 provides continuity because it intersects y=x+3 at (-1,2). There is no gap. You can draw the whole thing without lifting your pencil. If we had to draw y=x+3 and y=2x+3 starting at x=-1, then there is a gap and we would have to lift the pencil...discontinuous. See better now?.

 

[Kobe0824]

THANKS ALOT GALACTUS!!! I'M starting to understand continuity now a little bit. Could you please help me with 3 and 6 if possible?

 

[galactus]

3 and 6 are essentially the same as #1. For #3, do the same as I done for #1. For #6, just factor the denominator. It's similar to
#1.

Factor the difference of two squares and cancel any like terms in the numerator and denominator. What does a 'hole' in a graph
say about continuity?.

A 'hole' is known as a "removable discontinuity". We can redefine the function to make it continuous by
removing the discontinuity. Where is #6 discontinuous?.

 

[Kobe0824]

So for #3 i eventually got 5/(x^2 + 1)(x+1)(x-1) and so are the discontinuities just x=-1 and x=1? And for #6 i got to 1/x+2 which means the discontinuity is x=-2 with a removable discontinuity right?

Also, do you know anything about #'s 4,5 or 7 by any chance. Thanks alot for your help by the way!

 

[galactus]

Yep, you're gettin' the idea.

#3, works out to be \(\displaystyle \frac{5}{(x-1)(x+1)(x^{2}+1)}\). So, the discontinuities are x=1 and x=-1.

#6, factors to \(\displaystyle \frac{x-2}{(x+2)(x-2)}\). There is a hole at x=2. Thus, discontinuous.

We can redefine it so it is continuous:

\(\displaystyle f(x)=\left\{\begin{array}{rcl}\frac{x-2}{x^{2}-4}, \;\ x\neq 2\\ 1/4, \;\ x=2\end{array}\)

I have to go now. Be back later.

 

[Kobe0824]

For #6, did you mean there is a hole at x=-2 or x=2? And ok, talk to you later then

 

[mmm4444bot]

For #6, did you mean there is a hole at x = -2 or x = 2 ?

I think that galactus got turned around somewhere. There is a vertical asymptote at x = -2. The discontinuity is not removable, in this case, because asymptotes are lines (not points).

Have you used any graphs, to help you with these exercises?

Here is a graph of the interesting part, in exercise #6.

[attachment=0:3qcbyft8]Vertical Asymptote.JPG[/attachment:3qcbyft8]

Also, here are some notes about notation.

You typed: x - 2/x^2 - 4

That typing actually means: \(\displaystyle x - \frac{2}{x^2} - 4\)

(Remember something called "The Order of Operations"?)

To show that a polynomial belongs entirely in a numerator or denominator, you MUST type grouping symbols.

Typing (x - 2)/(x^2 - 4) instead means \(\displaystyle \frac{x - 2}{x^2 - 4}\)

Lastly, if you've forgotten how to factor a difference of squares, I do not think that you're ready to work with rational functions. Please review.

 

[galactus]

Yes, there is a VA at x=-2, thus discontinuous.

But, how about \(\displaystyle \frac{x-2}{(x-2)(x+2)}\)?. The x-2 cancels. There is a 'hole' at x=2.

 

[mmm4444bot]

But, how about \(\displaystyle \frac{x-2}{(x-2)(x+2)}\)?

There is a 'hole' at x=2.

Yes, that's correct, but I read your post as saying that the given function f in exercise #6 could be made continuous by piecewise-defining f(2) = 1/4.

I was thinking that you perhaps did not realize the function's asymptotic behavior at x = -2, when you made that statement.

I suppose a thorough response to #6 would report that the point-discontinuity can be removed and the asymptotic-discontinuity cannot be removed.

And, to be fair, I'll add that the original wording of the exercise is somewhat misleading because it uses a singular noun and a singular pronoun. :roll:

 

[galactus]

No problem, perhaps I was too vague.

 

[Kobe0824]

I am totally confused... Can someone explained what you guys were talking about???

 

[mmm4444bot]

The function in exercise #6 has two discontinuities, at x = -2 and at x = 2. galactus focused on x = 2 and I focused on x = -2.

The break in the graph at x = 2 is a point discontinuity. In other words, the point at (2, 1/4) is missing from the curve.

That discontinuity can be removed, by changing the function (galactus posted the piecewise definition, although it's not needed to answer the exercise).

The break in the graph at x = -2 is not a point discontinuity. The function has asymptotic behavior there, so that discontinuity cannot be removed. (See my graph above.)

Hence, the function cannot be made continuous.

The answer to exercise #6 is that the given function has two discontinuities, one can be removed and the other cannot.